Six moles of an ideal gas performs 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

Question

Dr. Abubacker · Accepted Answer

(c) Processes $A$ to $B$ and $C$ to $D$ are parts of straight line graphs of the form $y = mx$
Also $P = \frac{\mu R}{V}T\ (\mu = 6)$
$\Rightarrow P \propto T$. 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 $W_{BC} = P_{2}\left( V_{C} - V_{B} ight) = \mu R(T_{C} - T_{B})$
$= 6R(2200 - 800) = 6R 	imes 1400\ J$
Also $W_{DA} = P_{1}\left( V_{A} - V_{D} ight) = \mu R(T_{A} - T_{B})$
$= 6R(600 - 1200) = - 6R 	imes 600\ J$
Hence work done in complete cycle
$W = W_{AB} + W_{BC} + W_{CD} + W_{DA}$
$= 0 + 6R 	imes 1400 + 0 - 6R 	imes 600$
$= 6R 	imes 800 = 6 	imes 8.3 	imes 800 \approx 40\ kJ$

Q. Six moles of an ideal gas performs 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

Explanation

Reviewed by Dr. Abubacker

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