The temperature of a fixed mass of an ideal gas is changed from `T_(1)K` to `T_(2)K(T_(2) > T_(1))`. Calculate the work done if the change is brought about at (i) constant pressure (ii) constant volume. Hence find from the first law of thermodynamics, in which case heat absorbed will be greater.

Explanation

1st law of thermodynamics: `Q = (U_(2)-U_(1))+W`
Work done at constant volume, `W_(1) = 0, Q_(1) = C_(v)(T_(2)-T_(1))`
Change in internal energy `= U_(2)-U_(1) = C_(p)(T_(2)-T_(1))`
For the same change in temperature there will be same change in internal energy.
So the work done in this case,
`W_(2) = Q_(2)-(U_(2)-U_(1)) = C_(p)(T_(2)-T_(1)) - C_(v)(T_(2)-T_(1))`
`=(C_(p) - C_(v))(T_(2)-T_(1))`
As, `C_(p) > C_(v), W_(2)>0`. So, in this case `Q_(2) = W_(2) + C_(v)(T_(2)-T_(1))`
` Q_(2) > Q_(1)`
So in case of constant pressure heat absorbed will be greater.