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

For isothermal process :
`P_(1) V_(1) = P_(2) V_(2)`
i.e., `P_(1) = ((V_(2))/(V_(1))) P_(2)`
For adiabatic process :
`P_(3) V_(1)^(gamma) = P_(2) V_(2)^(gamma)`
i.e., `P_(3) = ((V_(2))/(V_(1)))^(gamma) P_(2)`
As `gamma gt 1`, hence `P_(3) gt P_(1)`,
Further, as slope of adiabatic curve is greater than that of isothermal process curve, adiabatic curve will lie above the isothermal curve. That is, area under adiabatic curve gt area under isothermal curve
i.e., Negative work gt Positive work
i.e., `W lt 0`