diff --git a/configs/datasets/scibench/lib_prompt/atkins_prompt.txt b/configs/datasets/scibench/lib_prompt/atkins_prompt.txt
new file mode 100644
index 00000000..7d1e9418
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/atkins_prompt.txt
@@ -0,0 +1,18 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: Suppose the concentration of a solute decays exponentially along the length of a container. Calculate the thermodynamic force on the solute at $25^{\circ} \mathrm{C}$ given that the concentration falls to half its value in $10 \mathrm{~cm}$.
+Answer: The answer is \boxed{17}.
+
+Promblem 2: Calculate the separation of the $\{123\}$ planes of an orthorhombic unit cell with $a=0.82 \mathrm{~nm}, b=0.94 \mathrm{~nm}$, and $c=0.75 \mathrm{~nm}$.
+Answer: The answer is \boxed{0.21}.
+
+Promblem 3: What is the mean speed, $\bar{c}$, of $\mathrm{N}_2$ molecules in air at $25^{\circ} \mathrm{C}$ ?
+Answer: The answer is \boxed{475}.
+
+Promblem 4: The data below show the temperature variation of the equilibrium constant of the reaction $\mathrm{Ag}_2 \mathrm{CO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{Ag}_2 \mathrm{O}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$. Calculate the standard reaction enthalpy of the decomposition.
+$\begin{array}{lllll}T / \mathrm{K} & 350 & 400 & 450 & 500 \\ K & 3.98 \times 10^{-4} & 1.41 \times 10^{-2} & 1.86 \times 10^{-1} & 1.48\end{array}$
+Answer: The answer is \boxed{+80}.
+
+Promblem 5: Calculate the moment of inertia of an $\mathrm{H}_2 \mathrm{O}$ molecule around the axis defined by the bisector of the $\mathrm{HOH}$ angle (3). The $\mathrm{HOH}$ bond angle is $104.5^{\circ}$ and the bond length is $95.7 \mathrm{pm}$.
+Answer: The answer is \boxed{1.91}.
+
diff --git a/configs/datasets/scibench/lib_prompt/atkins_sol.txt b/configs/datasets/scibench/lib_prompt/atkins_sol.txt
new file mode 100644
index 00000000..d1f5f60a
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/atkins_sol.txt
@@ -0,0 +1,101 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: A container is divided into two equal compartments (Fig. 5.8). One contains $3.0 \mathrm{~mol} \mathrm{H}_2(\mathrm{~g})$ at $25^{\circ} \mathrm{C}$; the other contains $1.0 \mathrm{~mol} \mathrm{~N}_2(\mathrm{~g})$ at $25^{\circ} \mathrm{C}$. Calculate the Gibbs energy of mixing when the partition is removed. Assume perfect behaviour.
+
+Explanation for Problem 1: Given that the pressure of nitrogen is $p$, the pressure of hydrogen is $3 p$; therefore, the initial Gibbs energy is
+$$
+G_{\mathrm{i}}=(3.0 \mathrm{~mol})\left\{\mu^{\ominus}\left(\mathrm{H}_2\right)+R T \ln 3 p\right\}+(1.0 \mathrm{~mol})\left\{\mu^{\ominus}\left(\mathrm{N}_2\right)+R T \ln p\right\}
+$$
+When the partition is removed and each gas occupies twice the original volume, the partial pressure of nitrogen falls to $\frac{1}{2} p$ and that of hydrogen falls to $\frac{3}{2} p$. Therefore, the Gibbs energy changes to
+$$
+G_{\mathrm{f}}=(3.0 \mathrm{~mol})\left\{\mu^{\ominus}\left(\mathrm{H}_2\right)+R T \ln \frac{3}{2} p\right\}+(1.0 \mathrm{~mol})\left\{\mu^{\ominus}\left(\mathrm{N}_2\right)+R T \ln \frac{1}{2} p\right\}
+$$
+The Gibbs energy of mixing is the difference of these two quantities:
+$$
+\begin{aligned}
+\Delta_{\text {mix }} G & =(3.0 \mathrm{~mol}) R T \ln \left(\frac{\frac{3}{2} p}{3 p}\right)+(1.0 \mathrm{~mol}) R T \ln \left(\frac{\frac{1}{2} p}{p}\right) \\
+& =-(3.0 \mathrm{~mol}) R T \ln 2-(1.0 \mathrm{~mol}) R T \ln 2 \\
+& =-(4.0 \mathrm{~mol}) R T \ln 2=-6.9 \mathrm{~kJ}
+\end{aligned}
+$$
+
+Therefore, the answer is \boxed{-6.9}.
+
+Promblem 2: The change in molar internal energy when $\mathrm{CaCO}_3(\mathrm{~s})$ as calcite converts to another form, aragonite, is $+0.21 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the difference between the molar enthalpy and internal energy changes when the pressure is 1.0 bar given that the densities of the polymorphs are $2.71 \mathrm{~g} \mathrm{~cm}^{-3}$ and $2.93 \mathrm{~g} \mathrm{~cm}^{-3}$, respectively.
+Explanation for Problem 2: The change in enthalpy when the transition occurs is
+$$
+\begin{aligned}
+\Delta H_{\mathrm{m}} & =H_{\mathrm{m}}(\text { aragonite })-H_{\mathrm{m}}(\text { calcite }) \\
+& =\left\{U_{\mathrm{m}}(\mathrm{a})+p V_{\mathrm{m}}(\mathrm{a})\right\}-\left\{U_{\mathrm{m}}(\mathrm{c})+p V_{\mathrm{m}}(\mathrm{c})\right\} \\
+& =\Delta U_{\mathrm{m}}+p\left\{V_{\mathrm{m}}(\mathrm{a})-V_{\mathrm{m}}(\mathrm{c})\right\}
+\end{aligned}
+$$
+where a denotes aragonite and c calcite. It follows by substituting $V_{\mathrm{m}}=M / \rho$ that
+$$
+\Delta H_{\mathrm{m}}-\Delta U_{\mathrm{m}}=p M\left(\frac{1}{\rho(\mathrm{a})}-\frac{1}{\rho(\mathrm{c})}\right)
+$$
+Substitution of the data, using $M=100 \mathrm{~g} \mathrm{~mol}^{-1}$, gives
+$$
+\begin{aligned}
+\Delta H_{\mathrm{m}}-\Delta U_{\mathrm{m}} & =\left(1.0 \times 10^5 \mathrm{~Pa}\right) \times\left(100 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times\left(\frac{1}{2.93 \mathrm{~g} \mathrm{~cm}^{-3}}-\frac{1}{2.71 \mathrm{~g} \mathrm{~cm}^{-3}}\right) \\
+& =-2.8 \times 10^5 \mathrm{~Pa} \mathrm{~cm}{ }^3 \mathrm{~mol}^{-1}=-0.28 \mathrm{~Pa} \mathrm{~m}^3 \mathrm{~mol}^{-1}
+\end{aligned}
+$$
+Therefore, the answer is \boxed{-0.28}.
+
+Promblem 3: The osmotic pressures of solutions of poly(vinyl chloride), PVC, in cyclohexanone at $298 \mathrm{~K}$ are given below. The pressures are expressed in terms of the heights of solution (of mass density $\rho=0.980 \mathrm{~g} \mathrm{~cm}^{-3}$ ) in balance with the osmotic pressure. Determine the molar mass of the polymer.
+$\begin{array}{llllll}c /\left(\mathrm{g} \mathrm{dm}^{-3}\right) & 1.00 & 2.00 & 4.00 & 7.00 & 9.00 \\ h / \mathrm{cm} & 0.28 & 0.71 & 2.01 & 5.10 & 8.00\end{array}$
+
+Explanation for Problem 3: The data give the following values for the quantities to plot:
+$$
+\begin{array}{llllll}
+c /\left(\mathrm{g} \mathrm{dm}^{-3}\right) & 1.00 & 2.00 & 4.00 & 7.00 & 9.00 \\
+(h / c) /\left(\mathrm{cm} \mathrm{g}^{-1} \mathrm{dm}^3\right) & 0.28 & 0.36 & 0.503 & 0.729 & 0.889
+\end{array}
+$$
+The points are plotted in Fig. 5.28. The intercept is at 0.21 . Therefore,
+$$
+\begin{aligned}
+M & =\frac{R T}{\rho g} \times \frac{1}{0.21 \mathrm{~cm} \mathrm{~g}^{-1} \mathrm{dm}^3} \\
+& =\frac{\left(8.3145 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(298 \mathrm{~K})}{\left(980 \mathrm{~kg} \mathrm{~m}^{-1}\right) \times\left(9.81 \mathrm{~m} \mathrm{~s}^{-2}\right)} \times \frac{1}{2.1 \times 10^{-3} \mathrm{~m}^4 \mathrm{~kg}^{-1}} \\
+& =1.2 \times 10^2 \mathrm{~kg} \mathrm{~mol}^{-1}
+\end{aligned}
+$$
+Therefore, the answer is \boxed{1.2}.
+
+Promblem 4: What is the mean speed, $\bar{c}$, of $\mathrm{N}_2$ molecules in air at $25^{\circ} \mathrm{C}$ ?
+Explanation for Problem 4: The integral required is
+$$
+\begin{aligned}
+\bar{c} & =4 \pi\left(\frac{M}{2 \pi R T}\right)^{3 / 2} \int_0^{\infty} v^3 \mathrm{e}^{-M v^2 / 2 R T} \mathrm{~d} v \\
+& =4 \pi\left(\frac{M}{2 \pi R T}\right)^{3 / 2} \times \frac{1}{2}\left(\frac{2 R T}{M}\right)^2=\left(\frac{8 R T}{\pi M}\right)^{1 / 2}
+\end{aligned}
+$$
+where we have used the standard result from tables of integrals (or software) that
+$$
+\int_0^{\infty} x^3 \mathrm{e}^{-a x^2} \mathrm{~d} x=\frac{1}{2 a^2}
+$$
+Substitution of the data then gives
+$$
+\bar{c}=\left(\frac{8 \times\left(8.3141 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(298 \mathrm{~K})}{\pi \times\left(28.02 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1}\right)}\right)^{1 / 2}=475 \mathrm{~m} \mathrm{~s}^{-1}
+$$
+where we have used $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}$.
+
+Therefore, the answer is \boxed{475}.
+
+Promblem 5: In an industrial process, nitrogen is heated to $500 \mathrm{~K}$ in a vessel of constant volume. If it enters the vessel at $100 \mathrm{~atm}$ and $300 \mathrm{~K}$, what pressure would it exert at the working temperature if it behaved as a perfect gas?
+Explanation for Problem 5: Cancellation of the volumes (because $V_1=V_2$ ) and amounts (because $\left.n_1=n_2\right)$ on each side of the combined gas law results in
+$$
+\frac{p_1}{T_1}=\frac{p_2}{T_2}
+$$
+which can be rearranged into
+$$
+p_2=\frac{T_2}{T_1} \times p_1
+$$
+Substitution of the data then gives
+$$
+p_2=\frac{500 \mathrm{~K}}{300 \mathrm{~K}} \times(100 \mathrm{~atm})=167 \mathrm{~atm}
+$$
+
+Therefore, the answer is \boxed{167}.
+
diff --git a/configs/datasets/scibench/lib_prompt/calculus_prompt.txt b/configs/datasets/scibench/lib_prompt/calculus_prompt.txt
new file mode 100644
index 00000000..08e0ea29
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/calculus_prompt.txt
@@ -0,0 +1,17 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: The region $\mathscr{R}$ enclosed by the curves $y=x$ and $y=x^2$ is rotated about the $x$-axis. Find the volume of the resulting solid.
+Answer: The answer is \boxed{0.41887902047}.
+
+Promblem 2: Find the average value of the function $f(x)=1+x^2$ on the interval $[-1,2]$.
+Answer: The answer is \boxed{2}.
+
+Promblem 3: Find the length of the cardioid $r=1+\sin \theta$.
+Answer: The answer is \boxed{8}.
+
+Promblem 4: How large should we take $n$ in order to guarantee that the Trapezoidal and Midpoint Rule approximations for $\int_1^2(1 / x) d x$ are accurate to within 0.0001 ?
+Answer: The answer is \boxed{41}.
+
+Promblem 5: A force of $40 \mathrm{~N}$ is required to hold a spring that has been stretched from its natural length of $10 \mathrm{~cm}$ to a length of $15 \mathrm{~cm}$. How much work is done in stretching the spring from $15 \mathrm{~cm}$ to $18 \mathrm{~cm}$ ?
+Answer: The answer is \boxed{1.56}.
+
diff --git a/configs/datasets/scibench/lib_prompt/calculus_sol.txt b/configs/datasets/scibench/lib_prompt/calculus_sol.txt
new file mode 100644
index 00000000..fb9c5619
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/calculus_sol.txt
@@ -0,0 +1,62 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: Find the average value of the function $f(x)=1+x^2$ on the interval $[-1,2]$.
+Explanation for Problem 1: With $a=-1$ and $b=2$ we have
+$$
+\begin{aligned}
+f_{\text {ave }} & =\frac{1}{b-a} \int_a^b f(x) d x=\frac{1}{2-(-1)} \int_{-1}^2\left(1+x^2\right) d x \\
+& =\frac{1}{3}\left[x+\frac{x^3}{3}\right]_{-1}^2=2
+\end{aligned}
+$$
+Therefore, the answer is \boxed{2}.
+
+Promblem 2: Estimate the volume of the solid that lies above the square $R=[0,2] \times[0,2]$ and below the elliptic paraboloid $z=16-x^2-2 y^2$. Divide $R$ into four equal squares and choose the sample point to be the upper right corner of each square $R_{i j}$. 
+Explanation for Problem 2: The squares are shown in Figure 6. The paraboloid is the graph of $f(x, y)=16-x^2-2 y^2$ and the area of each square is $\Delta A=1$. Approximating the volume by the Riemann sum with $m=n=2$, we have
+$$
+\begin{aligned}
+V & \approx \sum_{i=1}^2 \sum_{j=1}^2 f\left(x_i, y_j\right) \Delta A \\
+& =f(1,1) \Delta A+f(1,2) \Delta A+f(2,1) \Delta A+f(2,2) \Delta A \\
+& =13(1)+7(1)+10(1)+4(1)=34
+\end{aligned}
+$$
+Therefore, the answer is \boxed{34}.
+
+Promblem 3: Find the length of the cardioid $r=1+\sin \theta$.
