The entropy change in the isothermal reversible expansion of 2 moles of an ideal gas from 10 to 100 L at 300 K is

Options

A. `42.3 JK^(-1)`
B. `35.8 JK^(-1)`
C. `38.3 JK^(-1)`
D. `32.3 JK^(-1)`

Explanation

To find the entropy change (\( \Delta S \)) in the isothermal reversible expansion of 2 moles of an ideal gas from 10 L to 100 L at 300 K, we can use the formula for entropy change in an isothermal process:

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

where:
- \( n \) = number of moles
- \( R \) = ideal gas constant (8.314 J/(mol·K))
- \( V_f \) = final volume
- \( V_i \) = initial volume

### Step 1: Identify the given values
- Number of moles (\( n \)) = 2 moles
- Initial volume (\( V_i \)) = 10 L
- Final volume (\( V_f \)) = 100 L
- Temperature (\( T \)) = 300 K (not directly needed for this calculation)

### Step 2: Convert volumes to the same units if necessary
Since we are using the ideal gas constant \( R \) in J/(mol·K), we can keep the volumes in liters for the logarithmic calculation.

### Step 3: Substitute the values into the entropy change formula
\[ \Delta S = 2 \, \text{mol} \times 8.314 \, \text{J/(mol·K)} \times \ln\left(\frac{100 \, \text{L}}{10 \, \text{L}}\right) \]

### Step 4: Calculate the ratio of volumes
\[ \frac{100 \, \text{L}}{10 \, \text{L}} = 10 \]

### Step 5: Calculate the natural logarithm
\[ \ln(10) \approx 2.303 \]

### Step 6: Substitute the logarithm value back into the equation
\[ \Delta S = 2 \times 8.314 \times 2.303 \]

### Step 7: Perform the multiplication
\[ \Delta S \approx 2 \times 8.314 \times 2.303 \approx 38.29 \, \text{J/K} \]

### Final Answer
The entropy change (\( \Delta S \)) in the isothermal reversible expansion is approximately:
\[ \Delta S \approx 38.3 \, \text{J/K} \]