A gas expands with temperature according to the relation `V=KT^(2/3)`.Work done when the temperature changes by 60K is.

Options

A. `10 R`
B. `30 R`
C. `40 R`
D. `20 R`

Answer

c

Explanation

`dW = PdV = (RT)/V dV` ….(i)
As `V = KT^(2//3) :. dV = K 2/3 T^(-1//3) dT`
` :. (dV)/V = (K 2/3 T^(-1//3) dT)/(KT^(2//3)) = 2/3 (dT)/T`
From (i) , `W = underset(T_(1))overset(T_(2)) int RT (dV)/V = underset(T_(1)) overset(T_(2)) int RT 2/3 (dT)/T`
` W = 2/3 R(T_(2) - T_(1)) = 2/3 R xx 60 = 40 R`