Two liters of `N_(2)` at `0^(@)C` and `5` atm pressure is expanded isothermally against a constant external pressure of `1` atm untill the pressure of gas reaches `1` atm. Assuming gas to be ideal, claculate the work of expansion.

Options

A. `-202.6J`
B. `-202.6J`
C. `-810.4J`
D. `-303.9J`

Answer

D

Explanation

Work is done against constant external pressure is greatly different from the pressure of `2N_(2)` at `0^(@)C` and thus, process is irresible.
`becausep_(1)V_(1)=p_(2)V_(2)`
`thereforeV_(2)=(2xx5)//2=5L`.
`p_(ext)=1xx1.013xx10^(5)N//m^(2)`
`DeltaV=V_(2)-V_(1)=(5-2)L=3xx10^(-3)m^(3)`
`thereforeW=-p_(ext)(V_(2)-V_(1))`
`=-1xx1.013xx10^(5)xx3xx10^(-3)`
`=-303.9`J