Question

The magnitudes of enthalpy changes for irreversible adiabatic expansion of a gas from 1L to 2L is $\Delta H_1$ and for reversible adiabatic expansion for the same expansion is $\Delta H_2$. Then what is the relation between $\Delta H_1$ and $\Delta H_2$?

Answer 1

Solution:-

As we know that,

$\Delta H=\Delta U+P\Delta V$

For adiabatic process,

$\Delta H=\frac{f}{2}nR\Delta T+nR\Delta T=n{C}_P\Delta T$

$\therefore \Delta H\propto \Delta T.....\left(1\right)$

As the work done by the gas is maximum in reversible adiabatic process than irreversible adiabatic process.

Now,

$P\Delta V=W$

$\Rightarrow \Delta T\propto W.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\Delta H_1<\Delta H_2$