Question

An ideal gas is taken around the cycle `ABCA` shown in `P - V` diagram. The net work done by the gas during the cycle is equal to

A. `3P_(1)V_(1)`
B. `-3P_(1)V_(1)`
C. `6P_(1)V_(1)`
D. `12 P_(1)V_(1)`

Answer 1

Correct Answer - B
`W=-("area of " DeltaABC) =-1/2 2V_(1) 3P_(1) =-3P_(1) V_(1)`

Answer 2

HELLO,

For cyclic process net work done is area under PV diagram

So, here ABC is right angled triangle

Area = \(\frac{1}{2}*AB*AC\)

\(AB=4P_1-P_1=3P_1\)

\(AC=3V_1-V_1=2V_1\)

NET WORK DONE = \(\frac{1}{2}*3P_1*2V_1=3P_1V_1\) 

HENCE, THE CORRECT OPTION IS  A

I HOPE YOU WILL UNDERSTAND.