The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 a volume of 100 dm3 at `27^@C` is

Options

A. `38.3 J mol^(-1) k^(-1)`
B. `35.8 J mol^(-1) K^(-1)`
C. ` 32.3 J mol^(-1) K^(-1)`
D. `42.3 J mol^(-1) K^(-1)`

Explanation

To solve the problem of calculating the entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm³ to a volume of 100 dm³ at 27°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Number of moles (n) = 2 moles
- Initial volume (V1) = 10 dm³
- Final volume (V2) = 100 dm³
- Temperature (T) = 27°C = 27 + 273.15 = 300.15 K
- Gas constant (R) = 8.314 J/(mol·K)

2. **Use the Formula for Entropy Change:**
The formula for the change in entropy (ΔS) during an isothermal reversible expansion is given by:
\[ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) \]
Since we are using logarithm base 10 in the video transcript, we can convert it using the relation \( \ln x = 2.303 \log_{10} x \):
\[ \Delta S = nR \cdot 2.303 \log_{10}\left(\frac{V_2}{V_1}\right) \]

3. **Calculate the Volume Ratio:**
\[ \frac{V_2}{V_1} = \frac{100 \, \text{dm}^3}{10 \, \text{dm}^3} = 10 \]

4. **Calculate the Logarithm:**
\[ \log_{10}(10) = 1 \]

5. **Substitute the Values into the Entropy Change Formula:**
\[ \Delta S = 2 \, \text{moles} \times 8.314 \, \text{J/(mol·K)} \times 2.303 \times 1 \]

6. **Perform the Calculation:**
\[ \Delta S = 2 \times 8.314 \times 2.303 = 38.29 \, \text{J/(mol·K)} \]

7. **Round the Result:**
\[ \Delta S \approx 38.30 \, \text{J/(mol·K)} \]

### Final Answer:
The entropy change involved in the isothermal reversible expansion is approximately **38.30 J/(mol·K)**.

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