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Work done during isothermal reversible p...

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))`

Text Solution

Verified by Experts

The correct Answer is:
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))`
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