Chemistry - Why is the enthalpy change not zero in an adiabatic process?

Solution 1:

The Enthalpy $H$ is defined as $H=U+PV$. Therefore,

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

For an adiabatic process, $q=0$. therefore from the first law of thermodynamics, $$\Delta U = q +w =q-P\Delta V$$ $$\Delta U=w=-P\Delta V$$ Substituting this in the first equation you get, $$\Delta H=V\Delta P$$ If $\Delta P$ is zero during the process (Isobaric) then the enthalpy change will indeed be zero, but it will be non-zero if the process is not isobaric.

Solution 2:

If you have an ideal gas in a constant volume adiabatic chamber, with the gas initially occupying only half the chamber, and vacuum in the other half, with a barrier in between, and you remove the barrier and then let the system re-equilibrate (i.e., free expansion), the work done on the system will be zero (rigid container) and $\Delta U = 0$. Therefore, the temperature change will be zero, which also means that $\Delta (PV)$ will be zero. So, $\Delta H=\Delta U+\Delta(PV)=0$ even though the pressure change is not zero.

Even if there is gas present initially in the other half of the chamber, the $\Delta H$ for the system is still zero. Initially the enthalpies of the gas in each of the two chamber halves of volume $V/2$ are: $$H_1=n_1u(T)+P_1V/2=n_1u(T)+n_1RT$$ $$H_2=n_2u(T)+P_2V/2=n_2u(T)+n_2RT$$ where u(T) is the molar enthalpy at temperature T, and the $n$'s are the number of moles in the two chambers. So the initial enthalpy of the system is: $$H_{\text{init}}=H_1+H_2=(n_1+n_2)u(T)+(n_1+n_2)RT$$ But, since no work is done and no heat is transferred to the system, the final internal energy of the system is the same, as is the final temperature. And, of course, the total number of moles doesn't change. So, that means that $$H_{\text{final}}=H_{\text{init}}$$ and $\Delta H=0$. (Of course, the final pressure is $(P_1+P_2)/2$).