+Explanation for Problem 3: The cardioid is shown in Figure 8. (We sketched it in Example 7 in Section 10.3.) Its full length is given by the parameter interval $0 \leqslant \theta \leqslant 2 \pi$, so Formula 5 gives
+$$
+\begin{aligned}
+L & =\int_0^{2 \pi} \sqrt{r^2+\left(\frac{d r}{d \theta}\right)^2} d \theta=\int_0^{2 \pi} \sqrt{(1+\sin \theta)^2+\cos ^2 \theta} d \theta \\
+& =\int_0^{2 \pi} \sqrt{2+2 \sin \theta} d \theta
+\end{aligned}
+$$
+We could evaluate this integral by multiplying and dividing the integrand by $\sqrt{2-2 \sin \theta}$, or we could use a computer algebra system. In any event, we find that the length of the cardioid is $L=8$.
+
+Therefore, the answer is \boxed{8}.
+
+Promblem 4: The region $\mathscr{R}$ enclosed by the curves $y=x$ and $y=x^2$ is rotated about the $x$-axis. Find the volume of the resulting solid.
+Explanation for Problem 4: The curves $y=x$ and $y=x^2$ intersect at the points $(0,0)$ and $(1,1)$. The region between them, the solid of rotation, and a cross-section perpendicular to the $x$-axis are shown in Figure. A cross-section in the plane $P_x$ has the shape of a washer (an annular ring) with inner radius $x^2$ and outer radius $x$, so we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle:
+$$
+A(x)=\pi x^2-\pi\left(x^2\right)^2=\pi\left(x^2-x^4\right)
+$$
+Therefore we have
+$$
+\begin{aligned}
+V & =\int_0^1 A(x) d x=\int_0^1 \pi\left(x^2-x^4\right) d x \\
+& =\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1=\frac{2 \pi}{15}
+\end{aligned}
+$$
+Therefore, the answer is \boxed{0.41887902047}.
+
+Promblem 5: Use Simpson's Rule with $n=10$ to approximate $\int_1^2(1 / x) d x$.
+Explanation for Problem 5: Putting $f(x)=1 / x, n=10$, and $\Delta x=0.1$ in Simpson's Rule, we obtain
+$$
+\begin{aligned}
+\int_1^2 \frac{1}{x} d x & \approx S_{10} \\
+& =\frac{\Delta x}{3}[f(1)+4 f(1.1)+2 f(1.2)+4 f(1.3)+\cdots+2 f(1.8)+4 f(1.9)+f(2)] \\
+& =\frac{0.1}{3}\left(\frac{1}{1}+\frac{4}{1.1}+\frac{2}{1.2}+\frac{4}{1.3}+\frac{2}{1.4}+\frac{4}{1.5}+\frac{2}{1.6}+\frac{4}{1.7}+\frac{2}{1.8}+\frac{4}{1.9}+\frac{1}{2}\right) \\
+& \approx 0.693150
+\end{aligned}
+$$
+
+Therefore, the answer is \boxed{0.693150}.
+
diff --git a/configs/datasets/scibench/lib_prompt/chemmc_prompt.txt b/configs/datasets/scibench/lib_prompt/chemmc_prompt.txt
new file mode 100644
index 00000000..7876cd02
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/chemmc_prompt.txt
@@ -0,0 +1,17 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: The wave function $\Psi_2(1,2)$ given by Equation 9.39 is not normalized as it stands. Determine the normalization constant of $\Psi_2(1,2)$ given that the "1s" parts are normalized.
+Answer: The answer is \boxed{0.70710678}.
+
+Promblem 2: Find the bonding and antibonding Hückel molecular orbitals for ethene.
+Answer: The answer is \boxed{0.70710678}.
+
+Promblem 3: Given that the work function for sodium metal is $2.28 \mathrm{eV}$, what is the threshold frequency $v_0$ for sodium?
+Answer: The answer is \boxed{5.51}.
+
+Promblem 4: Calculate the de Broglie wavelength of an electron traveling at $1.00 \%$ of the speed of light.
+Answer: The answer is \boxed{243}.
+
+Promblem 5: To a good approximation, the microwave spectrum of $\mathrm{H}^{35} \mathrm{Cl}$ consists of a series of equally spaced lines, separated by $6.26 \times 10^{11} \mathrm{~Hz}$. Calculate the bond length of $\mathrm{H}^{35} \mathrm{Cl}$.
+Answer: The answer is \boxed{129}.
+
diff --git a/configs/datasets/scibench/lib_prompt/chemmc_sol.txt b/configs/datasets/scibench/lib_prompt/chemmc_sol.txt
new file mode 100644
index 00000000..9d67370e
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/chemmc_sol.txt
@@ -0,0 +1,108 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: Calculate the probability that a particle in a one-dimensional box of length $a$ is found between 0 and $a / 2$.
+Explanation for Problem 1: The probability that the particle will be found between 0 and $a / 2$ is
+$$
+\operatorname{Prob}(0 \leq x \leq a / 2)=\int_0^{a / 2} \psi^*(x) \psi(x) d x=\frac{2}{a} \int_0^{a / 2} \sin ^2 \frac{n \pi x}{a} d x
+$$
+If we let $n \pi x / a$ be $z$, then we find
+
+$$
+\begin{aligned}
+\operatorname{Prob}(0 \leq x \leq a / 2) & =\frac{2}{n \pi} \int_0^{n \pi / 2} \sin ^2 z d z=\frac{2}{n \pi}\left|\frac{z}{2}-\frac{\sin 2 z}{4}\right|_0^{n \pi / 2} \\
+& =\frac{2}{n \pi}\left(\frac{n \pi}{4}-\frac{\sin n \pi}{4}\right)=\frac{1}{2} \quad \text { (for all } n \text { ) }
+\end{aligned}
+$$
+Thus, the probability that the particle lies in one-half of the interval $0 \leq x \leq a$ is $\frac{1}{2}$.
+Therefore, the answer is \boxed{0.5}.
+
+Promblem 2: Calculate the de Broglie wavelength of an electron traveling at $1.00 \%$ of the speed of light.
+Explanation for Problem 2: The mass of an electron is $9.109 \times 10^{-31} \mathrm{~kg}$. One percent of the speed of light is
+$$
+v=(0.0100)\left(2.998 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)=2.998 \times 10^6 \mathrm{~m} \cdot \mathrm{s}^{-1}
+$$
+The momentum of the electron is given by
+$$
+\begin{aligned}
+p=m_{\mathrm{e}} v & =\left(9.109 \times 10^{-31} \mathrm{~kg}\right)\left(2.998 \times 10^6 \mathrm{~m} \cdot \mathrm{s}^{-1}\right) \\
+& =2.73 \times 10^{-24} \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}
+\end{aligned}
+$$
+The de Broglie wavelength of this electron is
+$$
+\begin{aligned}
+\lambda=\frac{h}{p} & =\frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{2.73 \times 10^{-24} \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}}=2.43 \times 10^{-10} \mathrm{~m} \\
+& =243 \mathrm{pm}
+\end{aligned}
+$$
+This wavelength is of atomic dimensions.
+
+Therefore, the answer is \boxed{243}.
+
+Promblem 3: Find the bonding and antibonding Hückel molecular orbitals for ethene.
+Explanation for Problem 3: The equations for $c_1$ and $c_2$ associated with Equation 11.7 are
+$$
+c_1(\alpha-E)+c_2 \beta=0 \quad \text { and } \quad c_1 \beta+c_2(\alpha-E)=0
+$$
+For $E=\alpha+\beta$, either equation yields $c_1=c_2$. Thus,
+$$
+\psi_{\mathrm{b}}=c_1\left(2 p_{z 1}+2 p_{z 2}\right)
+$$
+
+The value of $c_1$ can be found by requiring that the wave function be normalized. The normalization condition on $\psi_\pi$ gives $c_1^2(1+2 S+1)=1$. Using the Hückel assumption that $S=0$, we find that $c_1=1 / \sqrt{2}$.
+
+Substituting $E=\alpha-\beta$ into either of the equations for $c_1$ and $c_2$ yields $c_1=-c_2$, or
+$$
+\psi_{\mathrm{a}}=c_1\left(2 p_{z 1}-2 p_{z 2}\right)
+$$
+The normalization condition gives $c^2(1-2 S+1)=1$, or $c_1=1 / \sqrt{2}$.
+
+Therefore, the answer is \boxed{0.70710678}.
+
+Promblem 4: The wave function $\Psi_2(1,2)$ given by Equation 9.39 is not normalized as it stands. Determine the normalization constant of $\Psi_2(1,2)$ given that the "1s" parts are normalized.
+Explanation for Problem 4: We want to find the constant $c$ such that
+$$
+I=c^2\left\langle\Psi_2(1,2) \mid \Psi_2(1,2)\right\rangle=1
+$$
+First notice that $\Psi_2(1,2)$ can be factored into the product of a spatial part and a spin part:
+$$
+\begin{aligned}
+\Psi_2(1,2) & =1 s(1) 1 s(2)[\alpha(1) \beta(2)-\alpha(2) \beta(1)] \\
+& =1 s\left(\mathbf{r}_1\right) 1 s\left(\mathbf{r}_2\right)\left[\alpha\left(\sigma_1\right) \beta\left(\sigma_2\right)-\alpha\left(\sigma_2\right) \beta\left(\sigma_1\right)\right]
+\end{aligned}
+$$
+The normalization integral becomes the product of three integrals:
+$$
+I=c^2\langle 1 s(1) \mid 1 s(1)\rangle\langle 1 s(2) \mid 1 s(2)\rangle\langle\alpha(1) \beta(1)-\alpha(2) \beta(1) \mid \alpha(1) \beta(2)-\alpha(2) \beta(1)\rangle
+$$
+The spatial integrals are equal to 1 because we have taken the $1 s$ orbitals to be normalized. Now let's look at the spin integrals. When the two terms in the integrand of the spin integral are multiplied, we get four integrals. One of them is
+$$
+\begin{aligned}
+\iint \alpha^*\left(\sigma_1\right) \beta^*\left(\sigma_2\right) \alpha\left(\sigma_1\right) \beta\left(\sigma_2\right) d \sigma_1 d \sigma_2 & =\langle\alpha(1) \beta(2) \mid \alpha(1) \beta(2)\rangle \\
+& =\langle\alpha(1) \mid \alpha(1)\rangle\langle\beta(2) \mid \beta(2)\rangle=1
+\end{aligned}
+$$
+where once again we point out that integrating over $\sigma_1$ and $\sigma_2$ is purely symbolic; $\sigma_1$ and $\sigma_2$ are discrete variables. Another is
+$$
+\langle\alpha(1) \beta(2) \mid \alpha(2) \beta(1)\rangle=\langle\alpha(1) \mid \beta(1)\rangle\langle\beta(2) \mid \alpha(2)\rangle=0
+$$
+The other two are equal to 1 and 0 , and so
+$$
+I=c^2\left\langle\Psi_2(1,2) \mid \Psi_2(1,2)\right\rangle=2 c^2=1
+$$
+or $c=1 / \sqrt{2}$.
+Therefore, the answer is \boxed{0.70710678}.
+
+Promblem 5: Given that the work function for sodium metal is $2.28 \mathrm{eV}$, what is the threshold frequency $v_0$ for sodium?
+Explanation for Problem 5: We must first convert $\phi$ from electron volts to joules.
+$$
+\begin{aligned}
+\phi & =2.28 \mathrm{eV}=(2.28 \mathrm{eV})\left(1.602 \times 10^{-19} \mathrm{~J} \cdot \mathrm{eV}^{-1}\right) \\
+& =3.65 \times 10^{-19} \mathrm{~J}
+\end{aligned}
+$$
+Using Equation 1.11, we have
+$$
+v_0=\frac{3.65 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}=5.51 \times 10^{14} \mathrm{~Hz}$$
+Therefore, the answer is \boxed{5.51}.
+
diff --git a/configs/datasets/scibench/lib_prompt/class_prompt.txt b/configs/datasets/scibench/lib_prompt/class_prompt.txt
new file mode 100644
index 00000000..30ea60fa
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/class_prompt.txt
@@ -0,0 +1,17 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: Calculate the time needed for a spacecraft to make a Hohmann transfer from Earth to Mars
+Answer: The answer is \boxed{2.24}.
+
+Promblem 2: Next, we treat projectile motion in two dimensions, first without considering air resistance. Let the muzzle velocity of the projectile be $v_0$ and the angle of elevation be $\theta$ (Figure 2-7). Calculate the projectile's range.
+Answer: The answer is \boxed{72}.
+
+Promblem 3: Consider the first stage of a Saturn $V$ rocket used for the Apollo moon program. The initial mass is $2.8 \times 10^6 \mathrm{~kg}$, and the mass of the first-stage fuel is $2.1 \times 10^6$ kg. Assume a mean thrust of $37 \times 10^6 \mathrm{~N}$. The exhaust velocity is $2600 \mathrm{~m} / \mathrm{s}$. Calculate the final speed of the first stage at burnout. 
+Answer: The answer is \boxed{2.16}.
+
+Promblem 4: A particle of mass $m$ starts at rest on top of a smooth fixed hemisphere of radius $a$. Determine the angle at which the particle leaves the hemisphere.
+Answer: The answer is \boxed{48.189685}.
+
+Promblem 5: If the coefficient of static friction between the block and plane in the previous example is $\mu_s=0.4$, at what angle $\theta$ will the block start sliding if it is initially at rest?
+Answer: The answer is \boxed{22}.
+
diff --git a/configs/datasets/scibench/lib_prompt/class_sol.txt b/configs/datasets/scibench/lib_prompt/class_sol.txt
new file mode 100644
index 00000000..fc09e405
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/class_sol.txt
@@ -0,0 +1,169 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: If the coefficient of static friction between the block and plane in the previous example is $\mu_s=0.4$, at what angle $\theta$ will the block start sliding if it is initially at rest?
