What will be the entropy change when two moles of an ideal gas expand reversibly from initial volume of 1 litre to 10 litre at constant temperature of 300 K?

Options

A. `19.15 JK^(-1) mol^(-1)`
B. `38.27 JK^(-1) mol^(-1)`
C. `11.48 JK^(-1) mol^(-1)`
D. `5.74kJK^(-1) mol^(-1)`

Explanation

To solve the problem of finding the entropy change when two moles of an ideal gas expand reversibly from an initial volume of 1 liter to 10 liters at a constant temperature of 300 K, we can follow these steps:

### Step 1: Understand the formula for entropy change
The change in entropy (ΔS) for an ideal gas during a reversible isothermal expansion can be calculated using the formula:
\[ \Delta S = nR \ln \left( \frac{V_2}{V_1} \right) \]
where:
- \( n \) = number of moles of gas
- \( R \) = universal gas constant (8.314 J/(K·mol))
- \( V_1 \) = initial volume
- \( V_2 \) = final volume

### Step 2: Identify the values from the problem
From the problem statement, we have:
- \( n = 2 \) moles
- \( V_1 = 1 \) liter
- \( V_2 = 10 \) liters
- \( R = 8.314 \) J/(K·mol)

### Step 3: Substitute the values into the formula
Now, substituting the known values into the entropy change formula:
\[ \Delta S = 2 \times 8.314 \times \ln \left( \frac{10}{1} \right) \]

### Step 4: Calculate the natural logarithm
Calculate the natural logarithm:
\[ \ln(10) \approx 2.3026 \]

### Step 5: Complete the calculation
Now, substituting this value back into the equation:
\[ \Delta S = 2 \times 8.314 \times 2.3026 \]
\[ \Delta S \approx 2 \times 8.314 \times 2.3026 \approx 38.30 \text{ J/K} \]

### Step 6: Calculate per mole entropy change
Since the problem asks for the entropy change per mole, we divide the total entropy change by the number of moles:
\[ \Delta S_{\text{per mole}} = \frac{38.30}{2} \approx 19.15 \text{ J/(K·mol)} \]

### Final Answer
The entropy change per mole when two moles of an ideal gas expand reversibly from 1 liter to 10 liters at 300 K is approximately:
\[ \Delta S \approx 19.15 \text{ J/(K·mol)} \]

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