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An ideal gas expands from an initial tem...

An ideal gas expands from an initial temperature `T_(1)` to a final temperature `T_(2)`. Prove that the work done by the gas is `C_(V) (T_(2) - T_(2))`.

Text Solution

Verified by Experts

As `V prop T`
`(dx)/(dt) = KT` (K is constant)
`(dx)/(dt) = V " "(K = 1)`
dx = V dt
`dx = m sqrt(T) dt`
We known `(dT)/(dx) = (T_(2) - T_(1))/(L)`
where L is distance between X and Y.
`dx = (LdT)/(T_(2) - T_(1))`
Using (i)
`m sqrt(T) dt = (LdT)/(T_(2) - T_(1)) rArr dt = (LdT)/(m(T_(2) - T_(1)) sqrt(T))`
`underset(0)overset(t) int dt = underset(T_(1))overset(T_(2))int (L)/(m(T_(2) - T_(1)) sqrt(T)) dT`
`t = (L)/(m(T_(2) - T_(1))) underset(T_(1))overset(T_(2))int T^(-(1)/(2)) dT`
`= (L)/(m(T_(1) - T_(2))) |(T^(1/2))/((1)/(2))|_(T_(1))^(T_(2))`
`t = (2L)/(m(T_(2) - T_(1)))(sqrt(T_(2)) - sqrt(T_(1)))`
`rArr" "L = (mt(T_(2) - T_(1)))/(2(sqrt(T_(1)) - T_(1))) =(mt)/(2) ((T_(2) - T_(1)))/((sqrt(T_(2)) - T_(1)))`
`= (mt)/(2) ((sqrt(T_(2)) - sqrt(T_(1))) (sqrt(T_(2)) + sqrt(T_(1))))/((sqrt(T_(2)) - sqrt(T_(1))))`
`= (mt)/(2) (sqrt(T_(2)) + sqrt(T_(1)))`
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