A real gas within a closed chamber at 27°C undergoes the cyclic process as shown in figure. The gas obeys PV3 = RT equation for the path A to B. The net work done in the complete cycle is (assuming R = 8 J/mol K) :

Engineering

Physics

Thermodynamics

Cyclic Processes

Question

A real gas within a closed chamber at 27°C undergoes the cyclic process as shown in figure. The gas obeys PV³ = RT equation for the path A to B. The net work done in the complete cycle is (assuming R = 8 J/mol K) :

225 J

–20 J

205 J

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T_AB = 27°C + 273 = 300 K constant

$W_{AB} = \int_{V_{A}}^{V_{B}} PdV = \int_{V_{A}}^{V_{B}} \frac{RT}{V^{3}} dV$

$= RT {[- \frac{1}{2 V^{2}}]}_{V_{A}}^{V_{B}}$

$= \frac{300 R}{2} [\frac{1}{V_{A}^{2}} - \frac{1}{V_{B}^{2}}]$

$= \frac{300 R}{2} [\frac{1}{2^{2}} - \frac{1}{4^{2}}]$

$= \frac{300 R}{2} [\frac{3}{16}]$ R = 8J moll k given

$= \frac{300}{2} \times \frac{3}{16} \times 8 = 225 J$

W_BC = Area under BC from graph

= – 10 × 2 = –20 J

W_CA = 0

∴ W_cycle = W_AB + W_BC + W_CA

= 225 – 20 + 0 = 205 J