Gujarati
Hindi
11.Thermodynamics
normal

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

A

$4P_0V_0$

B

$2P_0V_0$

C

$3P_0V_0$

D

$P_0V_0$

Solution

$W_{AB} = -P_0V_0, W_{BC} = 0$ and  $W_{CD} = 4P_0V_0$

$\Rightarrow  W_{ABCD} = -P_0V_0 + 0 + 4P_0V_0 = 3P_0V_0$

Standard 11
Physics

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