Six moles of an ideal gas perfomrs a cycle shown in figure. If the temperature are $T_A = 600 K, T_B = 800 K, T_C = 2200 K \,and\, T_D = 1200 K$, the work done per cycle is ..... $kJ$
Diffcult
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(c) Processes $A$ to $B$ and $C$ to $D$ are parts of straight line graphs of the form $y = mx$
Also $P = \frac{{\mu R}}{V}T$ ($\mu = 6$)
==> $P \propto T.$ So volume remains constant for the graphs $AB$ and $CD$
So no work is done during processes for $A$ to $B$ and $C$ to $D$ i.e., $W_{AB} = W_{CD} = 0$ and $W_{BC} = P_2(V_C -V_B) = µR (TC -TB)$
$= 6R (2200 -800) = 6R \times 1400 J$
Also $ W_{DA} = P_1 (V_A -V_D) = µR(T_A -T_B)$
$= 6R (600 -1200)= -6R \times 600 J$
Hence work done in complete cycle
$W = W_{AB} + W_{BC} + W_{CD} + W_{DA}$
$= 0 + 6R \times 1400 + 0 -6R \times 600$
$= 6R \times 900 = 6 \times 8.3 \times 800 ≈ 40 kJ$
Also $P = \frac{{\mu R}}{V}T$ ($\mu = 6$)
==> $P \propto T.$ So volume remains constant for the graphs $AB$ and $CD$
So no work is done during processes for $A$ to $B$ and $C$ to $D$ i.e., $W_{AB} = W_{CD} = 0$ and $W_{BC} = P_2(V_C -V_B) = µR (TC -TB)$
$= 6R (2200 -800) = 6R \times 1400 J$
Also $ W_{DA} = P_1 (V_A -V_D) = µR(T_A -T_B)$
$= 6R (600 -1200)= -6R \times 600 J$
Hence work done in complete cycle
$W = W_{AB} + W_{BC} + W_{CD} + W_{DA}$
$= 0 + 6R \times 1400 + 0 -6R \times 600$
$= 6R \times 900 = 6 \times 8.3 \times 800 ≈ 40 kJ$
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