One mole of a perfect gas expands isothermally to ten times its original volume. The change in entropy is

Options

A. `0.1 R`
B. `2.303 R`
C. `10.0 R`
D. `100.0 R`

Explanation

To solve the problem of finding the change in entropy when one mole of a perfect gas expands isothermally to ten times its original volume, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- We have 1 mole of a perfect gas (n = 1).
- The gas expands isothermally to 10 times its original volume (V2 = 10V1).

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

3. **Substituting Values:**
- Since V2 = 10V1, we can substitute this into the formula:
\[ \Delta S = nR \ln\left(\frac{10V_1}{V_1}\right) = nR \ln(10) \]
- Given that n = 1, we have:
\[ \Delta S = R \ln(10) \]

4. **Calculate the Value of ΔS:**
- We can express \(\ln(10)\) in terms of base 10 logarithm:
\[ \ln(10) = 2.303 \log_{10}(10) = 2.303 \]
- Therefore, substituting this back into the equation gives:
\[ \Delta S = R \cdot 2.303 \]

5. **Final Result:**
- Thus, the change in entropy when one mole of a perfect gas expands isothermally to ten times its original volume is:
\[ \Delta S = 2.303R \]