Question

The magnitude of enthalpy changes for irreversible adiabatic expansion of a gas from 1L to 2L is $\Delta H$, and for reversible adiabatic expansion for the same expansion is $\Delta H_2$. Then:

• A. $\Delta H_1>\Delta H_2$
• B. $\Delta H_1<\Delta H_2$
• C. $\Delta H_1=\Delta H_2$ enthalpy being a state function ($\Delta H_1=\Delta H_2$)
• D. $\Delta H_1=\Delta E_1$ & $\Delta H_2=\Delta E_2$ where $\Delta E_1$ & $\Delta E_2$ are magnitude of change in internal energy of gas in these expansions respectively.

Answer 1

The correct option is B $\Delta H_1<\Delta H_2$

Solution:- (B) $\Delta H_1<\Delta H_2$

As we know that,

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

For adiabatic process,

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

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

Now 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

$\Delta H_1<\Delta H_2$