Please solve the above mentioned question with required explanation.

Question

Eshan · Answer

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 · Answer

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= $-2P_0V_0ln 2$

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

Please solve the above mentioned question with required explanation.

Please solve the above mentioned question with required explanation.

2 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on IIT JEE Entrance Exam

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship