If one mole of an ideal gas `C_p = 5/2 R` is expanded isothermally at 300 k until it's volume is tripled., if expansion is carried out freely `(P_(ext)=0), then DeltaS is :`

Options

A. zero
B. infinity
C. `R ln 3`
D. None

Explanation

To solve the problem, we need to calculate the change in entropy (ΔS) for one mole of an ideal gas that is expanded isothermally and freely at a constant temperature of 300 K, where the volume is tripled.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- 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)
- Temperature (T) = 300 K
- Since the process is isothermal, T1 = T2 = T = 300 K.

2. **Use the Entropy Change Formula:**
The change in entropy (ΔS) for an ideal gas during an isothermal process can be calculated using the formula:
\[ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) \]
where R is the universal gas constant.

3. **Substitute the Values:**
- Substitute n = 1, V2 = 3V, and V1 = V into the formula:
\[ \Delta S = 1 \cdot R \ln\left(\frac{3V}{V}\right) \]
- This simplifies to:
\[ \Delta S = R \ln(3) \]

4. **Final Result:**
Thus, the change in entropy (ΔS) for the process is:
\[ \Delta S = R \ln(3) \]

### Conclusion:
The final answer for the change in entropy (ΔS) when one mole of an ideal gas is expanded isothermally and freely until its volume is tripled is:
\[ \Delta S = R \ln(3) \]