+Explanation for Problem 1: We need a new sketch to indicate the additional frictional force $f$ (see Figure 2-2b). The static frictional force has the approximate maximum value
+$$
+f_{\max }=\mu_s N
+$$
+and Equation 2.7 becomes, in component form, $y$-direction
+$$
+-F_g \cos \theta+N=0
+$$
+$x$-direction
+$$
+-f_s+F_g \sin \theta=m \ddot{x}
+$$
+The static frictional force $f_s$ will be some value $f_s \leq f_{\max }$ required to keep $\ddot{x}=0$ -that is, to keep the block at rest. However, as the angle $\theta$ of the plane increases, eventually the static frictional force will be unable to keep the block at rest. At that angle $\theta^{\prime}, f_s$ becomes
+$$
+f_s\left(\theta=\theta^{\prime}\right)=f_{\max }=\mu_s N=\mu_s F_g \cos \theta
+$$
+and
+$$
+\begin{aligned}
+m \ddot{x} & =F_g \sin \theta-f_{\max } \\
+m \ddot{x} & =F_g \sin \theta-\mu_s F_g \cos \theta \\
+\ddot{x} & =g\left(\sin \theta-\mu_s \cos \theta\right)
+\end{aligned}
+$$
+Just before the block starts to slide, the acceleration $\ddot{x}=0$, so
+$$
+\begin{aligned}
+\sin \theta-\mu_s \cos \theta & =0 \\
+\tan \theta=\mu_s & =0.4 \\
+\theta=\tan ^{-1}(0.4) & =22^{\circ}
+\end{aligned}
+$$
+
+Therefore, the answer is \boxed{22}.
+
+Promblem 2: Consider the first stage of a Saturn $V$ rocket used for the Apollo moon program. The initial mass is $2.8 \times 10^6 \mathrm{~kg}$, and the mass of the first-stage fuel is $2.1 \times 10^6$ kg. Assume a mean thrust of $37 \times 10^6 \mathrm{~N}$. The exhaust velocity is $2600 \mathrm{~m} / \mathrm{s}$. Calculate the final speed of the first stage at burnout. 
+Explanation for Problem 2: From the thrust (Equation 9.157), we can determine the fuel burn rate:
+$$
+\frac{d m}{d t}=\frac{\text { thrust }}{-u}=\frac{37 \times 10^6 \mathrm{~N}}{-2600 \mathrm{~m} / \mathrm{s}}=-1.42 \times 10^4 \mathrm{~kg} / \mathrm{s}
+$$
+9.11 ROCKET MOTION
+377
+The final rocket mass is $\left(2.8 \times 10^6 \mathrm{~kg}-2.1 \times 10^6 \mathrm{~kg}\right)$ or $0.7 \times 10^6 \mathrm{~kg}$. We can determine the rocket speed at burnout $\left(v_b\right)$ using Equation 9.163.
+$$
+\begin{aligned}
+v_b & =-\frac{9.8 \mathrm{~m} / \mathrm{s}^2\left(2.1 \times 10^6 \mathrm{~kg}\right)}{1.42 \times 10^4 \mathrm{~kg} / \mathrm{s}}+(2600 \mathrm{~m} / \mathrm{s}) \ln \left[\frac{2.8 \times 10^6 \mathrm{~kg}}{0.7 \times 10^6 \mathrm{~kg}}\right] \\
+v_b & =2.16 \times 10^3 \mathrm{~m} / \mathrm{s}
+\end{aligned}
+$$
+
+Therefore, the answer is \boxed{2.16}.
+
+Promblem 3: Halley's comet, which passed around the sun early in 1986, moves in a highly elliptical orbit with an eccentricity of 0.967 and a period of 76 years. Calculate its minimum  distances from the Sun.
+Explanation for Problem 3: Equation 8.49 relates the period of motion with the semimajor axes. Because $m$ (Halley's comet) $\ll m_{\text {Sun }}$
+$$
+\begin{aligned}
+a & =\left(\frac{G m_{\text {Sun }} \tau^2}{4 \pi^2}\right)^{1 / 3} \\
+& =\left[\frac{\left.\left(6.67 \times 10^{-11} \frac{\mathrm{Nm}^2}{\mathrm{~kg}^2}\right)\left(1.99 \times 10^{30} \mathrm{~kg}\right)\left(76 \mathrm{yr} \frac{365 \mathrm{day}}{\mathrm{yr}} \frac{24 \mathrm{hr}}{\mathrm{day}} \frac{3600 \mathrm{~s}}{\mathrm{hr}}\right)^2\right]}{4 \pi^2}\right]^{1 / 3} \\
+a & =2.68 \times 10^{12} \mathrm{m}
+\end{aligned}
+$$
+Using Equation 8.44 , we can determine $r_{\min }$ and $r_{\max }$
+$$
+\begin{aligned}
+& r_{\min }=2.68 \times 10^{12} \mathrm{~m}(1-0.967)=8.8 \times 10^{10} \mathrm{~m} \\
+\end{aligned}
+$$
+Therefore, the answer is \boxed{8.8}.
+
+Promblem 4: Calculate the maximum height change in the ocean tides caused by the Moon.
+Explanation for Problem 4:  We continue to use our simple model of the ocean surrounding Earth. Newton proposed a solution to this calculation by imagining that two wells be dug, one along the direction of high tide (our $x$-axis) and one along the direction of low tide (our $y$-axis). If the tidal height change we want to determine is $h$, then the difference in potential energy of mass $m$ due to the height difference is $m g h$. Let's calculate the difference in work if we move the mass $m$ from point $c$ in Figure 5-12 to the center of Earth and then to point $a$. This work $W$ done by gravity must equal the potential energy change $m g h$. The work $W$ is
+$$
+W=\int_{r+\delta_1}^0 F_{T_y} d y+\int_0^{r+\delta_2} F_{T_x} d x
+$$
+where we use the tidal forces $F_{T_y}$ and $F_{T x}$ of Equations 5.54. The small distances $\delta_1$ and $\delta_2$ are to account for the small variations from a spherical Earth, but these values are so small they can be henceforth neglected. The value for $W$ becomes
+$$
+\begin{aligned}
+W & =\frac{G m M_m}{D^3}\left[\int_r^0(-y) d y+\int_0^r 2 x d x\right] \\
+& =\frac{G m M_m}{D^3}\left(\frac{r^2}{2}+r^2\right)=\frac{3 G m M_m r^2}{2 D^3}
+\end{aligned}
+$$
+Because this work is equal to $m g h$, we have
+$$
+\begin{aligned}
+m g h & =\frac{3 G m M_m r^2}{2 D^3} \\
+h & =\frac{3 G M_m r^2}{2 g D^3}
+\end{aligned}
+$$
+Note that the mass $m$ cancels, and the value of $h$ does not depend on $m$. Nor does it depend on the substance, so to the extent Earth is plastic, similar tidal effects should be (and are) observed for the surface land. If we insert the known values of the constants into Equation 5.55, we find
+$$
+h=\frac{3\left(6.67 \times 10^{-11} \mathrm{~m}^3 / \mathrm{kg} \cdot \mathrm{s}^2\right)\left(7.350 \times 10^{22} \mathrm{~kg}\right)\left(6.37 \times 10^6 \mathrm{~m}\right)^2}{2\left(9.80 \mathrm{~m} / \mathrm{s}^2\right)\left(3.84 \times 10^8 \mathrm{~m}\right)^3}=0.54 \mathrm{~m}
+$$
+
+Therefore, the answer is \boxed{0.54}.
+
+Promblem 5: Next, we treat projectile motion in two dimensions, first without considering air resistance. Let the muzzle velocity of the projectile be $v_0$ and the angle of elevation be $\theta$ (Figure 2-7). Calculate the projectile's range.
+Explanation for Problem 5: Next, we treat projectile motion in two dimensions, first without considering air resistance. Let the muzzle velocity of the projectile be $v_0$ and the angle of elevation be $\theta$ (Figure 2-7). Calculate the projectile's displacement, velocity, and range.
+Solution. Using $\mathbf{F}=m \mathrm{~g}$, the force components become
+$x$-direction
+$$
+0=m \ddot{x}
+$$
+y-direction
+$-m g=m \ddot{y}$
+$(2.31 b)$
+64
+2 / NEWTONIAN MECHANICS-SINGLE PARTICLE
+FIGURE 2-7 Example 2.6.
+Neglect the height of the gun, and assume $x=y=0$ at $t=0$. Then
+$$
+\begin{aligned}
+& \ddot{x}=0 \\
+& \dot{x}=v_0 \cos \theta \\
+& x=v_0 t \cos \theta \\
+& y=-\frac{-g t^2}{2}+v_0 t \sin \theta \\
+&
+\end{aligned}
+$$
+and
+$$
+\begin{aligned}
+& \ddot{y}=-g \\
+& \dot{y}=-g t+v_0 \sin \theta \\
+& y=\frac{-g t^2}{2}+v_0 t \sin \theta
+\end{aligned}
+$$
+
+We can find the range by determining the value of $x$ when the projectile falls back to ground, that is, when $y=0$.
+$$
+y=t\left(\frac{-g t}{2}+v_0 \sin \theta\right)=0
+$$
+One value of $y=0$ occurs for $t=0$ and the other one for $t=T$.
+$$
+\begin{aligned}
+\frac{-g T}{2}+v_0 \sin \theta & =0 \\
+T & =\frac{2 v_0 \sin \theta}{g}
+\end{aligned}
+$$
+2.4 THE EQUATION OF MOTION FOR A PARTICLE
+65
+The range $R$ is found from
+$$
+\begin{aligned}
+x(t=T) & =\text { range }=\frac{2 v_0^2}{g} \sin \theta \cos \theta \\
+R & =\text { range }=\frac{v_0^2}{g} \sin 2 \theta
+\end{aligned}
+$$
+Notice that the maximum range occurs for $\theta=45^{\circ}$.
+Let us use some actual numbers in these calculations. The Germans used a long-range gun named Big Bertha in World War I to bombard Paris. Its muzzle velocity was $1,450 \mathrm{~m} / \mathrm{s}$. Find its predicted range, maximum projectile height, and projectile time of flight if $\theta=55^{\circ}$. We have $v_0=1450 \mathrm{~m} / \mathrm{s}$ and $\theta=55^{\circ}$, so the range (from Equation 2.39) becomes
+$$
+R=\frac{(1450 \mathrm{~m} / \mathrm{s})^2}{9.8 \mathrm{~m} / \mathrm{s}^2}\left[\sin \left(110^{\circ}\right)\right]=202 \mathrm{~km}
+$$
+Big Bertha's actual range was $120 \mathrm{~km}$. The difference is a result of the real effect of air resistance.
+
+To find the maximum predicted height, we need to calculated $y$ for the time $T / 2$ where $T$ is the projectile time of flight:
+$$
+\begin{aligned}
+T & =\frac{(2)(1450 \mathrm{~m} / \mathrm{s})\left(\sin 55^{\circ}\right)}{9.8 \mathrm{~m} / \mathrm{s}^2}=242 \mathrm{~s} \\
+y_{\max }\left(t=\frac{T}{2}\right) & =\frac{-g T^2}{8}+\frac{v_0 T}{2} \sin \theta \\
+& =\frac{-(9.8 \mathrm{~m} / \mathrm{s})(242 \mathrm{~s})^2}{8}+\frac{(1450 \mathrm{~m} / \mathrm{s})(242 \mathrm{~s}) \sin \left(55^{\circ}\right)}{2} \\
+& =72 \mathrm{~km}
+\end{aligned}
+$$
+
+Therefore, the answer is \boxed{72}.
+
diff --git a/configs/datasets/scibench/lib_prompt/diff_prompt.txt b/configs/datasets/scibench/lib_prompt/diff_prompt.txt
new file mode 100644
index 00000000..cd50b23b
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/diff_prompt.txt
@@ -0,0 +1,17 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: At time $t=0$ a tank contains $Q_0 \mathrm{lb}$ of salt dissolved in 100 gal of water; see Figure 2.3.1. Assume that water containing $\frac{1}{4} \mathrm{lb}$ of salt/gal is entering the tank at a rate of $r \mathrm{gal} / \mathrm{min}$ and that the well-stirred mixture is draining from the tank at the same rate. Set up the initial value problem that describes this flow process. By finding the amount of salt $Q(t)$ in the tank at any time, and the limiting amount $Q_L$ that is present after a very long time, if $r=3$ and $Q_0=2 Q_L$, find the time $T$ after which the salt level is within $2 \%$ of $Q_L$. 
+Answer: The answer is \boxed{130.400766848}.
+
+Promblem 2: The logistic model has been applied to the natural growth of the halibut population in certain areas of the Pacific Ocean. ${ }^{12}$ Let $y$, measured in kilograms, be the total mass, or biomass, of the halibut population at time $t$. The parameters in the logistic equation are estimated to have the values $r=0.71 /$ year and $K=80.5 \times 10^6 \mathrm{~kg}$. If the initial biomass is $y_0=0.25 K$, find the biomass 2 years later. 
+Answer: The answer is \boxed{46.7}.
+
+Promblem 3: For instance, suppose that one opens an individual retirement account (IRA) at age 25 and makes annual investments of $\$ 2000$ thereafter in a continuous manner. Assuming a rate of return of $8 \%$, what will be the balance in the IRA at age 65 ?
+Answer: The answer is \boxed{588313}.
+
+Promblem 4: Suppose that a mass weighing $10 \mathrm{lb}$ stretches a spring $2 \mathrm{in}$. If the mass is displaced an additional 2 in. and is then set in motion with an initial upward velocity of $1 \mathrm{ft} / \mathrm{s}$, by determining the position of the mass at any later time, calculate the amplitude of the motion.
+Answer: The answer is \boxed{0.18162}.
+
+Promblem 5: Suppose that a mass weighing $10 \mathrm{lb}$ stretches a spring $2 \mathrm{in}$. If the mass is displaced an additional 2 in. and is then set in motion with an initial upward velocity of $1 \mathrm{ft} / \mathrm{s}$, by determining the position of the mass at any later time, calculate the phase of the motion.
+Answer: The answer is \boxed{-0.40864}.
+
diff --git a/configs/datasets/scibench/lib_prompt/diff_sol.txt b/configs/datasets/scibench/lib_prompt/diff_sol.txt
new file mode 100644
index 00000000..8810e31c
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/diff_sol.txt
@@ -0,0 +1,112 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: The logistic model has been applied to the natural growth of the halibut population in certain areas of the Pacific Ocean. ${ }^{12}$ Let $y$, measured in kilograms, be the total mass, or biomass, of the halibut population at time $t$. The parameters in the logistic equation are estimated to have the values $r=0.71 /$ year and $K=80.5 \times 10^6 \mathrm{~kg}$. If the initial biomass is $y_0=0.25 K$, find the biomass 2 years later. 
