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 derive the expression for the work done in reversible isothermal expansion of an ideal gas, we can follow these steps:

### Step 1: Understand the System
We have **n moles** of an ideal gas enclosed in a cylinder with a movable piston. The gas is expanding isothermally (at constant temperature) and reversibly.

### Step 2: Define the Initial Conditions
Let:
- \( P \) = pressure of the gas
- \( P_{ext} \) = external atmospheric pressure
- The external pressure is reduced by an infinitesimally small amount \( dp \), leading to an increase in volume \( dV \).

### Step 3: Write the Expression for Work Done
The work done \( dW \) during the expansion against the external pressure is given by:
\[ dW = -P_{ext} \, dV \]
Since the gas is expanding against a decreasing external pressure, we can express the work done as:
\[ dW = -P \, dV \]

### Step 4: Integrate the Work Done
To find the total work done during the expansion from an initial volume \( V_1 \) to a final volume \( V_2 \), we need to integrate:
\[ W = \int_{V_1}^{V_2} -P \, dV \]

### Step 5: Substitute the Ideal Gas Law
Using the ideal gas law, \( PV = nRT \), we can express pressure \( P \) in terms of volume:
\[ P = \frac{nRT}{V} \]
Substituting this into the work integral gives:
\[ W = \int_{V_1}^{V_2} -\frac{nRT}{V} \, dV \]

### Step 6: Perform the Integration
The integral becomes:
\[ W = -nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV \]
The integral of \( \frac{1}{V} \) is \( \ln V \), so we have:
\[ W = -nRT \left[ \ln V \right]_{V_1}^{V_2} \]
This evaluates to:
\[ W = -nRT (\ln V_2 - \ln V_1) = -nRT \ln \left( \frac{V_2}{V_1} \right) \]

### Step 7: Final Expression
Thus, the work done in reversible isothermal expansion is:
\[ W = -nRT \ln \left( \frac{V_2}{V_1} \right) \]
Alternatively, we can express this as:
\[ W = nRT \ln \left( \frac{V_1}{V_2} \right) \]

### Conclusion
The final expression for the work done in reversible isothermal expansion of an ideal gas is:
\[ W = -2.303 nRT \log \left( \frac{V_2}{V_1} \right) \]