Work done in reversible isothermal expansion is given by

Options

A. `-2.303 RT log V_(2)/V_(1)`
B. `nR/(lambda -1)(T_(2) - T_(1))`
C. `2.303 RT log V_(2)/V_(1)`
D. None of these

Answer

To determine the work done in a reversible isothermal expansion of an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Process**:
- In a reversible isothermal expansion, a gas expands at a constant temperature (T). The system does work on the surroundings as it expands.

2. **Setting Up the Work Equation**:
- The infinitesimal work done (dW) during an expansion against an external pressure (P) can be expressed as:
\[ dW = -P \, dV \]
- The negative sign indicates that work is done by the system (the gas) on the surroundings.

3. **Using the Ideal Gas Law**:
- According to the ideal gas law, we have:
\[ PV = nRT \]
- Rearranging this gives:
\[ P = \frac{nRT}{V} \]

4. **Substituting Pressure in the Work Equation**:
- Substitute the expression for P into the work equation:
\[ dW = -\frac{nRT}{V} \, dV \]

5. **Integrating the Work Done**:
- To find the total work done (W) during the expansion from an initial volume \(V_1\) to a final volume \(V_2\), we integrate:
\[ W = \int_{V_1}^{V_2} -\frac{nRT}{V} \, dV \]
- This simplifies to:
\[ W = -nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV \]

6. **Calculating the Integral**:
- The integral of \(\frac{1}{V}\) is:
\[ \int \frac{1}{V} \, dV = \ln V \]
- Therefore, we have:
\[ W = -nRT [\ln V]_{V_1}^{V_2} = -nRT (\ln V_2 - \ln V_1) \]

7. **Applying Logarithmic Properties**:
- Using the property of logarithms, we can combine the terms:
\[ W = -nRT \ln \left(\frac{V_2}{V_1}\right) \]

8. **Final Expression**:
- If we consider 1 mole of gas (n = 1), the expression for work done simplifies to:
\[ W = -RT \ln \left(\frac{V_2}{V_1}\right) \]
- If we want to express this in terms of base 10 logarithm, we can convert it using the conversion factor \(2.303\):
\[ W = -2.303 RT \log_{10} \left(\frac{V_2}{V_1}\right) \]

### Final Answer:
The work done in a reversible isothermal expansion is given by:
\[ W = -2.303 RT \log_{10} \left(\frac{V_2}{V_1}\right) \]