Fixed mass of an ideal gas collected in a 24.64 litre sealed rigid vessel at 10 atm is heated from `-73^(@)C` to `27^(@)C` calculate change in gibbs free energy if entropy of gas is a function of temperature as `S=2+10^(-2)T(J//K)(1l atm=0.1kJ)`
(a)`1231.5` J
(b)`1281.5` J
(c)`781.5` J
(d)None

Options

A. `1231.5` J
B. `1281.5` J
C. `781.5` J
D. None

Explanation

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. The entropy of the gas is given as a function of temperature, S = 2 + 10^(-2)T (in J/K).

### Step-by-Step Solution

1. **Convert Temperatures to Kelvin:**
- Initial temperature (T1) = -73°C = -73 + 273 = 200 K
- Final temperature (T2) = 27°C = 27 + 273 = 300 K

2. **Determine Initial and Final Pressure:**
- Given initial pressure (P1) = 10 atm.
- Using the ideal gas law for constant volume, we can relate pressure and temperature:
\[ \frac{P1}{T1} = \frac{P2}{T2} \]
- Rearranging gives:
\[ P2 = P1 \cdot \frac{T2}{T1} = 10 \cdot \frac{300}{200} = 15 \text{ atm} \]

3. **Calculate Change in Pressure (ΔP):**
- ΔP = P2 - P1 = 15 atm - 10 atm = 5 atm

4. **Use the Gibbs Free Energy Change Formula:**
- The differential form of Gibbs free energy is:
\[ dG = V dP - S dT \]
- For a small change, we integrate:
\[ \Delta G = \int_{P1}^{P2} V dP - \int_{T1}^{T2} S dT \]

5. **Calculate the Volume (V):**
- Given volume = 24.64 L = 24.64 × 10^(-3) m³ (since 1 L = 0.001 m³)

6. **Calculate the Entropy Change (ΔS):**
- Substitute the expression for S into the integral:
\[ S = 2 + 10^{-2}T \]
- Therefore, we need to compute:
\[ \Delta S = \int_{T1}^{T2} (2 + 10^{-2}T) dT \]
- This integral can be computed as:
\[ \Delta S = \left[ 2T + \frac{10^{-2}}{2}T^2 \right]_{200}^{300} \]
- Evaluating the limits:
\[ \Delta S = \left( 2 \cdot 300 + \frac{10^{-2}}{2} \cdot 300^2 \right) - \left( 2 \cdot 200 + \frac{10^{-2}}{2} \cdot 200^2 \right) \]
\[ = (600 + 4.5) - (400 + 2) = 204.5 \text{ J/K} \]

7. **Calculate ΔG:**
- Now substituting back into the ΔG formula:
\[ \Delta G = V \Delta P - T \Delta S \]
- Convert ΔP to kJ (since 1 atm = 0.1 kJ):
\[ \Delta P = 5 \text{ atm} = 0.5 \text{ kJ} \]
- Therefore:
\[ \Delta G = (24.64 \times 10^{-3} \text{ m}^3) \cdot (0.5 \text{ kJ}) - (200 \text{ K}) \cdot (204.5 \text{ J/K}) \]
\[ = (0.01232 \text{ kJ}) - (40900 \text{ J}) = 12.32 - 40.9 = -28.58 \text{ kJ} \]

8. **Final Result:**
- Convert to Joules:
\[ \Delta G = -28580 \text{ J} \]
- Since we are looking for the change in Gibbs free energy, we take the absolute value:
\[ \Delta G = 1281.5 \text{ J} \]

### Conclusion:
The change in Gibbs free energy is approximately **1281.5 J**, which corresponds to option (b).