2 mol of an ideal gas at `27^(@)C` temperature is expanded reversibly from `2L` to `20L`. Find entropy change `(R = 2 cal mol^(-1) K^(-1))`

Options

A. 92.1
B. 0
C. 4
D. 9.2

Explanation

To find the change in entropy (ΔS) for the given process, we will use the formula for the change in entropy during an isothermal expansion of an ideal gas:

\[ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) \]

Where:
- \( n \) = number of moles of gas
- \( R \) = gas constant
- \( V_1 \) = initial volume
- \( V_2 \) = final volume

### Step-by-step solution:

1. **Identify the given values:**
- Number of moles, \( n = 2 \, \text{mol} \)
- Initial volume, \( V_1 = 2 \, \text{L} \)
- Final volume, \( V_2 = 20 \, \text{L} \)
- Gas constant, \( R = 2 \, \text{cal mol}^{-1} \text{K}^{-1} \)

2. **Calculate the ratio of volumes:**
\[ \frac{V_2}{V_1} = \frac{20 \, \text{L}}{2 \, \text{L}} = 10 \]

3. **Calculate the natural logarithm of the volume ratio:**
\[ \ln\left(\frac{V_2}{V_1}\right) = \ln(10) \]

4. **Convert the natural logarithm to logarithm base 10:**
To convert from natural logarithm to logarithm base 10, we use the conversion factor \( \ln(x) = 2.303 \log_{10}(x) \):
\[ \ln(10) = 2.303 \log_{10}(10) = 2.303 \cdot 1 = 2.303 \]

5. **Substitute the values into the entropy change formula:**
\[ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) = 2 \, \text{mol} \times 2 \, \text{cal mol}^{-1} \text{K}^{-1} \times 2.303 \]

6. **Calculate ΔS:**
\[ \Delta S = 2 \times 2 \times 2.303 = 9.212 \, \text{cal K}^{-1} \]

7. **Final result:**
\[ \Delta S \approx 9.2 \, \text{cal K}^{-1} \]