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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:
$\mathrm{W}_{2-3}=\mathrm{P}_{2}\left(\mathrm{V}_{3}-\mathrm{V}_{2}\right)$
$w_{1-4}=P_{1}\left(V_{1}-v_{4}\right)$
Total work done $=\mathrm{P}_{2}\left(\mathrm{V}_{3}-\mathrm{V}_{2}\right)+\mathrm{P}_{1}\left(\mathrm{V}_{1}-\mathrm{V}_{4}\right)$
$\mathrm{W}_{\mathrm{T}}=\mathrm{P}_{2} \mathrm{V}_{3}-\mathrm{P}_{2} \mathrm{V}_{2}+\mathrm{P}_{1} \mathrm{V}_{1}-\mathrm{P}_{1} \mathrm{V}_{4}$
Three moles has been given, so
$\mathbf{P V}=n R T=3 R T$
$\mathrm{W}_{\mathrm{T}}=3 \mathrm{RT}_{3}-3 \mathrm{RT}_{2}+3 \mathrm{RT}_{1}-3 \mathrm{RT}_{4}$
$=3 \mathrm{R}\left[\mathrm{T}_{1}+\mathrm{T}_{3}-\mathrm{T}_{2}-\mathrm{T}_{4}\right]$
$=3 \mathrm{R}[400+2400-800-1200]$
$=3 \mathrm{R} \times 800=20 \times 10^{3} \mathrm{J}=20 \mathrm{kJ}$
