An ideal gas was expanded from these initial state to the volume `V` without any heat exchange with the surrounding bodies. Will the final gas pressure be the same in the case of
(a) a fast and in the case of
(b) a very slow expansion process ?

Explanation

In all adiabatic processes
`Q = U_f - U_i + A = 0`
by virtue of the first law of thermodynamics. Thus,
`U_f = U_i - A`
For a slow process, `A = int_(B_0)^V pdV` where for a quasistatic adiabatic process `pV^gamma = constant`.
On the other hand for a fast process the external work done is `A'' lt A'`.In fact `A'' = 0` for free expansion. Thus `U'_f (slow) lt U''_f (fast)`
Since `U` depends on temperature only , `T'_f lt T'_f`
Consequently, `p'' _f gt p'_f`
(From the ideal gas equation `pV = RT`).