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

Which of the following must be true for adiabatic processes?

• $$C_V = C_p$$
• $$\Delta H = 0$$
• $$\Delta U = 0$$
• $$\Delta S = 0$$
• $$q = 0$$

(Source: Chemistry GRE)

The answer is $$q = 0$$. From what I can find, an adiabatic process is when there is no transfer of heat, but then why is the enthalpy change not zero?

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.

• Isn't enthalpy defined for systems at constant pressure, so the V∆P part in the second line is itself zero. That gives us ∆H=0 for an adiabatic process. – jyoti proy Mar 23 '17 at 20:57
• @jyotiproy It is defined for all systems, and is a state variable. Change in enthalpy does depend on the change in pressure. Anyway, if we deal with infinitesimals, we need not worry about having P and $\Delta P$ in the same equation. – Satwik Pasani Mar 24 '17 at 6:07

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 $(P_1+P_2)/2$).

• (+1) I find that you have very aptly demonstrated the specific cases where $\Delta H=0$. However, in a general situation, we can't say that $\Delta H=0$, as Satwik demonstrated above. Is my interpretation of your answer correct? – Gaurang Tandon Mar 22 '18 at 7:04
• For an adiabatic process on a closed system, the enthalpy change from the initial thermodynamic equilibrium state to the final thermodynamic equilibrium state is $\Delta H=\Delta (PV)-\int{P_{ext}dV}$If the process is adiabatic and also reversible, this equation reduces to $\Delta H=\int{VdP}$. – Chet Miller Mar 22 '18 at 11:54
• How can we integrate V here with dp. I am asking this because I am solving a very similar question and I am stuck on using your expression. The volume is not constant in that reversible adiabatic – Aladdin Aug 18 '18 at 10:48
• $nC_pdT=\frac{nRT}{P}dP$ – Chet Miller Aug 18 '18 at 11:38

why is the enthalpy change not zero?

When only PV work is possible, and pressure is constant, we can write

$$q_p = \Delta H \tag{1}$$

that is, the enthalpy change at constant pressure is equal to the heat exchanged. If the pressure is not constant all bets are off and you should resort to some other expression such as

$$\Delta H = \int_1^2 d(U+PV)$$

For a reversible process,

$$\Delta H = \int_1^2 VdP$$

This integral cannot be zero unless $$V=0$$ (which is not possible) or $$dP=0$$ (which means $$P=$$constant, bringing us back to Eq. 1, which says that for an isobaric adiabatic process $$\Delta H =0$$).

Since we have found a case that violates the stated condition, $$\Delta H =0$$ need not be true.