# Is the enthalpy change of a reversible adiabatic expansion any different from enthalpy change of a irreversible adiabatic expansion?

Some sites say that adiabatic expansions of ideal gases in general have $$\triangle H =0$$ , whereas some say it is not.

I tried to find $$\triangle H$$ by considering a reversible adiabatic process where Initial conditions are $$P_1, V_1$$ and $$T_1$$ and final variables are $$P_2, T_2, V_2$$ also it implies $$PV^\gamma = k$$, hence $$\triangle U = \frac{P_2V_2 - P_1V_1}{\gamma-1}$$, $$\triangle (PV) =P_2V_2 - P_1V_1$$ and since $$\triangle H$$ = $$\triangle U$$ + $$\triangle (PV)$$ , $$\triangle H =\frac{\gamma (P_2V_2 - P_1V_1)}{\gamma -1}$$

What is wrong with this proof?

• Please do not post the same question twice. Make edits to the first instead. Please visit the help center for more information. Apr 28, 2023 at 6:19
• "Some sites say ..." Really? Which ones? Please provide references in support of such statements. Apr 28, 2023 at 6:21
• $\Delta H = 0$ statement for adiabatic expansion ($Q=0$) of an ideal gas would be true for expansion against vacuum in Joule-Thomson effect scenario ($W=0$), as $\Delta T = 0$, $\Delta U = 0$ and $\Delta (PV) = 0$ Apr 28, 2023 at 10:02
• Are we talking about a closed system or can we also include an adiabatic flow through an open system? Apr 28, 2023 at 10:47
• For a closed system, do you think it is possible to have an adiabatic reversible process between the same initial and final states as an adiabatic irreversible process? Apr 28, 2023 at 10:48

## 1 Answer

Is the enthalpy change of a reversible adiabatic expansion any different from enthalpy change of a irreversible adiabatic expansion?

Enthalpy is a state function, and its changes $$\Delta H$$ are independent of the path from state $$1$$ to state $$2$$. Thus, an adiabatic reversible process or adiabatic irreversible process, will yield an identical $$\Delta H$$.

So some sites say that adiabatic expansions of ideal gases in general have $$△H=0$$ , whereas some say it is not.

The enthalpy change for an ideal gas only depends on temperature. The only way that this change is zero is proved below \begin{align} \Delta H = \int_{T_1}^{T_2} c_p^\pu{ig}(T) \; dT = 0 \therefore \boxed{\Delta H = 0 \Leftrightarrow T_1 =T_2} \tag{1} \end{align}

In consequence, ideal gases suffer an isenthalpic process if and only if the process is isothermal. Adiabatic reversible or irreversible processes have a change in temperature, and thus, cannot be isenthalpic.

What is wrong with this proof?

The proof is fine. Remember it is only valid for:

1. An ideal gas.
2. It has constant-pressure and constant-volume heat capacities, so that $$\gamma$$ is also constant.

However, if you pay attention, there was no need to invoke any reversible/irreversible adiabatic process. It is a general expression, I will show you another way \begin{align} \Delta U &= \int_{T_1}^{T_2} c_V^\pu{ig} \; dT \\ \Delta U &= c_V^\pu{ig} \int_{T_1}^{T_2} \; dT \\ \Delta U &= c_V^\pu{ig} (T_2 - T_1) \\ \Delta U &= \frac{c_V^\pu{ig}}{R} (RT_2 - RT_1) \hspace{1 cm} (\text{use that R = c_P^\pu{ig} - c_V^\pu{ig}}) \\ \Delta U &= \frac{c_V^\pu{ig}}{c_P^\pu{ig} - c_V^\pu{ig}} (RT_2 - RT_1) \\ \Delta U &= \frac{1}{(c_P^\pu{ig}/ c_V^\pu{ig}) - 1} (RT_2 - RT_1) \hspace{1 cm} (\text{Use ideal gas law}) \\ \Delta U &= \frac{1}{\gamma - 1} (P_2V_2 - P_1V_1) \tag{2} \\ \end{align} And by the definition of enthalpy, using Eq. (2) \begin{align} \Delta H &= \Delta U + \Delta (PV) \\ \Delta H &= \frac{1}{\gamma - 1} (P_2V_2 - P_1V_1) + \Delta (PV) \\ \Delta H &= \Delta (PV)\left(\frac{1}{\gamma - 1} + 1\right) \rightarrow \Delta H = \left(\frac{\gamma}{\gamma - 1}\right)\Delta (PV) \tag{3} \end{align} So, it was not necessary to specify how state $$1$$ goes from state $$2$$. Nevertheless, it respects Eq. (1) \begin{align} \Delta H &= \left(\frac{\gamma}{\gamma - 1}\right) (P_2V_2 - P_1V_1) \\ \Delta H &= \left(\frac{\gamma}{\gamma - 1}\right) (RT_2 - RT_1) \\ \Delta H &= \left(\frac{R\gamma}{\gamma - 1}\right) (T_2 - T_1) = 0 \therefore T_2 = T_1 \tag{4} \end{align}

• Nice derivation. +1 // BTW the OP is not a proof, as it skips many deriving steps, taking the first equation for Delta U from thin air, with trivial following steps. Apr 28, 2023 at 6:13
• You could limit it to perfect gas = ideal gas with constant heat capacities. Apr 28, 2023 at 6:17
• @Poutnik Thanks Poutnik! Yes, those are the needed conditions to arrive to that equation. Apr 28, 2023 at 20:47