+Explanation for Problem 1: It is convenient to scale the solution (11) to the carrying capacity $K$; thus we write Eq. (11) in the form
+$$
+\frac{y}{K}=\frac{y_0 / K}{\left(y_0 / K\right)+\left[1-\left(y_0 / K\right)\right] e^{-r t}}
+$$
+Using the data given in the problem, we find that
+$$
+\frac{y(2)}{K}=\frac{0.25}{0.25+0.75 e^{-1.42}} \cong 0.5797 .
+$$
+Consequently, $y(2) \cong 46.7 \times 10^6 \mathrm{~kg}$.
+
+Therefore, the answer is \boxed{46.7}.
+
+Promblem 2: For instance, suppose that one opens an individual retirement account (IRA) at age 25 and makes annual investments of $\$ 2000$ thereafter in a continuous manner. Assuming a rate of return of $8 \%$, what will be the balance in the IRA at age 65 ?
+Explanation for Problem 2: We have $S_0=0, r=0.08$, and $k=\$ 2000$, and we wish to determine $S(40)$. From Eq. (16) we have
+$$
+S(40)=(25,000)\left(e^{3.2}-1\right)=\$ 588,313
+$$
+
+Therefore, the answer is \boxed{588313}.
+
+Promblem 3: At time $t=0$ a tank contains $Q_0 \mathrm{lb}$ of salt dissolved in 100 gal of water; see Figure 2.3.1. Assume that water containing $\frac{1}{4} \mathrm{lb}$ of salt/gal is entering the tank at a rate of $r \mathrm{gal} / \mathrm{min}$ and that the well-stirred mixture is draining from the tank at the same rate. Set up the initial value problem that describes this flow process. By finding the amount of salt $Q(t)$ in the tank at any time, and the limiting amount $Q_L$ that is present after a very long time, if $r=3$ and $Q_0=2 Q_L$, find the time $T$ after which the salt level is within $2 \%$ of $Q_L$. 
+Explanation for Problem 3: We assume that salt is neither created nor destroyed in the tank. Therefore variations in the amount of salt are due solely to the flows in and out of the tank. More precisely, the rate of change of salt in the tank, $d Q / d t$, is equal to the rate at which salt is flowing in minus the rate at which it is flowing out. In symbols,
+$$
+\frac{d Q}{d t}=\text { rate in }- \text { rate out }
+$$
+The rate at which salt enters the tank is the concentration $\frac{1}{4} \mathrm{lb} / \mathrm{gal}$ times the flow rate $r \mathrm{gal} / \mathrm{min}$, or $(r / 4) \mathrm{lb} / \mathrm{min}$. To find the rate at which salt leaves the tankl we need to multiply the concentration of salt in the tank by the rate of outflow, $r \mathrm{gal} / \mathrm{min}$. Since the rates of flow in and out are equal, the volume of water in the tank remains constant at $100 \mathrm{gal}$, and since the mixture is "well-stirred," the concentration throughout the tank is the same, namely, $[Q(t) / 100] \mathrm{lb} / \mathrm{gal}$.
+ith First Order Equations
+53
+Therefore the rate at which salt leaves the tank is $[r Q(t) / 100] \mathrm{lb} / \mathrm{min}$. Thus the differential equation governing this process is
+$$
+\frac{d Q}{d t}=\frac{r}{4}-\frac{r Q}{100}
+$$
+The initial condition is
+$$
+Q(0)=Q_0
+$$
+Upon thinking about the problem physically, we might anticipate that eventually the mixture originally in the tank will be essentially replaced by the mixture flowing in, whose concentration is $\frac{1}{4} \mathrm{lb} / \mathrm{gal}$. Consequently, we might expect that ultimately the amount of salt in the tank would be very close to $25 \mathrm{lb}$. We can also find the limiting amount $Q_L=25$ by setting $d Q / d t$ equal to zero in Eq. (2) and solving the resulting algebraic equation for $Q$.
+
+To solve the initial value problem (2), (3) analytically, note that Eq. (2) is both linear and separable. Rewriting it in the standard form for a linear equation, we have
+$$
+\frac{d Q}{d t}+\frac{r Q}{100}=\frac{r}{4}
+$$
+Thus the integrating factor is $e^{r t / 100}$ and the general solution is
+$$
+Q(t)=25+c e^{-r t / 100}
+$$
+where $c$ is an arbitrary constant. To satisfy the initial condition (3), we must choose $c=Q_0-25$. Therefore the solution of the initial value problem (2),(3) is
+$$
+Q(t)=25+\left(Q_0-25\right) e^{-r t / 100}
+$$
+or
+$$
+Q(t)=25\left(1-e^{-r t / 100}\right)+Q_0 e^{-r t / 100}
+$$
+From Eq. (6) or (7), you can see that $Q(t) \rightarrow 25$ (lb) as $t \rightarrow \infty$, so the limiting value $Q_L$ is 25 , confirming our physical intuition. Further, $Q(t)$ approaches the limit more rapidly as $r$ increases. In interpreting the solution (7), note that the second term on the right side is the portion of the original salt that remains at time $t$, while the first term gives the amount of salt in the tank due to the action of the flow processes. Plots of the solution for $r=3$ and for several values of $Q_0$ are shown in Figure 2.3.2.
+Now suppose that $r=3$ and $Q_0=2 Q_L=50$; then Eq. (6) becomes
+$$
+Q(t)=25+25 e^{-0.03 t}
+$$
+Since $2 \%$ of 25 is 0.5 , we wish to find the time $T$ at which $Q(t)$ has the value 25.5. Substituting $t=T$ and $Q=25.5$ in Eq. (8) and solving for $T$, we obtain
+$$
+T=(\ln 50) / 0.03 \cong 130.400766848(\mathrm{~min}) .
+$$
+
+Therefore, the answer is \boxed{130.400766848}.
+
+Promblem 4: Suppose that a mass weighing $10 \mathrm{lb}$ stretches a spring $2 \mathrm{in}$. If the mass is displaced an additional 2 in. and is then set in motion with an initial upward velocity of $1 \mathrm{ft} / \mathrm{s}$, by determining the position of the mass at any later time, calculate the phase of the motion.
+Explanation for Problem 4: The spring constant is $k=10 \mathrm{lb} / 2 \mathrm{in} .=60 \mathrm{lb} / \mathrm{ft}$, and the mass is $m=w / g=10 / 32 \mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}$. Hence the equation of motion reduces to
+$$
+u^{\prime \prime}+192 u=0
+$$
+and the general solution is
+$$
+u=A \cos (8 \sqrt{3} t)+B \sin (8 \sqrt{3} t)
+$$
+The solution satisfying the initial conditions $u(0)=1 / 6 \mathrm{ft}$ and $u^{\prime}(0)=-1 \mathrm{ft} / \mathrm{s}$ is
+$$
+u=\frac{1}{6} \cos (8 \sqrt{3} t)-\frac{1}{8 \sqrt{3}} \sin (8 \sqrt{3} t)
+$$
+The natural frequency is $\omega_0=\sqrt{192} \cong 13.856 \mathrm{rad} / \mathrm{s}$, so the period is $T=2 \pi / \omega_0 \cong 0.45345 \mathrm{~s}$. The amplitude $R$ and phase $\delta$ are found from Eqs. (17). We have
+$$
+R^2=\frac{1}{36}+\frac{1}{192}=\frac{19}{576}, \quad \text { so } \quad R \cong 0.18162 \mathrm{ft}
+$$
+The second of Eqs. (17) yields $\tan \delta=-\sqrt{3} / 4$. There are two solutions of this equation, one in the second quadrant and one in the fourth. In the present problem $\cos \delta>0$ and $\sin \delta<0$, so $\delta$ is in the fourth quadrant, namely,
+$$
+\delta=-\arctan (\sqrt{3} / 4) \cong-0.40864 \mathrm{rad}
+$$
+
+Therefore, the answer is \boxed{-0.40864}.
+
+Promblem 5: Suppose that a mass weighing $10 \mathrm{lb}$ stretches a spring $2 \mathrm{in}$. If the mass is displaced an additional 2 in. and is then set in motion with an initial upward velocity of $1 \mathrm{ft} / \mathrm{s}$, by determining the position of the mass at any later time, calculate the amplitude of the motion.
+Explanation for Problem 5: The spring constant is $k=10 \mathrm{lb} / 2 \mathrm{in} .=60 \mathrm{lb} / \mathrm{ft}$, and the mass is $m=w / g=10 / 32 \mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}$. Hence the equation of motion reduces to
+$$
+u^{\prime \prime}+192 u=0
+$$
+and the general solution is
+$$
+u=A \cos (8 \sqrt{3} t)+B \sin (8 \sqrt{3} t)
+$$
+The solution satisfying the initial conditions $u(0)=1 / 6 \mathrm{ft}$ and $u^{\prime}(0)=-1 \mathrm{ft} / \mathrm{s}$ is
+$$
+u=\frac{1}{6} \cos (8 \sqrt{3} t)-\frac{1}{8 \sqrt{3}} \sin (8 \sqrt{3} t)
+$$
+The natural frequency is $\omega_0=\sqrt{192} \cong 13.856 \mathrm{rad} / \mathrm{s}$, so the period is $T=2 \pi / \omega_0 \cong 0.45345 \mathrm{~s}$. The amplitude $R$ and phase $\delta$ are found from Eqs. (17). We have
+$$
+R^2=\frac{1}{36}+\frac{1}{192}=\frac{19}{576}, \quad \text { so } \quad R \cong 0.18162 \mathrm{ft}
+$$
+Therefore, the answer is \boxed{0.18162}.
+
diff --git a/configs/datasets/scibench/lib_prompt/fund_prompt.txt b/configs/datasets/scibench/lib_prompt/fund_prompt.txt
new file mode 100644
index 00000000..3bd9ae72
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/fund_prompt.txt
@@ -0,0 +1,20 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: The huge advantage of using the conservation of energy instead of Newton's laws of motion is that we can jump from the initial state to the final state without considering all the intermediate motion. Here is an example. In Figure, a child of mass $m$ is released from rest at the top of a water slide, at height $h=8.5 \mathrm{~m}$ above the bottom of the slide. Assuming that the slide is frictionless because of the water on it, find the child's speed at the bottom of the slide.
+Answer: The answer is \boxed{13}.
+
+Promblem 2: Let the disk in Figure start from rest at time $t=0$ and also let the tension in the massless cord be $6.0 \mathrm{~N}$ and the angular acceleration of the disk be $-24 \mathrm{rad} / \mathrm{s}^2$. What is its rotational kinetic energy $K$ at $t=2.5 \mathrm{~s}$ ?
+Answer: The answer is \boxed{90}.
+
+Promblem 3: A food shipper pushes a wood crate of cabbage heads (total mass $m=14 \mathrm{~kg}$ ) across a concrete floor with a constant horizontal force $\vec{F}$ of magnitude $40 \mathrm{~N}$. In a straight-line displacement of magnitude $d=0.50 \mathrm{~m}$, the speed of the crate decreases from $v_0=0.60 \mathrm{~m} / \mathrm{s}$ to $v=0.20 \mathrm{~m} / \mathrm{s}$. What is the increase $\Delta E_{\text {th }}$ in the thermal energy of the crate and floor?
+Answer: The answer is \boxed{22.2}.
+
+Promblem 4: A playful astronaut releases a bowling ball, of mass $m=$ $7.20 \mathrm{~kg}$, into circular orbit about Earth at an altitude $h$ of $350 \mathrm{~km}$.
+What is the mechanical energy $E$ of the ball in its orbit?
+Answer: The answer is \boxed{-214}.
+
+Promblem 5: If the particles in a system all move together, the com moves with them-no trouble there. But what happens when they move in different directions with different accelerations? Here is an example.
+
+The three particles in Figure are initially at rest. Each experiences an external force due to bodies outside the three-particle system. The directions are indicated, and the magnitudes are $F_1=6.0 \mathrm{~N}, F_2=12 \mathrm{~N}$, and $F_3=14 \mathrm{~N}$. What is the acceleration of the center of mass of the system?
+Answer: The answer is \boxed{ 1.16}.
+
diff --git a/configs/datasets/scibench/lib_prompt/fund_sol.txt b/configs/datasets/scibench/lib_prompt/fund_sol.txt
new file mode 100644
index 00000000..5399b46b
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/fund_sol.txt
@@ -0,0 +1,135 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: A food shipper pushes a wood crate of cabbage heads (total mass $m=14 \mathrm{~kg}$ ) across a concrete floor with a constant horizontal force $\vec{F}$ of magnitude $40 \mathrm{~N}$. In a straight-line displacement of magnitude $d=0.50 \mathrm{~m}$, the speed of the crate decreases from $v_0=0.60 \mathrm{~m} / \mathrm{s}$ to $v=0.20 \mathrm{~m} / \mathrm{s}$. What is the increase $\Delta E_{\text {th }}$ in the thermal energy of the crate and floor?
+Explanation for Problem 1: 
+We can relate $\Delta E_{\text {th }}$ to the work $W$ done by $\vec{F}$ with the energy statement of Eq. 8-33 for a system that involves friction:
+$$
+W=\Delta E_{\text {mec }}+\Delta E_{\text {th }} .
+$$
+Calculations: We know the value of $W$ from (a). The change $\Delta E_{\text {mec }}$ in the crate's mechanical energy is just the change in its kinetic energy because no potential energy changes occur, so we have
+$$
+\Delta E_{\mathrm{mec}}=\Delta K=\frac{1}{2} m v^2-\frac{1}{2} m v_0^2 .
+$$
+Substituting this into Eq. 8-34 and solving for $\Delta E_{\mathrm{th}}$, we find
+$$
+\begin{aligned}
+\Delta E_{\mathrm{th}} & =W-\left(\frac{1}{2} m v^2-\frac{1}{2} m v_0^2\right)=W-\frac{1}{2} m\left(v^2-v_0^2\right) \\
+& =20 \mathrm{~J}-\frac{1}{2}(14 \mathrm{~kg})\left[(0.20 \mathrm{~m} / \mathrm{s})^2-(0.60 \mathrm{~m} / \mathrm{s})^2\right] \\
+& =22.2 \mathrm{~J} \approx 22 \mathrm{~J} .
+\end{aligned}
+$$
+
+Therefore, the answer is \boxed{22.2}.
+
+Promblem 2: If the particles in a system all move together, the com moves with them-no trouble there. But what happens when they move in different directions with different accelerations? Here is an example.
+
+The three particles in Figure are initially at rest. Each experiences an external force due to bodies outside the three-particle system. The directions are indicated, and the magnitudes are $F_1=6.0 \mathrm{~N}, F_2=12 \mathrm{~N}$, and $F_3=14 \mathrm{~N}$. What is the acceleration of the center of mass of the system?
