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.

Explanation

Since the external pressure is gretely different from the pressure of `N_(2)` and thus, process is irreversible.
`W=-P_(ext)(V_(2)-V_(1))`
`W=-1xx(V_(2)-V_(1))`
given `V_(1)=2 liter, V_(2)=?,T=273 K`
`P_(1)=5 atm,P_(2)=1 atm`
`:. P_(1)V_(1)=P_(2)V_(2)`
`:. V_(2)=(2xx5)/1=10 litre`
`:. W=-1xx(10xx2)=-8 litre`
`:. =-8xx101.3J`
`=-810.4` joule