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Starting from the same initial conditions, an ideal gas expands from volume `V_(1) to V_(2)` in three different ways. The work done by the gas is `W_(1)` if he process is purely isothermal, `W_(2)` if purely isobaric and `W_(3)` if purely adiabatic. Then

A

`W_(2) gt W_(1) gt W_(3)`

B

`W_(2) gt W_(3) gt W_(1)`

C

`W_(1) gt W_(2) gt W_(3)`

D

`W_(1) gt W_(3) gt W_(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Area is largest for isobaric process and least for adiabatic.
Workdown = Area.
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