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 find the entropy change when two moles of an ideal gas expand reversibly from an initial volume of 1 liter to a final volume of 10 liters at a constant temperature of 300 K, we can use the formula for the change in entropy (ΔS) for an ideal gas during an isothermal expansion:

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

Where:
- \(N\) = number of moles of the gas
- \(R\) = universal gas constant (8.314 J/(K·mol))
- \(V_1\) = initial volume
- \(V_2\) = final volume

### Step 1: Identify the values
- Number of moles, \(N = 2\) moles
- Initial volume, \(V_1 = 1\) L
- Final volume, \(V_2 = 10\) L
- Gas constant, \(R = 8.314 \, \text{J/(K·mol)}\)

### Step 2: Calculate the ratio of volumes
Calculate the ratio of the final volume to the initial volume:

\[ \frac{V_2}{V_1} = \frac{10 \, \text{L}}{1 \, \text{L}} = 10 \]

### Step 3: Calculate the natural logarithm of the volume ratio
Now, calculate the natural logarithm of the volume ratio:

\[ \ln \left(\frac{V_2}{V_1}\right) = \ln(10) \approx 2.3026 \]

### Step 4: Substitute values into the entropy change formula
Substituting the values into the entropy change formula:

\[ \Delta S = N R \ln \left(\frac{V_2}{V_1}\right) = 2 \times 8.314 \, \text{J/(K·mol)} \times 2.3026 \]

### Step 5: Perform the calculation
Now, perform the multiplication:

\[ \Delta S = 2 \times 8.314 \times 2.3026 \approx 38.28 \, \text{J/K} \]

### Final Result
Thus, the entropy change when two moles of an ideal gas expand reversibly from 1 liter to 10 liters at 300 K is:

\[ \Delta S \approx 38.28 \, \text{J/K} \]