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Work in a cyclic process
Appears in
Concepts tested here
- Area of P–V cycle
- Area of P–V loops
- First law for a cycle
All Questions
2014 AIPMT 1 question
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A thermodynamic system undergoes cyclic process ABCDA as shown in fig. The work done by the system in the cycle is:
Work done in the cycle = area enclosed by the P–V loops; the two triangular loops are traversed in opposite senses.
$W = \frac{1}{2}(2P_0 - P_0)(2V_0 - V_0) + \left[-\frac{1}{2}(3P_0 - 2P_0)(2V_0 - V_0)\right]$
$W = \frac{P_0V_0}{2} - \frac{P_0V_0}{2} = 0$
2013 NEET 1 question
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A gas is taken through the cycle A → B → C → A, as shown. What is the net work done by the gas?
Net work done = area of triangle ABC
$= \frac{1}{2}\times[(7 - 2)\times10^{-3}][(6 - 2)\times10^5] = 1000$ J
2012 AIPMT-PRE 1 question
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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
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.