Question

The work done during isothermal expansion of an ideal gas is:

• A. Zero
• B. \( P\Delta V \)
• C. \( nRT \ln \left( \frac{V_2}{V_1} \right) \)
• D. \( nC_v \Delta T \)

Hint:

In an isothermal process, the work done by or on the gas can be calculated using the formula \( W = nRT \ln \left( \frac{V_2}{V_1} \right) \), where the temperature is constant.

Answer 1

The Correct Option is C

In an isothermal expansion of an ideal gas, the temperature remains constant. The work done \( W \) during such an expansion is given by the formula: \[ W = \int_{V_1}^{V_2} P \, dV \] Using the ideal gas law \( P = \frac{nRT}{V} \), where \( n \) is the number of moles of gas, \( R \) is the gas constant, and \( T \) is the temperature, we get: \[ W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV \] Since the temperature is constant (\( T = {constant} \)), the work done becomes: \[ W = nRT \ln \left( \frac{V_2}{V_1} \right) \] This is the correct expression for the work done during an isothermal expansion.
- Zero work would apply only if there is no volume change, which is not the case in this scenario.
- \( P\Delta V \) is the work formula for an ideal gas in an isobaric process, not an isothermal process.
- \( nC_v \Delta T \) represents the heat or work done in an isochoric process (constant volume), not an isothermal one.
Thus, the correct answer is \( nRT \ln \left( \frac{V_2}{V_1} \right) \).