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 \( R = 8 \, \text{J/mol K} \)):
- 225 J
- 205 J
- 20 J
- -20 J
The Correct Option is B
Approach Solution - 1
\[ W_{AB} = \int P\,dV \quad \text{(Assuming T to be constant)} \] \[ = \int \frac{RT\,dV}{V^3} \] \[ = RT \int_{2}^{4} V^{-3}\,dV \] \[ = 8 \times 300 \times \left( -\frac{1}{2} \left[ \frac{1}{4^2} - \frac{1}{2^2} \right] \right) \] \[ = 225\,J \] \[ W_{BC} = P \int_{4}^{2} dV = 10(2 - 4) = -20\,J \] \[ W_{CA} = 0 \] \[ \therefore W_{\text{cycle}} = 205\,J \]
Approach Solution -2
We are given that the gas obeys the equation PV³ = RT during the path from A to B, and the process is cyclic.
To calculate the work done in the cycle, we need to analyze the area enclosed by the cycle in the P – V diagram, which represents the work done.
1. Work Done in a Cycle:
The work done in a cyclic process is given by the area enclosed by the cycle on the P – V diagram. The work is the integral of pressure with respect to volume along the path of the cycle.
Since the gas obeys the equation PV³ = RT, the work done can be calculated by finding the area under the curve from A to B and then calculating the work done along the other parts of the cycle.
2. Net Work Done:
By calculating the areas based on the graph provided, the total net work done over the complete cycle is found to be:
Wtotal = 205 J.
Thus, the net work done in the complete cycle is 205 J, and the correct answer is Option (2).
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