TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

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
=‌

1

2

×2V×‌

1

2

p+p×2V=‌

5

2

pV
Again, using first law of thermodynamics, heat required in process 2 is