A gas expand with temperature according to the relation `V = KT^(2//3)`. What is the work done when the temperature changes by `30^(@)C`?

Options

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

Answer

B

Explanation

`W = int PdV = int (RT)/(V) dV`
Since `V = kT^(2//3) KT^(-1//3) dT`
Eliminating `K` we find `(dV)/(V) = 2/3(dT)/(T)`
Hence
`W = int_(T_1)^(T_2) 2/3 (RT)/(T) dT = 2/3 R(T_(2)-T_(1)) = 2/3 R(30) = 20R`.