One mole of oxygen is expanded from a volume `V_1 = 1.00 1` to `V_2 = 5.0 1` at a constant temperature `T = 280 K`. Calculate :
(i) the increment of te internal energy of the gas :
(b) the amount of the absorbed heat.
The gas is assumed to be a Van der Walls gas.

Explanation

(a) The increment in the internal energy is
`Delta U = int_(V_(1))^(V_1) ((del U)/(del V))_T dV`
But from second law
`((del U)/(del V))_T = T((del S)/(del V))_T -p = T ((del p)/(del T))_V - p`
On the other hand `p = (RT)/(V - b) - (a)/(V^2)`
or, `T((del T)/(del T))_V = (RT)/(V - b)` and `((del U)/(del V))_t = "(a)/(V^2)`
So, `Delta U = a((1)/(V_1) - (1)/(V_2))`
(b) From the first law
`Q = A + Delta U = RT 1n (V_2 - b)/(V_1 - b)`.