2 mol of an ideal gas at `27^(@)C` temperature is expanded reversibly from `1L` to `10L`. 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 entropy change for the expansion of an ideal gas, we can use the formula for the change in entropy (ΔS) during a reversible isothermal expansion:

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

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

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Number of moles, \( n = 2 \, \text{mol} \)
- Initial volume, \( V_1 = 1 \, \text{L} \)
- Final volume, \( V_2 = 10 \, \text{L} \)
- Gas constant, \( R = 2 \, \text{cal mol}^{-1} \text{K}^{-1} \)

**Step 2: Substitute the values into the entropy change formula.**
\[ \Delta S = nR \ln \left( \frac{V_2}{V_1} \right) \]
\[ \Delta S = 2 \, \text{mol} \times 2 \, \text{cal mol}^{-1} \text{K}^{-1} \times \ln \left( \frac{10 \, \text{L}}{1 \, \text{L}} \right) \]

**Step 3: Calculate the ratio of volumes.**
\[ \frac{V_2}{V_1} = \frac{10}{1} = 10 \]

**Step 4: Calculate the natural logarithm of the volume ratio.**
\[ \ln(10) \approx 2.303 \]

**Step 5: Substitute the logarithm back into the equation.**
\[ \Delta S = 2 \times 2 \times 2.303 \]

**Step 6: Perform the multiplication.**
\[ \Delta S = 4 \times 2.303 = 9.212 \, \text{cal K}^{-1} \]

**Step 7: Round to appropriate significant figures.**
\[ \Delta S \approx 9.2 \, \text{cal K}^{-1} \]

### Final Answer:
The change in entropy (ΔS) is approximately \( 9.2 \, \text{cal K}^{-1} \).

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