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 K
D. 20 K

Answer

C

Explanation

According to the question `V=KT^(2//3)`
`therefore " " "dV"=K(2)/(3)T^(-1//3)"dT"` (after differentainting )
`therefore " " ("dV")/(V)=((2)/(3)KT^(-1//3)"dT")/(KT^(2//3))=(2)/(3)(dT)/(T)`
`"Work done" , " " W=int_(T_(1))^(T_(2))RT(dT)/(V)=int_(T_(1))^(T_(2))RT(2)/(3)(dT)/(T)`
`W=(2)/(3)R(T_(1)-T_(1))(2)/(3)Rxx60=40R`