By how much does the entropy of `3` mole of an ideal gas change in going from a pressure of `2` bar to a pressure of `1` bar without any change in Temperature . If the surrounding is at `1` bar and `300` K (Expansion is again of the cosntant extenal pressure of surrounding).

Options

A. `+7.29 J-K^(-1)`
B. `+4.82 J-K^(-1)`
C. `-5.29 J-K^(-1)`
D. `-8.35J-K^(-1)`

Explanation

To calculate the change in entropy (ΔS) of 3 moles of an ideal gas when it expands isothermally from a pressure of 2 bar to 1 bar, we can use the formula for entropy change during an isothermal process:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles (n) = 3 moles
- Initial pressure (P1) = 2 bar
- Final pressure (P2) = 1 bar
- Ideal gas constant (R) = 8.314 J/(K·mol)

2. **Use the Entropy Change Formula:**
The formula for the change in entropy for an isothermal process is given by:
\[ \Delta S = nR \ln\left(\frac{P_1}{P_2}\right) \]

3. **Substitute the Values into the Formula:**
\[ \Delta S = 3 \, \text{moles} \times 8.314 \, \text{J/(K·mol)} \times \ln\left(\frac{2 \, \text{bar}}{1 \, \text{bar}}\right) \]

4. **Calculate the Natural Logarithm:**
\[ \ln\left(\frac{2}{1}\right) = \ln(2) \approx 0.693 \]

5. **Perform the Calculation:**
\[ \Delta S = 3 \times 8.314 \times 0.693 \]
\[ \Delta S \approx 3 \times 8.314 \times 0.693 \approx 17.3 \, \text{J/K} \]

6. **Final Result:**
The change in entropy (ΔS) for the process is approximately:
\[ \Delta S \approx 17.3 \, \text{J/K} \]