# Why is latent heat of water vapor included in specific enthalpy of moist air?

In a lecture book we read about latent heat of moist air.

I would have guessed, that the total specific enthalpy of moist air, comprised of $$m_d$$ kg of dry air and $$m_v$$ kg of water vapor is (we have just a mixture of two gases...)

$$h_{total} = \frac{c_p^d \cdot T \cdot m_d+c_p^v \cdot T \cdot m_v }{m_d+m_v}$$

Defining absolute humidity as $$q = \frac{m_v}{m_d+m_v}$$

this is written as:

$$h_{total} = c_p^d\cdot (1-q)\cdot T + c_p^v\cdot q\cdot T$$

In the lecture, however, we learned

$$h_{total} = c_p^d\cdot (1-q)\cdot T + L \cdot q$$

with L denoting the enthalpy of vaporization (aka latent heat). I know, what latent heat physically means, but anyway, I do not understand why it is included in h for the gas.

My questions:

Q1: The lecture book says literally, that the enthalpy of pure water vapor is

$$H_v = m_v \cdot L$$

so, in contrast to my equation, the $$c_p^v$$ is not included at all. This would mean, water vapor just contributes a constant to h which doesn't depend anymore on T - for me this sounds weird, since we are still talking about a gas...

Q2: Why is it required at all, to include L into the enthalpy of moist air? After liquid water has vaporized (for whatever reason) it is just a "regular" gas with another specific gas constant and also the mixture of those two gases behaves like an ideal gas to a good approximation. So, to make my question more clear: If L for water is included in h, why isn't the enthalpy of vaporization of liquid dry air also included? Physically there is no difference, but it would be somehow ridiculous to include latent heat of liquid air into the enthalpy of the atmospheric gas.

I see no reason to do so - although I believe, that good arguments exist.

This is the copy from the book: Unfortunately, since it is a german book, D and L denote here vapor ("Dampf") and dry air ("Luft"), respectively. 