A real gas within a closed chamber at \( 27^\circ \text{C} \) undergoes the cyclic process as shown in the figure. The gas obeys the equation \( PV^3 = RT \) for the path A to B. The net work done in the complete cycle is (assuming \( R = 8 \, \text{J/molK} \)):
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For cyclic processes, calculate the work done for each stage and combine the results. Always account for the nature of each segment—whether it is isochoric, isobaric, or follows a specific relation between pressure and volume.
The cyclic process includes three distinct stages: AC (isochoric process), BC (isobaric process), and AB (defined by \( PV^3 = RT \)).
Step 1: Isochoric process (AC) In an isochoric process, the volume remains constant, so no work is done: \[ W_{\text{AC}} = 0. \]
Step 2: Isobaric process (BC) The work done in an isobaric process is calculated using: \[ W_{\text{BC}} = P \Delta V = 10 (2 - 4) = -20 \, \text{J}. \] Step 3: Process AB (\( PV^3 = RT \)) The work done in this process is: \[ W_{\text{AB}} = \int P \, dV = \int_{V_1}^{V_2} \frac{RT}{V^3} \, dV. \] Substitute \( V_1 = 2 \, \text{m}^3 \) and \( V_2 = 4 \, \text{m}^3 \): \[ W_{\text{AB}} = RT \int_2^4 V^{-3} \, dV = -\frac{RT}{2} \left[ \frac{1}{V^2} \right]_2^4. \] Simplify the expression: \[ W_{\text{AB}} = -\frac{RT}{2} \left( \frac{1}{4^2} - \frac{1}{2^2} \right) = -\frac{RT}{2} \left( \frac{1}{16} - \frac{1}{4} \right). \] \[ W_{\text{AB}} = -\frac{RT}{2} \left( \frac{1 - 4}{16} \right) = \frac{RT}{2} \cdot \frac{3}{16}. \] Substitute \( R = 8 \, \text{J/molK} \) and \( T = 27^\circ \text{C} = 300 \, \text{K} \): \[ W_{\text{AB}} = \frac{8 \cdot 300}{2} \cdot \frac{3}{16} = 225 \, \text{J}. \] Net Work Done: The total work done is the sum of work from all segments of the process: \[ W_{\text{net}} = W_{\text{AC}} + W_{\text{BC}} + W_{\text{AB}} = 0 - 20 + 225 = 205 \, \text{J}. \] Final Answer: \[ \boxed{205 \, \text{J}} \]
