If one mole of gas doubles its volume at temperature T isothermally then work done by the gas is

Options

A. `-RT log_e 4`
B. `RT log_e 2`
C. `-RT log_(10) 2`
D. RT

Answer

To solve the problem of calculating the work done by one mole of gas that doubles its volume at a constant temperature \( T \) during an isothermal process, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Isothermal Process**:
- An isothermal process is one in which the temperature remains constant. For an ideal gas, the work done during an isothermal expansion or compression can be calculated using the formula:
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]
where:
- \( W \) = work done by the gas,
- \( n \) = number of moles of gas,
- \( R \) = universal gas constant,
- \( T \) = absolute temperature,
- \( V_1 \) = initial volume,
- \( V_2 \) = final volume.

2. **Identify Initial and Final Volumes**:
- According to the problem, the gas doubles its volume. If we let the initial volume \( V_1 \) be \( V \), then the final volume \( V_2 \) will be:
\[ V_2 = 2V \]

3. **Substitute Values into the Work Formula**:
- We know that \( n = 1 \) mole (as given in the problem). Now substituting \( n \), \( R \), \( T \), \( V_1 \), and \( V_2 \) into the work done formula:
\[ W = 1 \cdot R \cdot T \cdot \ln\left(\frac{2V}{V}\right) \]

4. **Simplify the Expression**:
- The \( V \) terms in the logarithm cancel out:
\[ W = RT \cdot \ln(2) \]

5. **Final Result**:
- Therefore, the work done by the gas when it doubles its volume is:
\[ W = RT \ln(2) \]

### Final Answer:
The work done by the gas is \( W = RT \ln(2) \).

---