Is the enthalpy change in a reversible adiabatic expansion of an ideal gas 0?

Question

I would like to know what happens to the enthalpy in a reversible adiabatic expansion for an ideal gas. Will it be greater than or equal to zero?

Answer 1

Will it (the change in enthalpy) be greater than or equal to zero?

It will be less than zero.

The change in enthalpy is

$$\Delta H=\Delta U +\Delta (PV)$$

The first law for a closed system (no mass transfer), any process, is

$$\Delta U=Q-W$$

For any adiabatic process, $Q=0$ and the first law becomes

$$\Delta U=-W$$

so the response to your following comment

delta U is zero because it is an adiabatic process, but the second part of the expression can either be zero if delta P = 0 or positive if delta P > zero. Will that be correct?

is $\Delta U$ is not zero for an adiabatic expansion process. It is negative since expansion work $W$ is positive. Since there is no heat transfer to the gas all the expansion work done by the gas is at the expense of its internal energy. Furthermore $\Delta P \ne 0$ and it not greater than zero. It is negative because pressure decreases as the the volume increases in the expansion.

To go on:

For a reversible adiabatic process we have

$$P_{2}V_{2}^{k}=P_{1}V_{1}^{k}$$

$$k=\frac{C_p}{C_v}$$

For an ideal gas, $C_{p}>C_{v}$, therefore $k>1$ making $P_{1}v_{1}>P_{2}v_2$ and $\Delta (Pv)<0$

Since $\Delta U<0$ and $\Delta (Pv)<0$, then $\Delta H<0$

Hope this helps.