Question:medium

A real gas within a closed chamber at \( 27^\circ \text{C} \) undergoes the cyclic process as shown in the figure. The gas obeys \( PV^3 = RT \) equation for the path \( A \) to \( B \). The net work done in the complete cycle is (assuming \( R = 8 \, \text{J/mol K} \)):
Graph cycle

Updated On: Jan 31, 2026
  • 225 J
  • 205 J
  • 20 J
  • -20 J
Show Solution

The Correct Option is B

Solution and Explanation

\[ W_{AB} = \int P\,dV \quad \text{(Assuming T to be constant)} \] \[ = \int \frac{RT\,dV}{V^3} \] \[ = RT \int_{2}^{4} V^{-3}\,dV \] \[ = 8 \times 300 \times \left( -\frac{1}{2} \left[ \frac{1}{4^2} - \frac{1}{2^2} \right] \right) \] \[ = 225\,J \] \[ W_{BC} = P \int_{4}^{2} dV = 10(2 - 4) = -20\,J \] \[ W_{CA} = 0 \] \[ \therefore W_{\text{cycle}} = 205\,J \]

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