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

Question

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?

Answer 1

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}