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11.Thermodynamics
easy
A thermodynamic system is taken through the cycle $ABCD$ as shown in figure. Heat rejected by the gas during the cycle is
A
$2PV$ $\;$
B
$\;4{\rm{\;PV}}$
C
$\;{\rm{PV}}$
D
$\;\frac{{PV}}{2}$
(AIPMT-2012)
Solution
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 anticlockwiase.
$\therefore \,\,\,\Delta W = – Area\,of\,rectangle\,ABCD = – P\left( {2V} \right)$
$ = – 2PV$
According to first law of thermodynamics
$\Delta Q = \Delta u + \Delta W\,or\,\Delta Q = \Delta W\,\,\left( {As\,\Delta u = 0} \right)$
$i.e.,$ heat supplied to the system is equal to the work done
So heat absorbed,$\Delta Q = \Delta W = – 2PV$
$\therefore $ Heat rejected by the gas $ = 2PV$
Standard 11
Physics
