An ideal gas expands isothermally from volume `V_(1)` to `V_(2)` and is then compressed to original volume `V_(1)` adiabatically. Initialy pressure is `P_(1)` and final pressure is `P_(3)`. The total work done is `W`. Then

Options

A. `P_(3) gt P_(1),W gt 0`
B. `P_(3) lt P_(1),W lt 0`
C. `P_(3) gt P_(1),W lt 0`
D. `P_(3)=P_(1),W=0`

Answer

C

Explanation

The slope of adiabatic process, at a given state `(P,V,T)` is greater than the slope of isothermal and `BC` is adiabatic Workdone in a process is given by area under the curve and V-axis.

The workdone `W` is positive if volume increases in the process and `W` is negative if volume decreases in the process. Hence `W_(AB)` is positive as volume increases from `V_(1) to V_(2)` , `W_(BC)` is negative as volume decrases from `V_(2) to V_(1)` .
Further `|W_(BC)|gt|W_(AB)|`
Net workdone `=W_(AB)+(-W_(BC))`
`W =A` negative value `( :.|W_(BC)|gt|W_(AB)|)or Wlt0` From the graph itself, `P_(3)gtP_(1)`
`:. Wlt0 and P_(3)gtP_(1)`
Hence option (c) repesents correct answer.