Entropy is a measure of randomess of system. When a liquid is converted to the vapour state entropy of the system increases. Entropy in the phase transformation is calculated using `Delta S = (Delta H)/(T)` but in reversible adiabatic process `Delta`S will be zero. The rise in temperature in isobaric or isochoric process increases the randomness of system, which is given by
`Delta S = "2.303 n C log"((T_(2))/(T_(1)))`
`C = C_(P) or C_(V)`
The change in entropy when 1 mole `O_(2)` gas expands isothermally and reversibly from an initial volume 1 litre to a final volume 100 litre at `27^(@)C`

Options

A. `20.5JK^(-1(mol^(-1)`
B. `38.29 JK^(-1)mol^(-1)`
C. `42.50 JK^(-1)mol^(-1)`
D. `50.65 JK^(-1)mol^(-1)`

Answer

To solve the problem of calculating the change in entropy (ΔS) when 1 mole of oxygen gas expands isothermally and reversibly from an initial volume of 1 liter to a final volume of 100 liters at 27°C, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin:**
\[ T = 27°C + 273 = 300 \, K \]

2. **Identify Given Values:**
- Number of moles (n) = 1 mole
- Initial volume (V1) = 1 liter
- Final volume (V2) = 100 liters
- Universal gas constant (R) = 8.314 J/(K·mol)

3. **Use the Formula for Change in Entropy:**
For an isothermal and reversible expansion, the change in entropy can be calculated using the formula:
\[ \Delta S = -nR \ln\left(\frac{V_2}{V_1}\right) \]

4. **Substitute the Values:**
\[ \Delta S = -1 \times 8.314 \times \ln\left(\frac{100}{1}\right) \]

5. **Calculate the Natural Logarithm:**
\[ \ln(100) = \ln(10^2) = 2 \ln(10) \approx 2 \times 2.303 = 4.606 \]

6. **Complete the Calculation:**
\[ \Delta S = -1 \times 8.314 \times 4.606 \]
\[ \Delta S \approx -38.29 \, J/(K \cdot mol) \]

7. **Interpret the Result:**
The negative sign indicates that the entropy of the system decreases during the expansion.

### Final Answer:
\[ \Delta S \approx -38.29 \, J/(K \cdot mol) \]