The temperature of 20 g of oxygen gas is raised from `10^(@)C` to `90^(@)C`. Find out the heat supplied, the rise in internal energy and the work done by the gas, if the temperature rises at
(i) constant volume. (ii) constant pressure. Given the specific heats of oxygen are `0.155 cal cdot g^(-1) cdot^(@)C^(-1)` at constant volume, and `0.218 cal cdot g^(-1) cdot^(@)C^(-1)` at constant pressure.

Explanation

(i) Heat supplied at constant volume is
`Q_(v) = mc_(v) (t_(f) - t-(i))`
`=20 xx 0.155 xx (90-10) = 248 cal`
As the volume is constant, the work done, W = 0.
From the first law of thermodynamics, rise in internal energy is `U_(f) - U_(i) = Q - W = 248 - 0 = 248 cal`.
(ii) Heat supplied at constant pressure is
`Q_(p) = mc_(p)(t_(f)-t_i)`
`=20 xx 0.218 xx (90-10) = 348.8 cal`
Now, oxygen may be considered as an ideal gas. The temperature rise is the same in both cases. As internal energy is a function of temperature only, the rise in internal energy will also be the same.
`:. U_(f) - U_(i) = 248 cal`
Then, `U_(f) - U_(i) = Q - W`
or, `W = Q - (U_(f) - U_(i)) = 348.8 - 248 = 100.8 cal`.