Question:

A gas goes through: (i) isothermal expansion \( W = 50J \), (ii) adiabatic \( W = ? \), (iii) isothermal expansion \( W = 20J \). Total \( \Delta U = -30J \). Find work in adiabatic.

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In cyclic processes, apply energy conservation across all paths. \( \Delta U = Q - W \)
Updated On: May 19, 2025
  • 40 J
  • 100 J
  • 30 J
  • 20 J
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The Correct Option is A

Solution and Explanation

From First Law: \[ \Delta U = Q - W \Rightarrow \Delta U = (Q_1 + Q_2 + Q_3) - (W_1 + W_2 + W_3) \] For isothermal: \( \Delta U = 0 \), so \( Q = W \) Given: \[ W_1 = 50, W_3 = 20, W_2 = ?, \quad \Delta U = -30 = Q - (50 + W_2 + 20) \Rightarrow Q = 70 + W_2 \] Also \( Q = W_1 + W_3 = 70 \), so \[ -30 = 70 - (70 + W_2) \Rightarrow W_2 = 40\ \text{J} \]
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