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

Options

A. KRIn 60
B. Rin 60
C. 40 KR
D. 120 R

Answer

D

Explanation

For given gas ,
` pV = RT and V = KT^(2)`
`pV = R ((V)/(K))^((1)/(2))`
`pV^((1)/(2)) = a` constant
`:. gamma = (1)/(2) ` for this process.
`:.` work done in this process,
`W = (nR(T_(i)-T_(f)))/(gamma - 1)`
`= (R (60))/((1)/(2))=120 R `