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Three moles of an ideal monoatomic gas perform a cycle as shown in the figure. The gas temperature in different states are: $T_1 = 400\, K,\, T_2 = 800\, K,\, T_3 = 2400\, K$ and $T_4 = 1200\,K$ . The work done by the gas during the cycle is .... $kJ$
$10$
$20 $
$5$
$8.3$
Solution
In the curves $1-2$ and $3-4,$ we find that the pressure is directly proportional to temperature. So, the volume remains unchanged, i.e., gas does not work. The work done during the isobaric processes $2-3$ and $1-4$ are as follows:
${\mathrm{W}_{2-3}=\mathrm{P}_{2}\left(\mathrm{V}_{3}-\mathrm{V}_{2}\right)} $
${\mathrm{W}_{1-4}=\mathrm{P}_{1}\left(\mathrm{V}_{1}-\mathrm{V}_{4}\right)}$
Total work done $=\mathrm{P}_{2}\left(\mathrm{V}_{3}-\mathrm{V}_{2}\right)+\mathrm{P}_{1}\left(\mathrm{V}_{1}-\mathrm{V}_{4}\right)$
$\therefore $ $\mathrm{W}_{\mathrm{T}}=\mathrm{P}_{2} \mathrm{V}_{3}-\mathrm{P}_{2} \mathrm{V}_{2}+\mathrm{P}_{1} \mathrm{V}_{1}-\mathrm{P}_{1} \mathrm{V}_{4}$
Three moles has been given, so
$ \mathrm{PV} =\mathrm{nRT}=3 \mathrm{RT} $
$\therefore $ $ \mathrm{W}_{\mathrm{T}} =3 \mathrm{RT}_{3}-3 \mathrm{RT}_{2}+3 \mathrm{RT}_{1}-3 \mathrm{RT}_{4} $
$=3 \mathrm{R}\left[\mathrm{T}_{1}+\mathrm{T}_{3}-\mathrm{T}_{2}-\mathrm{T}_{4}\right] $
$=3 \mathrm{R}[400+2400-800-1200] $
$=3 \mathrm{R} \times 800=20 \times 10^{3} \mathrm{J}=20 \mathrm{kJ}. $
