When 5 moles of an ideal gas is expended at 350 K reversibly from 5 L to 40 L, then the entropy change for the process is (R = 2 cal `mol^-1 K^-1`)

Options

A. `9.01 cal K^-1`
B. `20.8 cal K^-1`
C. `-15.2 cal K^-1`
D. `22.3 cal K^-1`

Answer

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

\[ \Delta S = nR \ln \left( \frac{V_f}{V_i} \right) \]

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

### Step 1: Identify the given values
- Number of moles, \( n = 5 \, \text{moles} \)
- Initial volume, \( V_i = 5 \, \text{L} \)
- Final volume, \( V_f = 40 \, \text{L} \)
- Temperature, \( T = 350 \, \text{K} \)
- Universal gas constant, \( R = 2 \, \text{cal} \, \text{mol}^{-1} \, \text{K}^{-1} \)

### Step 2: Substitute the values into the entropy change formula
\[ \Delta S = 5 \, \text{mol} \times 2 \, \text{cal} \, \text{mol}^{-1} \, \text{K}^{-1} \times \ln \left( \frac{40 \, \text{L}}{5 \, \text{L}} \right) \]

### Step 3: Calculate the ratio of volumes
\[ \frac{V_f}{V_i} = \frac{40}{5} = 8 \]

### Step 4: Calculate the natural logarithm of the volume ratio
\[ \ln(8) \approx 2.0794 \]

### Step 5: Substitute back into the entropy change equation
\[ \Delta S = 5 \times 2 \times 2.0794 \]

### Step 6: Perform the multiplication
\[ \Delta S = 10 \times 2.0794 = 20.794 \, \text{cal} \, \text{K}^{-1} \]

### Final Answer
The change in entropy for the process is approximately:
\[ \Delta S \approx 20.794 \, \text{cal} \, \text{K}^{-1} \]

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