A gas expands with temperature according to the relation `V = KT^(2//3)` where K is a constant. The work done by the gas when the temperature changes by 60 K 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)` ….(ii)
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) = (2)/(3)R[T]_(T_(1))^(T_(2))` (Using (ii))
`= (2)/(3)R(T_(2) - T_(1)) = (2)/(3) R xx 60 = 40R`