Question 968: Thermodynamics & KTG

Question

A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas  during the cycle is
![](https://neetcare-image.cdn.examgoal.net/1qpq2yck7l7o194/4ac1e382-3aaf-494a-a613-f47fcffed0df/c9e64490-626d-11ea-bddc-5f310cab17d0/file-1qpq2yck7l7o195-298w.jpg)

Accepted Answer

In a cyclic process,
$\Delta U = 0$
In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.
$ 	herefore \Delta $W = Ã¢ÂÂ Area of rectangle ABCD = Ã¢ÂÂ P(2V)  = Ã¢ÂÂ 2PV
According to first law of thermodynamics
$\Delta Q = \Delta U + \Delta W \Rightarrow \Delta Q = \Delta W$
(As $\Delta U = 0$)
i.e., heat supplied to the system is equal to the work done
So heat absorbed, $\Delta $Q = $\Delta $W = Ã¢ÂÂ 2PV
$ 	herefore $  Heat rejected by the gas = 2PV
![](https://neetcare-image.cdn.examgoal.net/1qpq2yck7l7o194/4ac1e382-3aaf-494a-a613-f47fcffed0df/c9e64490-626d-11ea-bddc-5f310cab17d0/file-1qpq2yck7l7o195-298w.jpg)

Answer

2PV

Answer

4PV

Answer

${1 \over 2}$ PV

Answer

PV

Thermodynamics & KTG

A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is

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