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# Please solve the above mentioned question with required explanation.

Eshan
3 years ago
Dear student,

Work done by gas in a process is area under the P-V curve.

Work done by gas from A to B=$P_0(2V_0-V_0)=P_0V_0$
Work done by gas from B to C (isothermal process)=$nRTln(\dfrac{4V_0}{2V_0})=(2P_0V_0)ln2=2P_0V_0ln2$
Work done by gas from C to D =$(\dfrac{P_0}{2})(2V_0-4V_0)=-P_0V_0$
Hence total work done by gas=$2P_0V_0ln 2$
Hence work done on the gas=$-2P_0V_0ln 2$
Eshan
3 years ago
Dear student,

Work done by gas in a process is area under the P-V curve.

Work done by gas from A to B=$P_0(2V_0-V_0)=P_0V_0$
Work done by gas from B to C (isothermal process)=$nRTln(\dfrac{4V_0}{2V_0})=(2P_0V_0)ln2=2P_0V_0ln2$
Work done by gas from C to D =$(\dfrac{P_0}{2})(2V_0-4V_0)=-P_0V_0$
Hence total work done by gas=$2P_0V_0ln 2$
Hence work done on the gas=$-2P_0V_0ln 2$