+Explanation for Problem 2: The position of the center of mass is marked by a dot in the figure. We can treat the center of mass as if it were a real particle, with a mass equal to the system's total mass $M=16 \mathrm{~kg}$. We can also treat the three external forces as if they act at the center of mass (Fig. 9-7b).
+
+Calculations: We can now apply Newton's second law $\left(\vec{F}_{\text {net }}=m \vec{a}\right)$ to the center of mass, writing
+$$
+\vec{F}_{\text {net }}=M \vec{a}_{\mathrm{com}}
+$$
+or
+$$
+\begin{aligned}
+& \vec{F}_1+\vec{F}_2+\vec{F}_3=M \vec{a}_{\mathrm{com}} \\
+& \vec{a}_{\mathrm{com}}=\frac{\vec{F}_1+\vec{F}_2+\vec{F}_3}{M} .
+\end{aligned}
+$$
+Equation 9-20 tells us that the acceleration $\vec{a}_{\text {com }}$ of the center of mass is in the same direction as the net external force $\vec{F}_{\text {net }}$ on the system (Fig. 9-7b). Because the particles are initially at rest, the center of mass must also be at rest. As the center of mass then begins to accelerate, it must move off in the common direction of $\vec{a}_{\text {com }}$ and $\vec{F}_{\text {net }}$.
+
+We can evaluate the right side of Eq. 9-21 directly on a vector-capable calculator, or we can rewrite Eq. 9-21 in component form, find the components of $\vec{a}_{\text {com }}$, and then find $\vec{a}_{\text {com }}$. Along the $x$ axis, we have
+$$
+\begin{aligned}
+a_{\mathrm{com}, x} & =\frac{F_{1 x}+F_{2 x}+F_{3 x}}{M} \\
+& =\frac{-6.0 \mathrm{~N}+(12 \mathrm{~N}) \cos 45^{\circ}+14 \mathrm{~N}}{16 \mathrm{~kg}}=1.03 \mathrm{~m} / \mathrm{s}^2 .
+\end{aligned}
+$$
+Along the $y$ axis, we have
+$$
+\begin{aligned}
+a_{\mathrm{com}, y} & =\frac{F_{1 y}+F_{2 y}+F_{3 y}}{M} \\
+& =\frac{0+(12 \mathrm{~N}) \sin 45^{\circ}+0}{16 \mathrm{~kg}}=0.530 \mathrm{~m} / \mathrm{s}^2 .
+\end{aligned}
+$$
+From these components, we find that $\vec{a}_{\mathrm{com}}$ has the magnitude
+$$
+\begin{aligned}
+a_{\mathrm{com}} & =\sqrt{\left(a_{\mathrm{com}, x}\right)^2+\left(a_{\text {com }, y}\right)^2} \\
+& =1.16 \mathrm{~m} / \mathrm{s}^2 
+\end{aligned}
+$$
+
+Therefore, the answer is \boxed{ 1.16}.
+
+Promblem 3: While you are operating a Rotor (a large, vertical, rotating cylinder found in amusement parks), you spot a passenger in acute distress and decrease the angular velocity of the cylinder from $3.40 \mathrm{rad} / \mathrm{s}$ to $2.00 \mathrm{rad} / \mathrm{s}$ in $20.0 \mathrm{rev}$, at constant angular acceleration. (The passenger is obviously more of a "translation person" than a "rotation person.")
+What is the constant angular acceleration during this decrease in angular speed?
+Explanation for Problem 3: Because the cylinder's angular acceleration is constant, we can relate it to the angular velocity and angular displacement via the basic equations for constant angular acceleration (Eqs. 10-12 and 10-13).
+
+Calculations: Let's first do a quick check to see if we can solve the basic equations. The initial angular velocity is $\omega_0=3.40$
+$\mathrm{rad} / \mathrm{s}$, the angular displacement is $\theta-\theta_0=20.0 \mathrm{rev}$, and the angular velocity at the end of that displacement is $\omega=2.00$ $\mathrm{rad} / \mathrm{s}$. In addition to the angular acceleration $\alpha$ that we want, both basic equations also contain time $t$, which we do not necessarily want.
+
+To eliminate the unknown $t$, we use Eq. 10-12 to write
+$$
+t=\frac{\omega-\omega_0}{\alpha}
+$$
+which we then substitute into Eq. 10-13 to write
+$$
+\theta-\theta_0=\omega_0\left(\frac{\omega-\omega_0}{\alpha}\right)+\frac{1}{2} \alpha\left(\frac{\omega-\omega_0}{\alpha}\right)^2 .
+$$
+Solving for $\alpha$, substituting known data, and converting 20 rev to $125.7 \mathrm{rad}$, we find
+$$
+\begin{aligned}
+\alpha & =\frac{\omega^2-\omega_0^2}{2\left(\theta-\theta_0\right)}=\frac{(2.00 \mathrm{rad} / \mathrm{s})^2-(3.40 \mathrm{rad} / \mathrm{s})^2}{2(125.7 \mathrm{rad})} \\
+& =-0.0301 \mathrm{rad} / \mathrm{s}^2
+\end{aligned}
+$$
+
+Therefore, the answer is \boxed{-0.0301}.
+
+Promblem 4: An astronaut whose height $h$ is $1.70 \mathrm{~m}$ floats "feet down" in an orbiting space shuttle at distance $r=6.77 \times 10^6 \mathrm{~m}$ away from the center of Earth. What is the difference between the gravitational acceleration at her feet and at her head?
+Explanation for Problem 4: We can approximate Earth as a uniform sphere of mass $M_E$. Then, from Eq. 13-11, the gravitational acceleration at any distance $r$ from the center of Earth is
+$$
+a_g=\frac{G M_E}{r^2}
+$$
+We might simply apply this equation twice, first with $r=$ $6.77 \times 10^6 \mathrm{~m}$ for the location of the feet and then with $r=6.77 \times 10^6 \mathrm{~m}+1.70 \mathrm{~m}$ for the location of the head. However, a calculator may give us the same value for $a_g$ twice, and thus a difference of zero, because $h$ is so much smaller than $r$. Here's a more promising approach: Because we have a differential change $d r$ in $r$ between the astronaut's feet and head, we should differentiate Eq. 13-15 with respect to $r$.
+Calculations: The differentiation gives us
+$$
+d a_g=-2 \frac{G M_E}{r^3} d r
+$$
+where $d a_g$ is the differential change in the gravitational acceleration due to the differential change $d r$ in $r$. For the astronaut, $d r=h$ and $r=6.77 \times 10^6 \mathrm{~m}$. Substituting data into Eq.13-16, we find
+$$
+\begin{aligned}
+d a_g & =-2 \frac{\left(6.67 \times 10^{-11} \mathrm{~m}^3 / \mathrm{kg} \cdot \mathrm{s}^2\right)\left(5.98 \times 10^{24} \mathrm{~kg}\right)}{\left(6.77 \times 10^6 \mathrm{~m}\right)^3}(1.70 \mathrm{~m}) \\
+& =-4.37 \times 10^{-6} \mathrm{~m} / \mathrm{s}^2, \quad \text { (Answer) }
+\end{aligned}
+$$
+Therefore, the answer is \boxed{-4.37 }.
+
+Promblem 5: A $2.00 \mathrm{~kg}$ particle moves along an $x$ axis in one-dimensional motion while a conservative force along that axis acts on it. The potential energy $U(x)$ associated with the force is plotted in Fig. 8-10a. That is, if the particle were placed at any position between $x=0$ and $x=7.00 \mathrm{~m}$, it would have the plotted value of $U$. At $x=6.5 \mathrm{~m}$, the particle has velocity $\vec{v}_0=(-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}$
+From Figure, determine the particle's speed at $x_1=4.5 \mathrm{~m}$.
+Explanation for Problem 5: 
+(1) The particle's kinetic energy is given by Eq. 7-1 $\left(K=\frac{1}{2} m v^2\right)$. (2) Because only a conservative force acts on the particle, the mechanical energy $E_{\mathrm{mec}}(=K+U)$ is conserved as the particle moves. (3) Therefore, on a plot of $U(x)$ such as Fig. 8-10a, the kinetic energy is equal to the difference between $E_{\mathrm{mec}}$ and $U$.
+Calculations: At $x=6.5 \mathrm{~m}$, the particle has kinetic energy
+$$
+\begin{aligned}
+K_0 & =\frac{1}{2} m v_0^2=\frac{1}{2}(2.00 \mathrm{~kg})(4.00 \mathrm{~m} / \mathrm{s})^2 \\
+& =16.0 \mathrm{~J} .
+\end{aligned}
+$$
+Because the potential energy there is $U=0$, the mechanical energy is
+$$
+E_{\text {mec }}=K_0+U_0=16.0 \mathrm{~J}+0=16.0 \mathrm{~J} .
+$$
+This value for $E_{\mathrm{mec}}$ is plotted as a horizontal line in Fig. 8-10a. From that figure we see that at $x=4.5 \mathrm{~m}$, the potential energy is $U_1=7.0 \mathrm{~J}$. The kinetic energy $K_1$ is the difference between $E_{\text {mec }}$ and $U_1$ :
+$$
+K_1=E_{\text {mec }}-U_1=16.0 \mathrm{~J}-7.0 \mathrm{~J}=9.0 \mathrm{~J} .
+$$
+Because $K_1=\frac{1}{2} m v_1^2$, we find
+$$
+v_1=3.0 \mathrm{~m} / \mathrm{s}
+$$
+
+Therefore, the answer is \boxed{3.0}.
+
diff --git a/configs/datasets/scibench/lib_prompt/matter_prompt.txt b/configs/datasets/scibench/lib_prompt/matter_prompt.txt
new file mode 100644
index 00000000..cc8022f8
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/matter_prompt.txt
@@ -0,0 +1,21 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: Using the Planck distribution
+Compare the energy output of a black-body radiator (such as an incandescent lamp) at two different wavelengths by calculating the ratio of the energy output at $450 \mathrm{~nm}$ (blue light) to that at $700 \mathrm{~nm}$ (red light) at $298 \mathrm{~K}$.
+
+Answer: The answer is \boxed{2.10}.
+
+Promblem 2: Using the perfect gas equation
+Calculate the pressure in kilopascals exerted by $1.25 \mathrm{~g}$ of nitrogen gas in a flask of volume $250 \mathrm{~cm}^3$ at $20^{\circ} \mathrm{C}$.
+Answer: The answer is \boxed{435}.
+
+Promblem 3: Calculate the typical wavelength of neutrons after reaching thermal equilibrium with their surroundings at $373 \mathrm{~K}$. For simplicity, assume that the particles are travelling in one dimension.
+Answer: The answer is \boxed{226}.
+
+Promblem 4: Determine the energies and degeneracies of the lowest four energy levels of an ${ }^1 \mathrm{H}^{35} \mathrm{Cl}$ molecule freely rotating in three dimensions. What is the frequency of the transition between the lowest two rotational levels? The moment of inertia of an ${ }^1 \mathrm{H}^{35} \mathrm{Cl}$ molecule is $2.6422 \times 10^{-47} \mathrm{~kg} \mathrm{~m}^2$.
+
+Answer: The answer is \boxed{635.7}.
+
+Promblem 5: The single electron in a certain excited state of a hydrogenic $\mathrm{He}^{+}$ion $(Z=2)$ is described by the wavefunction $R_{3,2}(r) \times$ $Y_{2,-1}(\theta, \phi)$. What is the energy of its electron?
+Answer: The answer is \boxed{ -6.04697}.
+
diff --git a/configs/datasets/scibench/lib_prompt/matter_sol.txt b/configs/datasets/scibench/lib_prompt/matter_sol.txt
new file mode 100644
index 00000000..908037a9
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/matter_sol.txt
@@ -0,0 +1,120 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: The single electron in a certain excited state of a hydrogenic $\mathrm{He}^{+}$ion $(Z=2)$ is described by the wavefunction $R_{3,2}(r) \times$ $Y_{2,-1}(\theta, \phi)$. What is the energy of its electron?
+Explanation for Problem 1: Replacing $\mu$ by $m_{\mathrm{e}}$ and using $\hbar=h / 2 \pi$, we can write the expression for the energy (eqn 17.7) as
+$$
+E_n=-\frac{Z^2 m_e e^4}{8 \varepsilon_0^2 h^2 n^2}=-\frac{Z^2 h c \tilde{R}_{\infty}}{n^2}
+$$
+with
+$$
+\begin{aligned}
+& \times \underbrace{2.997926 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}}_c \\
+& =109737 \mathrm{~cm}^{-1} \\
+&
+\end{aligned}
+$$
+and
+$$
+\begin{aligned}
+h c \tilde{R}_{\infty}= & \left(6.62608 \times 10^{-34} \mathrm{Js}\right) \times\left(2.997926 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right) \\
+& \times\left(109737 \mathrm{~cm}^{-1}\right) \\
+= & 2.17987 \times 10^{-18} \mathrm{~J}
+\end{aligned}
+$$
+Therefore, for $n=3$, the energy is
+$$
+\begin{aligned}
+& E_3=-\frac{\overbrace{4}^{Z^2} \times \overbrace{2.17987 \times 10^{-18} \mathrm{~J}}^{h c \tilde{R}_{\infty}}}{\underset{\tilde{n}^2}{9}} \\
+& =-9.68831 \times 10^{-19} \mathrm{~J} \\
+&
+\end{aligned}
+$$
+or $-0.968831 \mathrm{aJ}$ (a, for atto, is the prefix that denotes $10^{-18}$ ). In some applications it is useful to express the energy in electronvolts $\left(1 \mathrm{eV}=1.602176 \times 10^{-19} \mathrm{~J}\right)$; in this case, $E_3=-6.04697 \mathrm{eV}$
+Therefore, the answer is \boxed{ -6.04697}.
+
+Promblem 2: Using the Planck distribution
+Compare the energy output of a black-body radiator (such as an incandescent lamp) at two different wavelengths by calculating the ratio of the energy output at $450 \mathrm{~nm}$ (blue light) to that at $700 \mathrm{~nm}$ (red light) at $298 \mathrm{~K}$.
