Question

Work done during isothermal reversible process is given
A. `w = (RT)/(n) ln (V_(f))/(V_(i))`
B. `w = - (nRT)/(n) ln (V_(f))/(V_(i))`
C. `w = - nRT ln (V_(f))/(V_(i))`
D. `w = - nRT ln (V_(f))/(V_(i))`

Answer 1

Correct Answer - C
Under reversible coditions, the external pressure and the internal pressure differ only by `dP`. So substituting the ideall gas law into `dw = - PdV` gives
`w = - int_(V_(i))^(V_(f)) (nRT)/(V) = nRT ln (V_(f))/(V_(i))`