Question

Consider two different paths in which the volume of an ideal gas doubles isothermally: (i) Reversible expansion (work done = \( w_{\text{rev}} \))
(ii) Irreversible expansion, with the external pressure equal to the final pressure of the gas (work done = \( w_{\text{irrev}} \)) Here, \[ \dfrac{w_{\text{rev}}}{w_{\text{irrev}}} = ? \]

• A. \(2 \ln 2\)
• B. \(\tfrac{1}{2} \ln 2\)
• C. \(\tfrac{1}{2} \ln \tfrac{1}{2}\)
• D. \(2 \ln \tfrac{1}{2}\)

Hint:

Reversible work is always maximum because the system remains in equilibrium at each step. Irreversible expansion against a constant external pressure always produces less work, hence the ratio is greater than 1.

Answer 1

The Correct Option is A

Step 1: Reversible isothermal expansion work
For an isothermal reversible expansion, \[ w_{\text{rev}} = nRT \ln \left( \frac{V_f}{V_i} \right). \] Here, the gas volume doubles \(⇒ V_f = 2V_i\).
So, \[ w_{\text{rev}} = nRT \ln 2. \] Step 2: Irreversible isothermal expansion work
For irreversible work, external pressure is equal to the final pressure of the gas: \[ P_{\text{ext}} = P_f. \] Since \( P V = nRT \), the final pressure is \[ P_f = \frac{nRT}{V_f} = \frac{nRT}{2V_i}. \] Now the work in irreversible expansion is \[ w_{\text{irrev}} = P_{\text{ext}} (V_f - V_i). \] Substitute values: \[ w_{\text{irrev}} = \frac{nRT}{2V_i} (2V_i - V_i). \] \[ w_{\text{irrev}} = \frac{nRT}{2}. \] Step 3: Ratio of works
\[ \frac{w_{\text{rev}}}{w_{\text{irrev}}} = \frac{nRT \ln 2}{\tfrac{nRT}{2}} = 2 \ln 2. \] \fbox{\(\dfrac{w_{\text{rev}}}{w_{\text{irrev}}} = 2 \ln 2\)}