5 mol of an ideal gas expand reversibly from a volume of `8 dm^(3)` to `80 dm^(3)` at an temperature of `27^(@)C`. Calculate the change in entropy.

Options

A. `-95.73 JK^(-1)`
B. `95.93 JK^(-1)`
C. `-41.57 JK^(-1)`
D. `41.57JK^(-1)`

Answer

To calculate the change in entropy (ΔS) for the reversible expansion of an ideal gas, we can use the formula for an isothermal process:

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

Where:
- \( n \) = number of moles of 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, \( n = 5 \, \text{mol} \)
- Initial volume, \( V_1 = 8 \, \text{dm}^3 \)
- Final volume, \( V_2 = 80 \, \text{dm}^3 \)
- Temperature, \( T = 27^\circ C = 27 + 273.15 = 300.15 \, \text{K} \)
- Universal gas constant, \( R = 8.314 \, \text{J/(mol K)} \)

**Step 2: Calculate the ratio of volumes.**
\[ \frac{V_2}{V_1} = \frac{80 \, \text{dm}^3}{8 \, \text{dm}^3} = 10 \]

**Step 3: Substitute the values into the entropy change formula.**
Using the formula for entropy change:
\[ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) \]
Substituting the values:
\[ \Delta S = 5 \, \text{mol} \times 8.314 \, \text{J/(mol K)} \times \ln(10) \]

**Step 4: Calculate \( \ln(10) \).**
Using a calculator, we find:
\[ \ln(10) \approx 2.303 \]

**Step 5: Substitute \( \ln(10) \) back into the equation.**
\[ \Delta S = 5 \times 8.314 \times 2.303 \]

**Step 6: Perform the calculation.**
\[ \Delta S \approx 5 \times 8.314 \times 2.303 \approx 95.93 \, \text{J/K} \]

### Final Answer:
The change in entropy (ΔS) is approximately \( 95.93 \, \text{J/K} \).

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