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An ideal gas expands isothermally from v...

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

A

`P_(3) gt O_(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 `

Text Solution

Verified by Experts

The correct Answer is:
c
The two processes are shown in the following P - V diagram .

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 is done more than the positive work 1.e., `W lt 0` .
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