Work done in a thermodynamic process
= Area under
p−V graph
For process 1 ,
Work done,
W1=p∆V=p(3V−V)=2pV Let change of internal energy in process 1 (point
A to point
B ) is
U1. Then, by first law of thermodynamics, heat required is
∆Q1=∆U+∆W=U1+W1 Here,
Q1=5pV (given)
or
5pV=U1+2pV Now for process 2 ,
⇒U1=3pV Change of internal energy
=U2=U1=3pV As process 2 and process 1 lies between same isotherms.
Also, work done for process 2 is
W2= Area under process 2
=×2V×p+p×2V=pV Again, using first law of thermodynamics, heat required in process 2 is