+
+Explanation for Problem 2: At a temperature $T$, the ratio of the spectral density of states at a wavelength $\lambda_1$ to that at $\lambda_2$ is given by
+$$
+\frac{\rho\left(\lambda_1, T\right)}{\rho\left(\lambda_2, T\right)}=\left(\frac{\lambda_2}{\lambda_1}\right)^5 \times \frac{\left(\mathrm{e}^{h c / \lambda_2 k T}-1\right)}{\left(\mathrm{e}^{h c / \lambda_1 k T}-1\right)}
+$$
+Insert the data and evaluate this ratio.
+Answer With $\lambda_1=450 \mathrm{~nm}$ and $\lambda_2=700 \mathrm{~nm}$,
+$$
+\begin{aligned}
+\frac{h c}{\lambda_1 k T} & =\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(2.998 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(450 \times 10^{-9} \mathrm{~m}\right) \times\left(1.381 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right) \times(298 \mathrm{~K})}=107.2 \ldots \\
+\frac{h c}{\lambda_2 k T} & =\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(2.998 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(700 \times 10^{-9} \mathrm{~m}\right) \times\left(1.381 \times 10^{-23} \mathrm{JK}^{-1}\right) \times(298 \mathrm{~K})}=68.9 \ldots
+\end{aligned}
+$$
+and therefore
+$$
+\begin{aligned}
+& \frac{\rho(450 \mathrm{~nm}, 298 \mathrm{~K})}{\rho(700 \mathrm{~nm}, 298 \mathrm{~K})}=\left(\frac{700 \times 10^{-9} \mathrm{~m}}{450 \times 10^{-9} \mathrm{~m}}\right)^5 \times \frac{\left(\mathrm{e}^{68.9 \cdots}-1\right)}{\left(\mathrm{e}^{107.2 \cdots}-1\right)} \\
+& =9.11 \times\left(2.30 \times 10^{-17}\right)=2.10 \times 10^{-16}
+\end{aligned}
+$$
+Therefore, the answer is \boxed{2.10}.
+
+Promblem 3: Determine the energies and degeneracies of the lowest four energy levels of an ${ }^1 \mathrm{H}^{35} \mathrm{Cl}$ molecule freely rotating in three dimensions. What is the frequency of the transition between the lowest two rotational levels? The moment of inertia of an ${ }^1 \mathrm{H}^{35} \mathrm{Cl}$ molecule is $2.6422 \times 10^{-47} \mathrm{~kg} \mathrm{~m}^2$.
+
+Explanation for Problem 3: First, note that
+$$
+\frac{\hbar^2}{2 I}=\frac{\left(1.055 \times 10^{-34} \mathrm{Js}^2\right.}{2 \times\left(2.6422 \times 10^{-47} \mathrm{~kg} \mathrm{~m}^2\right)}=2.106 \ldots \times 10^{-22} \mathrm{~J}
+$$
+or $0.2106 \ldots$ zJ. We now draw up the following table, where the molar energies are obtained by multiplying the individual energies by Avogadro's constant:
+\begin{tabular}{llll}
+\hline$J$ & $E / z J$ & $E /\left(\mathrm{J} \mathrm{mol}^{-1}\right)$ & Degeneracy \\
+\hline 0 & 0 & 0 & 1 \\
+1 & 0.4212 & 253.6 & 3 \\
+2 & 1.264 & 760.9 & 5 \\
+3 & 2.527 & 1522 & 7 \\
+\hline
+\end{tabular}
+
+The energy separation between the two lowest rotational energy levels $\left(J=0\right.$ and 1 ) is $4.212 \times 10^{-22} \mathrm{~J}$, which corresponds to a photon frequency of
+$$
+\nu=\frac{\Delta E}{h}=\frac{4.212 \times 10^{-22} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}=6.357 \times 10^{11} \mathrm{~s}^{-1}=635.7 \mathrm{GHz}
+$$
+Therefore, the answer is \boxed{635.7}.
+
+Promblem 4: Calculate the shielding constant for the proton in a free $\mathrm{H}$ atom.
+Explanation for Problem 4: The wavefunction for a hydrogen 1 s orbital is
+$$
+\psi=\left(\frac{1}{\pi a_0^3}\right)^{1 / 2} \mathrm{e}^{-r / a_0}
+$$
+so, because $\mathrm{d} \tau=r^2 \mathrm{~d} r \sin \theta \mathrm{d} \theta \mathrm{d} \phi$, the expectation value of $1 / r$ is written as
+$$
+\begin{aligned}
+\left\langle\frac{1}{r}\right\rangle & =\int \frac{\psi^* \psi}{r} \mathrm{~d} \tau=\frac{1}{\pi a_0^3} \int_0^{2 \pi} \mathrm{d} \phi \int_0^\pi \sin \theta \mathrm{d} \theta \int_0^{\infty} r \mathrm{e}^{-2 r / a_0} \mathrm{~d} r \\
+& =\frac{4}{a_0^3} \overbrace{\int_0^{\infty} r \mathrm{e}^{-2 r / a_0} \mathrm{~d} r}^{a_0^2 / 4 \text { (Integral E.1) }}=\frac{1}{a_0}
+\end{aligned}
+$$
+where we used the integral listed in the Resource section. Therefore,
+$$
+\begin{aligned}
+& =\frac{\left(1.602 \times 10^{-19} \mathrm{C}\right)^2 \times(4 \pi \times 10^{-7} \overbrace{\mathrm{J}}^{\mathrm{Jg} \mathrm{m}^2 \mathrm{~s}^{-2}} \mathrm{~s}^2 \mathrm{C}^{-2} \mathrm{~m}^{-1})}{12 \pi \times\left(9.109 \times 10^{-31} \mathrm{~kg}\right) \times\left(5.292 \times 10^{-11} \mathrm{~m}\right)} \\
+& =1.775 \times 10^{-5} \\
+&
+\end{aligned}
+$$
+
+Therefore, the answer is \boxed{1.775}.
+
+Promblem 5: Estimate the molar volume of $\mathrm{CO}_2$ at $500 \mathrm{~K}$ and 100 atm by treating it as a van der Waals gas.
+Explanation for Problem 5: According to Table 36.3, $a=3.610 \mathrm{dm}^6$ atm $\mathrm{mol}^{-2}$ and $b=4.29 \times 10^{-2} \mathrm{dm}^3 \mathrm{~mol}^{-1}$. Under the stated conditions, $R T / p=0.410 \mathrm{dm}^3 \mathrm{~mol}^{-1}$. The coefficients in the equation for $V_{\mathrm{m}}$ are therefore
+$$
+\begin{aligned}
+b+R T / p & =0.453 \mathrm{dm}^3 \mathrm{~mol}^{-1} \\
+a / p & =3.61 \times 10^{-2}\left(\mathrm{dm}^3 \mathrm{~mol}^{-1}\right)^2 \\
+a b / p & =1.55 \times 10^{-3}\left(\mathrm{dm}^3 \mathrm{~mol}^{-1}\right)^3
+\end{aligned}
+$$
+Therefore, on writing $x=V_{\mathrm{m}} /\left(\mathrm{dm}^3 \mathrm{~mol}^{-1}\right)$, the equation to solve is
+$$
+x^3-0.453 x^2+\left(3.61 \times 10^{-2}\right) x-\left(1.55 \times 10^{-3}\right)=0
+$$
+The acceptable root is $x=0.366$, which implies that $V_{\mathrm{m}}=0.366$ $\mathrm{dm}^3 \mathrm{~mol}^{-1}$. The molar volume of a perfect gas under these conditions is $0.410 \mathrm{dm}^3 \mathrm{~mol}^{-1}$.
+
+Therefore, the answer is \boxed{0.366}.
+
diff --git a/configs/datasets/scibench/lib_prompt/quan_prompt.txt b/configs/datasets/scibench/lib_prompt/quan_prompt.txt
new file mode 100644
index 00000000..45b05595
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/quan_prompt.txt
@@ -0,0 +1,17 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: A one-particle, one-dimensional system has $\Psi=a^{-1 / 2} e^{-|x| / a}$ at $t=0$, where $a=1.0000 \mathrm{~nm}$. At $t=0$, the particle's position is measured.  (b) Find the probability that the measured value is between $x=0$ and $x=2 \mathrm{~nm}$.
+Answer: The answer is \boxed{0.4908}.
+
+Promblem 2: Calculate the ground-state energy of the hydrogen atom using SI units and convert the result to electronvolts.
+Answer: The answer is \boxed{-13.598 }.
+
+Promblem 3: The strongest infrared band of ${ }^{12} \mathrm{C}^{16} \mathrm{O}$ occurs at $\widetilde{\nu}=2143 \mathrm{~cm}^{-1}$. Find the force constant of ${ }^{12} \mathrm{C}^{16} \mathrm{O}$. 
+Answer: The answer is \boxed{1855}.
+
+Promblem 4: Find the probability that the electron in the ground-state $\mathrm{H}$ atom is less than a distance $a$ from the nucleus.
+Answer: The answer is \boxed{0.323}.
+
+Promblem 5: A one-particle, one-dimensional system has $\Psi=a^{-1 / 2} e^{-|x| / a}$ at $t=0$, where $a=1.0000 \mathrm{~nm}$. At $t=0$, the particle's position is measured. (a) Find the probability that the measured value lies between $x=1.5000 \mathrm{~nm}$ and $x=1.5001 \mathrm{~nm}$.
+Answer: The answer is \boxed{4.979}.
+
diff --git a/configs/datasets/scibench/lib_prompt/quan_sol.txt b/configs/datasets/scibench/lib_prompt/quan_sol.txt
new file mode 100644
index 00000000..73ccfc89
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/quan_sol.txt
@@ -0,0 +1,75 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: A one-particle, one-dimensional system has $\Psi=a^{-1 / 2} e^{-|x| / a}$ at $t=0$, where $a=1.0000 \mathrm{~nm}$. At $t=0$, the particle's position is measured. (a) Find the probability that the measured value lies between $x=1.5000 \mathrm{~nm}$ and $x=1.5001 \mathrm{~nm}$.
+Explanation for Problem 1: (a) In this tiny interval, $x$ changes by only $0.0001 \mathrm{~nm}$, and $\Psi$ goes from $e^{-1.5000} \mathrm{~nm}^{-1 / 2}=0.22313 \mathrm{~nm}^{-1 / 2}$ to $e^{-1.5001} \mathrm{~nm}^{-1 / 2}=0.22311 \mathrm{~nm}^{-1 / 2}$, so $\Psi$ is nearly constant in this interval, and it is a very good approximation to consider this interval as infinitesimal. The desired probability is given by (1.15) as
+$$
+\begin{aligned}
+|\Psi|^2 d x=a^{-1} e^{-2|x| / a} d x & =(1 \mathrm{~nm})^{-1} e^{-2(1.5 \mathrm{~nm}) /(1 \mathrm{~nm})}(0.0001 \mathrm{~nm}) \\
+& =4.979 \times 10^{-6}
+\end{aligned}
+$$
+Therefore, the answer is \boxed{4.979}.
+
+Promblem 2: The lowest-frequency pure-rotational absorption line of ${ }^{12} \mathrm{C}^{32} \mathrm{~S}$ occurs at $48991.0 \mathrm{MHz}$. Find the bond distance in ${ }^{12} \mathrm{C}^{32} \mathrm{~S}$.
+Explanation for Problem 2: The lowest-frequency rotational absorption is the $J=0 \rightarrow 1$ line. Equations (1.4), $(6.52)$, and $(6.51)$ give
+$$
+h \nu=E_{\mathrm{upper}}-E_{\mathrm{lower}}=\frac{1(2) \hbar^2}{2 \mu d^2}-\frac{0(1) \hbar^2}{2 \mu d^2}
+$$
+which gives $d=\left(h / 4 \pi^2 \nu \mu\right)^{1 / 2}$. Table A.3 in the Appendix gives
+$$
+\mu=\frac{m_1 m_2}{m_1+m_2}=\frac{12(31.97207)}{(12+31.97207)} \frac{1}{6.02214 \times 10^{23}} \mathrm{~g}=1.44885 \times 10^{-23} \mathrm{~g}
+$$
+The SI unit of mass is the kilogram, and
+$$
+\begin{aligned}
+d=\frac{1}{2 \pi}\left(\frac{h}{\nu_{0 \rightarrow 1} \mu}\right)^{1 / 2} & =\frac{1}{2 \pi}\left[\frac{6.62607 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left(48991.0 \times 10^6 \mathrm{~s}^{-1}\right)\left(1.44885 \times 10^{-26} \mathrm{~kg}\right)}\right]^{1 / 2} \\
+& =1.5377 \times 10^{-10} \mathrm{~m}
+\end{aligned}
+$$
+
+Therefore, the answer is \boxed{1.5377}.
+
+Promblem 3: Find the probability that the electron in the ground-state $\mathrm{H}$ atom is less than a distance $a$ from the nucleus.
+Explanation for Problem 3: We want the probability that the radial coordinate lies between 0 and $a$. This is found by taking the infinitesimal probability (6.116) of being between $r$ and $r+d r$ and summing it over the range from 0 to $a$. This sum of infinitesimal quantities is the definite integral
+$$
+\begin{aligned}
+\int_0^a R_{n l}^2 r^2 d r & =\frac{4}{a^3} \int_0^a e^{-2 r / a} r^2 d r=\left.\frac{4}{a^3} e^{-2 r / a}\left(-\frac{r^2 a}{2}-\frac{2 r a^2}{4}-\frac{2 a^3}{8}\right)\right|_0 ^a \\
+& =4\left[e^{-2}(-5 / 4)-(-1 / 4)\right]=0.323
+\end{aligned}
+$$
+Therefore, the answer is \boxed{0.323}.
+
+Promblem 4: A one-particle, one-dimensional system has $\Psi=a^{-1 / 2} e^{-|x| / a}$ at $t=0$, where $a=1.0000 \mathrm{~nm}$. At $t=0$, the particle's position is measured.  (b) Find the probability that the measured value is between $x=0$ and $x=2 \mathrm{~nm}$.
+Explanation for Problem 4: (b) Use of Eq. (1.23) and $|x|=x$ for $x \geq 0$ gives
+$$
+\begin{aligned}
+\operatorname{Pr}(0 \leq x \leq 2 \mathrm{~nm}) & =\int_0^{2 \mathrm{~nm}}|\Psi|^2 d x=a^{-1} \int_0^{2 \mathrm{~nm}} e^{-2 x / a} d x \\
+& =-\left.\frac{1}{2} e^{-2 x / a}\right|_0 ^{2 \mathrm{~nm}}=-\frac{1}{2}\left(e^{-4}-1\right)=0.4908
+\end{aligned}
+$$
+Therefore, the answer is \boxed{0.4908}.
