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A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is
A
PV
B
2PV
C
4PV
D
$\frac{1}{2}PV$
Detailed Solution
The cycle is a rectangle with pressures P and 2P and volumes V and 3V.
Area enclosed $= (2P - P)(3V - V) = 2PV$
The cycle is traversed anticlockwise, so the work done by the gas is negative: $W = -2PV$
For a complete cycle $\Delta E = 0$, so $Q = W + \Delta E = -2PV$
So the heat rejected by the gas is 2PV.
Area enclosed $= (2P - P)(3V - V) = 2PV$
The cycle is traversed anticlockwise, so the work done by the gas is negative: $W = -2PV$
For a complete cycle $\Delta E = 0$, so $Q = W + \Delta E = -2PV$
So the heat rejected by the gas is 2PV.