If one mole of an ideal gas `(C_(p.m.)=(5)/(2)R)` is expanded isothermally at 300K until it’s volume is tripled, then change in entropy of gas is:

Options

A. zero
B. infinity
C. `(5)/(2)R ln 3`
D. `R ln 3`

Explanation

To find the change in entropy of one mole of an ideal gas that is expanded isothermally at 300 K until its volume is tripled, we can use the formula for the change in entropy (ΔS) during an isothermal process:

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles (n) = 1 mole
- Initial volume (V1) = V (let's assume it as V)
- Final volume (V2) = 3V (since the volume is tripled)
- Gas constant (R) = 8.314 J/(mol·K)
- Temperature (T) = 300 K (isothermal process)

2. **Calculate the Change in Entropy:**
- Since the volume is tripled, we can write:
\[ \frac{V_2}{V_1} = \frac{3V}{V} = 3 \]
- Substitute the values into the entropy change formula:
\[ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) = 1 \cdot R \cdot \ln(3) \]
- Therefore, we have:
\[ \Delta S = R \ln(3) \]

3. **Substituting the Value of R:**
- Using R = 8.314 J/(mol·K):
\[ \Delta S = 8.314 \ln(3) \]

4. **Calculate ln(3):**
- The natural logarithm of 3 is approximately 1.0986.
- Thus:
\[ \Delta S \approx 8.314 \cdot 1.0986 \approx 9.13 \, \text{J/K} \]

5. **Final Answer:**
- The change in entropy of the gas is approximately:
\[ \Delta S \approx 9.13 \, \text{J/K} \]