+
+Promblem 5: In this example, $2.50 \mathrm{~mol}$ of an ideal gas with $C_{V, m}=12.47 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ is expanded adiabatically against a constant external pressure of 1.00 bar. The initial temperature and pressure of the gas are $325 \mathrm{~K}$ and $2.50 \mathrm{bar}$, respectively. The final pressure is 1.25 bar. Calculate the final temperature, $q, w, \Delta U$.
+Explanation for Problem 5: Because the process is adiabatic, $q=0$, and $\Delta U=w$. Therefore,
+$$
+\Delta U=n C_{\mathrm{v}, m}\left(T_f-T_i\right)=-P_{e x t e r n a l}\left(V_f-V_i\right)
+$$
+Using the ideal gas law,
+$$
+\begin{aligned}
+& n C_{\mathrm{v}, m}\left(T_f-T_i\right)=-n R P_{\text {external }}\left(\frac{T_f}{P_f}-\frac{T_i}{P_i}\right) \\
+& T_f\left(n C_{\mathrm{v}, m}+\frac{n R P_{\text {external }}}{P_f}\right)=T_i\left(n C_{\mathrm{v}, m}+\frac{n R P_{\text {external }}}{P_i}\right) \\
+& T_f=T_i\left(\frac{C_{\mathrm{v}, m}+\frac{R P_{\text {external }}}{P_i}}{C_{\mathrm{v}, m}+\frac{R P_{\text {external }}}{P_f}}\right) \\
+& =325 \mathrm{~K} \times\left(\frac{12.47 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}+\frac{8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 1.00 \mathrm{bar}}{2.50 \mathrm{bar}}}{12.47 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}+\frac{8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 1.00 \mathrm{bar}}{1.25 \mathrm{bar}}}\right)=268 \mathrm{~K} \\
+&
+\end{aligned}
+$$
+We calculate $\Delta U=w$ from
+$$
+\begin{aligned}
+\Delta U & =n C_{V, m}\left(T_f-T_i\right)=2.5 \mathrm{~mol} \times 12.47 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times(268 \mathrm{~K}-325 \mathrm{~K}) \\
+& =-1.78 \mathrm{~kJ}
+\end{aligned}
+$$
+Therefore, the answer is \boxed{-1.78}.
+
diff --git a/configs/datasets/scibench/lib_prompt/stat_prompt.txt b/configs/datasets/scibench/lib_prompt/stat_prompt.txt
new file mode 100644
index 00000000..e304df93
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/stat_prompt.txt
@@ -0,0 +1,17 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: In an orchid show, seven orchids are to be placed along one side of the greenhouse. There are four lavender orchids and three white orchids. How many ways are there to lineup these orchids?
+Answer: The answer is \boxed{35}.
+
+Promblem 2: What is the number of ways of selecting a president, a vice president, a secretary, and a treasurer in a club consisting of 10 persons?
+Answer: The answer is \boxed{5040}.
+
+Promblem 3: From an ordinary deck of playing cards, cards are to be drawn successively at random and without replacement. What is the probability that the third spade appears on the sixth draw?
+Answer: The answer is \boxed{0.064}.
+
+Promblem 4: What is the number of possible 5-card hands (in 5-card poker) drawn from a deck of 52 playing cards?
+Answer: The answer is \boxed{2598960}.
+
+Promblem 5: A grade school boy has five blue and four white marbles in his left pocket and four blue and five white marbles in his right pocket. If he transfers one marble at random from his left to his right pocket, what is the probability of his then drawing a blue marble from his right pocket?
+Answer: The answer is \boxed{0.444444444444444 }.
+
diff --git a/configs/datasets/scibench/lib_prompt/stat_sol.txt b/configs/datasets/scibench/lib_prompt/stat_sol.txt
new file mode 100644
index 00000000..75490f2a
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/stat_sol.txt
@@ -0,0 +1,48 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: A rocket has a built-in redundant system. In this system, if component $K_1$ fails, it is bypassed and component $K_2$ is used. If component $K_2$ fails, it is bypassed and component $K_3$ is used. (An example of a system with these kinds of components is three computer systems.) Suppose that the probability of failure of any one component is 0.15 , and assume that the failures of these components are mutually independent events. Let $A_i$ denote the event that component $K_i$ fails for $i=1,2,3$. What is the probability that the system fails?
+Explanation for Problem 1: 
+Because the system fails if $K_1$ fails and $K_2$ fails and $K_3$ fails, the probability that the system does not fail is given by
+$$
+\begin{aligned}
+P\left[\left(A_1 \cap A_2 \cap A_3\right)^{\prime}\right] & =1-P\left(A_1 \cap A_2 \cap A_3\right) \\
+& =1-P\left(A_1\right) P\left(A_2\right) P\left(A_3\right) \\
+& =1-(0.15)^3 \\
+& =0.9966 .
+\end{aligned}
+$$
+Therefore, the answer is \boxed{0.9966}.
+
+Promblem 2: At a county fair carnival game there are 25 balloons on a board, of which 10 balloons 1.3-5 are yellow, 8 are red, and 7 are green. A player throws darts at the balloons to win a prize and randomly hits one of them. Given that the first balloon hit is yellow, what is the probability that the next balloon hit is also yellow?
+Explanation for Problem 2: Of the 24 remaining balloons, 9 are yellow, so a natural value to assign to this conditional probability is $9 / 24$.
+Therefore, the answer is \boxed{0.375}.
+
+Promblem 3: A certain food service gives the following choices for dinner: $E_1$, soup or tomato 1.2-2 juice; $E_2$, steak or shrimp; $E_3$, French fried potatoes, mashed potatoes, or a baked potato; $E_4$, corn or peas; $E_5$, jello, tossed salad, cottage cheese, or coleslaw; $E_6$, cake, cookies, pudding, brownie, vanilla ice cream, chocolate ice cream, or orange sherbet; $E_7$, coffee, tea, milk, or punch. How many different dinner selections are possible if one of the listed choices is made for each of $E_1, E_2, \ldots$, and $E_7$ ?
+Explanation for Problem 3:  By the multiplication principle, there are
+$(2)(2)(3)(2)(4)(7)(4)=2688$
+different combinations.
+
+Therefore, the answer is \boxed{2688}.
+
+Promblem 4: A grade school boy has five blue and four white marbles in his left pocket and four blue and five white marbles in his right pocket. If he transfers one marble at random from his left to his right pocket, what is the probability of his then drawing a blue marble from his right pocket?
+Explanation for Problem 4: For notation, let $B L, B R$, and $W L$ denote drawing blue from left pocket, blue from right pocket, and white from left pocket, respectively. Then
+$$
+\begin{aligned}
+P(B R) & =P(B L \cap B R)+P(W L \cap B R) \\
+& =P(B L) P(B R \mid B L)+P(W L) P(B R \mid W L) \\
+& =\frac{5}{9} \cdot \frac{5}{10}+\frac{4}{9} \cdot \frac{4}{10}=\frac{41}{90}
+\end{aligned}
+$$
+is the desired probability.
+Therefore, the answer is \boxed{0.444444444444444 }.
+
+Promblem 5: In an orchid show, seven orchids are to be placed along one side of the greenhouse. There are four lavender orchids and three white orchids. How many ways are there to lineup these orchids?
+Explanation for Problem 5: Considering only the color of the orchids, we see that the number of lineups of the orchids is
+$$
+\left(\begin{array}{l}
+7 \\
+4
+\end{array}\right)=\frac{7 !}{4 ! 3 !}=35 \text {. }
+$$
+Therefore, the answer is \boxed{35}.
+
diff --git a/configs/datasets/scibench/lib_prompt/thermo_prompt.txt b/configs/datasets/scibench/lib_prompt/thermo_prompt.txt
new file mode 100644
index 00000000..d67e0d83
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/thermo_prompt.txt
@@ -0,0 +1,20 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: Imagine tossing a coin 50 times. What are the probabilities of observing heads 25 times (i.e., 25 successful experiments)?
+Answer: The answer is \boxed{0.11}.
+
+Promblem 2: One mole of $\mathrm{CO}$ gas is transformed from an initial state characterized by $T_i=320 . \mathrm{K}$ and $V_i=80.0 \mathrm{~L}$ to a final state characterized by $T_f=650 . \mathrm{K}$ and $V_f=120.0 \mathrm{~L}$. Using Equation (5.22), calculate $\Delta S$ for this process. Use the ideal gas values for $\beta$ and $\kappa$. For CO,
+$$
+\frac{C_{V, m}}{\mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}}=31.08-0.01452 \frac{T}{\mathrm{~K}}+3.1415 \times 10^{-5} \frac{T^2}{\mathrm{~K}^2}-1.4973 \times 10^{-8} \frac{T^3}{\mathrm{~K}^3}
+$$
+Answer: The answer is \boxed{24.4}.
+
+Promblem 3: At $298.15 \mathrm{~K}, \Delta G_f^{\circ}(\mathrm{C}$, graphite $)=0$, and $\Delta G_f^{\circ}(\mathrm{C}$, diamond $)=2.90 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Therefore, graphite is the more stable solid phase at this temperature at $P=P^{\circ}=1$ bar. Given that the densities of graphite and diamond are 2.25 and $3.52 \mathrm{~kg} / \mathrm{L}$, respectively, at what pressure will graphite and diamond be in equilibrium at $298.15 \mathrm{~K}$ ?
+Answer: The answer is \boxed{1.51}.
+
+Promblem 4: The reactant 1,3-cyclohexadiene can be photochemically converted to cis-hexatriene. In an experiment, $2.5 \mathrm{mmol}$ of cyclohexadiene are converted to cis-hexatriene when irradiated with 100. W of 280. nm light for $27.0 \mathrm{~s}$. All of the light is absorbed by the sample. What is the overall quantum yield for this photochemical process?
+Answer: The answer is \boxed{0.396}.
+
+Promblem 5: Gas cylinders of $\mathrm{CO}_2$ are sold in terms of weight of $\mathrm{CO}_2$. A cylinder contains $50 \mathrm{lb}$ (22.7 $\mathrm{kg}$ ) of $\mathrm{CO}_2$. How long can this cylinder be used in an experiment that requires flowing $\mathrm{CO}_2$ at $293 \mathrm{~K}(\eta=146 \mu \mathrm{P})$ through a 1.00-m-long tube (diameter $\left.=0.75 \mathrm{~mm}\right)$ with an input pressure of $1.05 \mathrm{~atm}$ and output pressure of $1.00 \mathrm{~atm}$ ? The flow is measured at the tube output.
+Answer: The answer is \boxed{4.49}.
+
diff --git a/configs/datasets/scibench/lib_prompt/thermo_sol.txt b/configs/datasets/scibench/lib_prompt/thermo_sol.txt
new file mode 100644
index 00000000..fff183bc
--- /dev/null
+++ b/configs/datasets/scibench/lib_prompt/thermo_sol.txt
@@ -0,0 +1,112 @@
+Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].
+
+Promblem 1: For an ensemble consisting of 1.00 moles of particles having two energy levels separated by $h v=1.00 \times 10^{-20} \mathrm{~J}$, at what temperature will the internal energy of this system equal $1.00 \mathrm{~kJ}$ ?
+Explanation for Problem 1: Using the expression for total energy and recognizing that $N=n N_A$,
+$$
+U=-\left(\frac{\partial \ln Q}{\partial \beta}\right)_V=-n N_A\left(\frac{\partial \ln q}{\partial \beta}\right)_V
+$$
+Evaluating the preceding expression and paying particular attention to units, we get
+$$
+\begin{aligned}
+& U=-n N_A\left(\frac{\partial}{\partial \beta} \ln q\right)_V=-\frac{n N_A}{q}\left(\frac{\partial q}{\partial \beta}\right)_V \\
+& \frac{U}{n N_A}=\frac{-1}{\left(1+e^{-\beta h \nu}\right)}\left(\frac{\partial}{\partial \beta}\left(1+e^{-\beta h \nu}\right)\right)_V \\
+&=\frac{h \nu e^{-\beta h \nu}}{1+e^{-\beta h \nu}}=\frac{h \nu}{e^{\beta h \nu}+1} \\
+& \frac{n N_A h \nu}{U}-1=e^{\beta h \nu} \\
+& \ln \left(\frac{n N_A h \nu}{U}-1\right)=\beta h \nu=\frac{h \nu}{k T}
+\end{aligned}
+$$
+$$
+\begin{aligned}
+T & =\frac{h \nu}{k \ln \left(\frac{n N_A h \nu}{U}-1\right)} \\
+= & \frac{1.00 \times 10^{-20} \mathrm{~J}}{\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right) \ln \left(\frac{(1.00 \mathrm{~mol})\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)\left(1.00 \times 10^{-20} \mathrm{~J}\right)}{\left(1.00 \times 10^3 \mathrm{~J}\right)}-1\right)} \\
+& =449 \mathrm{~K}
+\end{aligned}
+$$
+Therefore, the answer is \boxed{449}.
+
+Promblem 2: At $298.15 \mathrm{~K}, \Delta G_f^{\circ}(\mathrm{C}$, graphite $)=0$, and $\Delta G_f^{\circ}(\mathrm{C}$, diamond $)=2.90 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Therefore, graphite is the more stable solid phase at this temperature at $P=P^{\circ}=1$ bar. Given that the densities of graphite and diamond are 2.25 and $3.52 \mathrm{~kg} / \mathrm{L}$, respectively, at what pressure will graphite and diamond be in equilibrium at $298.15 \mathrm{~K}$ ?
+Explanation for Problem 2: At equilibrium $\Delta G=G(\mathrm{C}$, graphite $)-G(\mathrm{C}$, diamond $)=0$. Using the pressure dependence of $G,\left(\partial G_m / \partial P\right)_T=V_m$, we establish the condition for equilibrium:
+$$
+\begin{gathered}
+\Delta G=\Delta G_f^{\circ}(\mathrm{C}, \text { graphite })-\Delta G_f^{\circ}(\mathrm{C}, \text { diamond }) \\
++\left(V_m^{\text {graphite }}-V_m^{\text {diamond }}\right)(\Delta P)=0 \\
+0=0-2.90 \times 10^3+\left(V_m^{\text {graphite }}-V_m^{\text {diamond }}\right)(P-1 \mathrm{bar}) \\
+P=1 \mathrm{bar}+\frac{2.90 \times 10^3}{M_C\left(\frac{1}{\rho_{\text {graphite }}}-\frac{1}{\rho_{\text {diamond }}}\right)} \\
+=1 \mathrm{bar}+\frac{2.90 \times 10^3}{12.00 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1} \times\left(\frac{1}{2.25 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}-\frac{1}{3.52 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}\right)}\\
+=10^5 \mathrm{~Pa}+1.51 \times 10^9 \mathrm{~Pa}=1.51 \times 10^4 \mathrm{bar}
+\end{gathered}
+$$
+Fortunately for all those with diamond rings, although the conversion of diamond to graphite at $1 \mathrm{bar}$ and $298 \mathrm{~K}$ is spontaneous, the rate of conversion is vanishingly small.
