One mole of a gas expands such that its volume 'V' changes with absolute temperature 'T' in accordance with the relation `V = KT^2` where 'K' is a constant. If the temperature of the gas changes by `60^@C`, then work done by the gas (R is universal gas constant)

Options

A. 120R
B. Rln60
C. KRln 60
D. 40 KR

Answer

A

Explanation

Answer(1) pV = RT, V = `KT^2`
`Rightarrow pV = Rsqrt(V/K)`.
`pV^1/2` = constant .
for this process, `gamma = 1/2` .
The workdone in this process is, `W = nR(T_i -T_f)/gamma-1`.
W = R(60)/1/2 = 120R`.