The volume of `N_(2)` gas increases from `1 cm^(3)` "to" `100 cm^(3) ` at a constant temperature of 300K. If there are 10 moles of gas present, then calculate the amount of work done by `N_(2)` on the surroundings.

Explanation

`V_(1)=1xx10^(-6)m^(3).V_(2)=100xx10^(-6) m^(3)`
`DeltaV=(100-1)10^(-6)=1000xx10^(-6)=10^(-3)m^(3)`
`mu=10,R=8.312J "mol"^(-1)K^(-1)`
Work done in an isothermal process `W= muRTlog, ((V_(2))/(V_(1)))`
`=2.303mu RT log ((V_(2))/(V_(1)))=2.303xx108.3123xx300xx3`
`W=1.723xx10^(5)J`