+
+Therefore, the answer is \boxed{1.51}.
+
+Promblem 3: The vibrational frequency of $I_2$ is $208 \mathrm{~cm}^{-1}$. What is the probability of $I_2$ populating the $n=2$ vibrational level if the molecular temperature is $298 \mathrm{~K}$ ?
+Explanation for Problem 3: Molecular vibrational energy levels can be modeled as harmonic oscillators; therefore, this problem can be solved by employing a strategy identical to the one just presented. To evaluate the partition function $q$, the "trick" used earlier was to write the partition function as a series and use the equivalent series expression:
+$$
+\begin{aligned}
+q & =\sum_n e^{-\beta \varepsilon_n}=1+e^{-\beta h c \widetilde{\nu}}+e^{-2 \beta h c \tilde{\nu}}+e^{-3 \beta h c \widetilde{\nu}}+\ldots \\
+& =\frac{1}{1-e^{-\beta h c \widetilde{\nu}}}
+\end{aligned}
+$$
+Since $\tilde{\nu}=208 \mathrm{~cm}^{-1}$ and $T=298 \mathrm{~K}$, the partition function is
+$$
+\begin{aligned}
+q & =\frac{1}{1-e^{-\beta h c \widetilde{\nu}}} \\
+& =\frac{1}{1-e^{-h c \widetilde{\nu} / k T}} \\
+& =\frac{1}{1-\exp \left[-\left(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.00 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right)\left(208 \mathrm{~cm}^{-1}\right)}{\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)(298 \mathrm{~K})}\right)\right]} \\
+& =\frac{1}{1-e^{-1}}=1.58
+\end{aligned}
+$$
+This result is then used to evaluate the probability of occupying the second vibrational state $(n=2)$ as follows:
+$$
+\begin{aligned}
+p_2 & =\frac{e^{-2 \beta h c \tilde{\nu}}}{q} \\
+& =\frac{\exp \left[-2\left(\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}^{-1}\right)\left(3.00 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right)\left(208 \mathrm{~cm}^{-1}\right)}{\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)(298 \mathrm{~K})}\right)\right]}{1.58} \\
+& =0.086
+\end{aligned}
+$$
+Therefore, the answer is \boxed{0.086}.
+
+Promblem 4: In a rotational spectrum of $\operatorname{HBr}\left(B=8.46 \mathrm{~cm}^{-1}\right)$, the maximum intensity transition in the R-branch corresponds to the $J=4$ to 5 transition. At what temperature was the spectrum obtained?
+Explanation for Problem 4: The information provided for this problem dictates that the $J=4$ rotational energy level was the most populated at the temperature at which the spectrum was taken. To determine the temperature, we first determine the change in occupation number for the rotational energy level, $a_J$, versus $J$ as follows:
+$$
+\begin{aligned}
+a_J & =\frac{N(2 J+1) e^{-\beta h c B J(J+1)}}{q_R}=\frac{N(2 J+1) e^{-\beta h c B J(J+1)}}{\left(\frac{1}{\beta h c B}\right)} \\
+& =N \beta h c B(2 J+1) e^{-\beta h c B J(J+1)}
+\end{aligned}
+$$
+Next, we take the derivative of $a_J$ with respect to $J$ and set the derivative equal to zero to find the maximum of the function:
+$$
+\begin{aligned}
+\frac{d a_J}{d J} & =0=\frac{d}{d J} N \beta h c B(2 J+1) e^{-\beta h c B J(J+1)} \\
+0 & =\frac{d}{d J}(2 J+1) e^{-\beta h c B J(J+1)} \\
+0 & =2 e^{-\beta h c B J(J+1)}-\beta h c B(2 J+1)^2 e^{-\beta h c B J(J+1)} \\
+0 & =2-\beta h c B(2 J+1)^2 \\
+2 & =\beta h c B(2 J+1)^2=\frac{h c B}{k T}(2 J+1)^2 \\
+T & =\frac{(2 J+1)^2 h c B}{2 k}
+\end{aligned}
+$$
+Substitution of $J=4$ into the preceding expression results in the following temperature at which the spectrum was obtained:
+$$
+\begin{aligned}
+T & =\frac{(2 J+1)^2 h c B}{2 k} \\
+& =\frac{(2(4)+1)^2\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3.00 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right)\left(8.46 \mathrm{~cm}^{-1}\right)}{2\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)} \\
+& =4943 \mathrm{~K}
+\end{aligned}
+$$
+Therefore, the answer is \boxed{4943}.
+
+Promblem 5: Determine the diffusion coefficient for Ar at $298 \mathrm{~K}$ and a pressure of $1.00 \mathrm{~atm}$.
+Explanation for Problem 5: Using Equation (17.10) and the collisional cross section for Ar provided in Table 17.1,
+ $$
+\begin{aligned}
+D_{Ar} &= \frac{1}{3} \nu_{ave, Ar} \lambda_{Ar} \\
+&= \frac{1}{3} \left(\frac{8RT}{\pi M_{Ar}}\right)^{\frac{1}{2}} \left(\frac{RT}{PN_A\sqrt{2}\sigma_{Ar}}\right) \\
+&= \frac{1}{3} \left(\frac{8(8.314~\mathrm{J~mol^{-1}~K^{-1}}) \times 298~\mathrm{K}}{\pi(0.040~\mathrm{kg~mol^{-1}})}\right)^{\frac{1}{2}} \\
+&\quad \times \left(\frac{(8.314~\mathrm{J~mol^{-1}~K^{-1}}) \times 298~\mathrm{K}}{(101,325~\mathrm{Pa}) \times (6.022 \times 10^{23}~\mathrm{mol^{-1}})} \times \frac{1}{\sqrt{2}(3.6 \times 10^{-19}~\mathrm{m^2})}\right) \\
+&= \frac{1}{3} \times (397~\mathrm{m~s^{-1}}) \times (7.98 \times 10^{-8}~\mathrm{m}) \\
+&= 1.1 \times 10^{-5}~\mathrm{m^2~s^{-1}}
+\end{aligned}
+$$
+Therefore, the answer is \boxed{1.1}.
+
diff --git a/configs/datasets/scibench/scibench_gen.py b/configs/datasets/scibench/scibench_gen.py
new file mode 100644
index 00000000..82f842cf
--- /dev/null
+++ b/configs/datasets/scibench/scibench_gen.py
@@ -0,0 +1,4 @@
+from mmengine.config import read_base
+
+with read_base():
+    from .scibench_gen_2b21f3 import scibench_datasets  # noqa: F401, F403
diff --git a/configs/datasets/scibench/scibench_gen_2b21f3.py b/configs/datasets/scibench/scibench_gen_2b21f3.py
new file mode 100644
index 00000000..cbd66202
--- /dev/null
+++ b/configs/datasets/scibench/scibench_gen_2b21f3.py
@@ -0,0 +1,71 @@
+import os
+from opencompass.openicl.icl_prompt_template import PromptTemplate
+from opencompass.openicl.icl_retriever import ZeroRetriever
+from opencompass.openicl.icl_inferencer import GenInferencer
+from opencompass.openicl.icl_evaluator import AccEvaluator
+from opencompass.datasets import ScibenchDataset, scibench_postprocess
+
+scibench_reader_cfg = dict(input_columns=['question'], output_column='answer')
+
+scibench_subsets = [
+    "atkins",
+    "calculus",
+    "chemmc",
+    "class",
+    "diff",
+    "fund",
+    "matter",
+    "quan",
+    "stat",
+    "thermo"
+]
+
+scibench_datasets = []
+for prompt_type in ["zs", "zs-cot", "fs", "fs-cot"]:
+    for _name in scibench_subsets:
+        if prompt_type == "fs":
+            prompt_path = os.path.join(os.path.dirname(__file__), 'lib_prompt', f'{_name}_prompt.txt')
+        elif prompt_type == "fs-cot":
+            prompt_path = os.path.join(os.path.dirname(__file__), 'lib_prompt', f'{_name}_sol.txt')
+        else:
+            prompt_path = None
+        if prompt_path is not None:
+            with open(prompt_path, 'r') as f:
+                _hint = f.read()
+        else:
+            _hint = ""
+
+        human_prompt = {
+            'zs': "Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \\boxed[ANSWER].'\n\nProblem: {question}\nAnswer:",
+            'zs-cot': "Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \\boxed[ANSWER].'\n\nProblem: {question}\nAnswer:Let’s think step by step.",
+            'fs': f"{_hint}\n\nProblem 6: {{question}}\nAnswer: ",
+            'fs-cot': f"{_hint}\n\nProblem 6: {{question}}\nExplanation for Problem 6: ",
+        }[prompt_type]
+
+        scibench_infer_cfg = dict(
+            prompt_template=dict(
+                type=PromptTemplate,
+                template=dict(round=[
+                    dict(role="HUMAN", prompt=human_prompt)
+                ])
+            ),
+            retriever=dict(type=ZeroRetriever),
+            inferencer=dict(type=GenInferencer, max_out_len=512)
+        )
+
+        scibench_eval_cfg = dict(
+            evaluator=dict(type=AccEvaluator),
+            pred_postprocessor=dict(type=scibench_postprocess))
+
+
+        scibench_datasets.append(
+            dict(
+                type=ScibenchDataset,
+                path="./data/scibench",
+                name=_name,
+                abbr= f"scibench-{_name}" if prompt_type == 'zs' else f"scibench-{_name}_{prompt_type}",
+                reader_cfg=scibench_reader_cfg,
+                infer_cfg=scibench_infer_cfg.copy(),
+                eval_cfg=scibench_eval_cfg.copy()
+            )
+        )
diff --git a/configs/summarizers/example.py b/configs/summarizers/example.py
index 3ceb0f6d..c2a52c06 100644
--- a/configs/summarizers/example.py
+++ b/configs/summarizers/example.py
@@ -10,6 +10,7 @@ with read_base():
     from .groups.flores import flores_summary_groups
     from .groups.tydiqa import tydiqa_summary_groups
     from .groups.xiezhi import xiezhi_summary_groups
+    from .groups.scibench import scibench_summary_groups
 
 summarizer = dict(
     summary_groups=sum([v for k, v in locals().items() if k.endswith("_summary_groups")], []),
diff --git a/configs/summarizers/groups/scibench.py b/configs/summarizers/groups/scibench.py
new file mode 100644
index 00000000..e9562bf3
--- /dev/null
+++ b/configs/summarizers/groups/scibench.py
@@ -0,0 +1,6 @@
+scibench_summary_groups = []
+
+scibench_tasks = ["atkins", "calculus", "chemmc", "class", "diff", "fund", "matter", "quan", "stat", "thermo"]
+for suffix in ["", "_zs-cot", "_fs", "_fs-cot"]:
+    subsets = [f"scibench-{subset}{suffix}" for subset in scibench_tasks]
+    scibench_summary_groups.append({'name': f'scibench{suffix}', 'subsets': subsets})
diff --git a/opencompass/datasets/__init__.py b/opencompass/datasets/__init__.py
index 8585d2b2..566e8bdc 100644
--- a/opencompass/datasets/__init__.py
+++ b/opencompass/datasets/__init__.py
@@ -56,6 +56,7 @@ from .race import *  # noqa: F401, F403
 from .realtoxicprompts import *  # noqa: F401, F403
 from .record import *  # noqa: F401, F403
 from .safety import *  # noqa: F401, F403
+from .scibench import ScibenchDataset, scibench_postprocess  # noqa: F401, F403
 from .siqa import *  # noqa: F401, F403
 from .squad20 import SQuAD20Dataset, SQuAD20Evaluator  # noqa: F401, F403
 from .storycloze import *  # noqa: F401, F403
diff --git a/opencompass/datasets/scibench.py b/opencompass/datasets/scibench.py
new file mode 100644
index 00000000..2403b632
--- /dev/null
+++ b/opencompass/datasets/scibench.py
@@ -0,0 +1,50 @@
+import json
+import os.path as osp
+import re
+
+from datasets import Dataset
+
+from opencompass.registry import LOAD_DATASET, TEXT_POSTPROCESSORS
+
+from .base import BaseDataset
+
+
+@LOAD_DATASET.register_module()
+class ScibenchDataset(BaseDataset):
+
+    @staticmethod
+    def load(path: str, name: str):
+        train_data = []
+
+        filename = osp.join(path, f'{name}.json')
+        with open(filename, 'r') as infile:
+            raw_data = json.load(infile)
+
+        for entry in raw_data:
+            train_data.append({
+                'question': entry['problem_text'].strip(),
+                'answer': entry['answer_number'].strip()
+            })
+
+        dataset = Dataset.from_list(train_data)
+        return dataset
+
+
+@TEXT_POSTPROCESSORS.register_module('scibench')
+def scibench_postprocess(text: str) -> str:
+    ans = text
+    ans_line = ans.split('answer is')
+    if len(ans_line) != 1:
+        ans = ans_line[1].strip()
+
+    match = re.search(r'\\boxed\{(.+?)\}', ans)
+    if match:
+        extracted_content = match.group(1)
+        return extracted_content
+
+    output = re.sub(r'(\d),(\d)', r'\1\2', ans)
+    numbers = re.findall(r'-?\d*\.?\d+|\d+', output)
+    if numbers:
+        return numbers[-1]
+
+    return ans
diff --git a/opencompass/tasks/openicl_eval.py b/opencompass/tasks/openicl_eval.py
index 424c713e..1bf76919 100644
--- a/opencompass/tasks/openicl_eval.py
+++ b/opencompass/tasks/openicl_eval.py
@@ -141,7 +141,7 @@ class OpenICLEvalTask(BaseTask):
                     filename = root + f'_{i}' + ext
                     i += 1
 
-            preds = {k: [pred[k] for pred in preds] for k in preds[0]}
+            preds = {k: [pred.get(k) for pred in preds] for k in preds[0]}
 
             pred_strs = preds.pop('prediction')