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

A gas expands with temperature according to the relation, V=kT^(2//3) where k is a constant Word done when the temperature changes by 60 K is (R= Universal gas constant).

A gas expands with temperature according to the relation V=KT^(2/3) .Work done when the temperature changes by 60K is.

A gas expands with temperature according to the relation V=KT^(2/3) .Work done when the temperature changes by 60K is.

The volume V of a given mass of monoatomic gas changes with temperature T according to the relation V=KT^(2//3) . The workdone when temperature changes by 90 K will be xR. The value of x is [R = universal gas constant]

One mole of a gas expands with temperature T such thaht its volume, V= KT^(2) , where K is a constant. If the temperature of the gas changes by 60^(@)C then the work done by the gas is

One mole of a gas expands with temperature T such thaht its volume, V= KT^(2) , where K is a constant. If the temperature of the gas changes by 60^(@)C then the work done by the gas is 120R