Gujarati
11.Thermodynamics
medium

$P-V$ diagram of an ideal gas is as shown in figure. Work done by the gas in process $ABCD$ is

A

$4\,{P_0}{V_0}$

B

$2\,{P_0}{V_0}$

C

$3\,{P_0}{V_0}$

D

${P_0}{V_0}$

Solution

(c) ${W_{AB}} = – \,{P_0}{V_0}$, ${W_{BC}} = 0$ and ${W_{CD}} = 4{P_0}{V_0}$

==> ${W_{ABCD}} = – \,{P_0}{V_0} + 0 + 4{P_0}{V_0} = 3{P_0}{V_0}$

Standard 11
Physics

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