To solve the problem, we need to calculate the change in Gibbs free energy (ΔG) for a fixed mass of an ideal gas in a sealed rigid vessel as it is heated from -73°C to 27°C. We are given the entropy function of the gas as \( S = 2 + 10^{-2}T \) (in J/K) and the pressure and volume of the gas.
### Step-by-Step Solution:
1. **Convert Temperatures to Kelvin:**
- Initial temperature \( T_1 = -73°C = -73 + 273 = 200 \, K \)
- Final temperature \( T_2 = 27°C = 27 + 273 = 300 \, K \)
2. **Calculate Initial and Final Pressures:**
- Initial pressure \( P_1 = 10 \, atm \)
- Using the ideal gas law at constant volume, we can find the final pressure \( P_2 \) using the relation:
\[
\frac{P_1}{T_1} = \frac{P_2}{T_2}
\]
- Rearranging gives:
\[
P_2 = P_1 \cdot \frac{T_2}{T_1} = 10 \, atm \cdot \frac{300 \, K}{200 \, K} = 15 \, atm
\]
3. **Calculate Change in Pressure:**
- Change in pressure \( \Delta P = P_2 - P_1 = 15 \, atm - 10 \, atm = 5 \, atm \)
4. **Calculate the Volume in Appropriate Units:**
- Volume \( V = 24.64 \, L = 24.64 \times 10^{-3} \, m^3 \) (not needed directly for ΔG calculation but useful for understanding)
5. **Integrate to Find ΔG:**
- The change in Gibbs free energy at constant volume is given by:
\[
dG = V dP - S dT
\]
- Integrating gives:
\[
\Delta G = \int V dP - \int S dT
\]
- Since the volume is constant, we can take it outside the integral:
\[
\Delta G = V \Delta P - \int_{T_1}^{T_2} S(T) dT
\]
6. **Calculate the Entropy Change:**
- The entropy function is given as \( S(T) = 2 + 10^{-2}T \).
- We need to calculate:
\[
\int_{200}^{300} S(T) dT = \int_{200}^{300} (2 + 10^{-2}T) dT
\]
- This can be computed as:
\[
= \left[ 2T + \frac{10^{-2}}{2}T^2 \right]_{200}^{300}
\]
- Evaluating this:
\[
= \left[ 2(300) + 0.005(300^2) \right] - \left[ 2(200) + 0.005(200^2) \right]
\]
\[
= (600 + 0.005 \times 90000) - (400 + 0.005 \times 40000)
\]
\[
= (600 + 450) - (400 + 200) = 1050 - 600 = 450 \, J/K
\]
7. **Calculate ΔG:**
- Now substituting back into the ΔG equation:
\[
\Delta G = V \Delta P - \Delta S
\]
- Convert \( \Delta P \) to kJ:
\[
\Delta P = 5 \, atm = 5 \times 0.1 \, kJ = 0.5 \, kJ
\]
- Therefore:
\[
\Delta G = (24.64 \, L)(0.5 \, kJ) - 450 \, J/K
\]
- Convert \( 450 \, J/K \) to kJ:
\[
450 \, J/K = 0.45 \, kJ/K
\]
- Finally, substituting values:
\[
\Delta G = 12.32 \, kJ - 0.45 \, kJ = 11.87 \, kJ
\]
### Final Answer:
\[
\Delta G \approx 11.87 \, kJ
\]
