Wikipedia defines wet bulb temperature as:
the temperature of a parcel of air cooled to saturation (100% relative humidity) by the evaporation of water into it, with the latent heat supplied by the parcel.
I would like to calculate this temperature for:
$$\begin{align*} T & = 300\;\mathrm{K} \\ RH & = 0.5 \\ P & = 1\;\mathrm{atm} \approx 10^5\;\mathrm{Pa} \end{align*} $$
where $RH$ is the relative humidity.
The vapor pressure under these conditions is about $P_V \approx 3.5 \times 10^3\;\mathrm{Pa}$ (source)
So I estimate
$$\Delta T = \dfrac{Q}{n_{air}c_P}$$
where $\Delta T$ is the difference between wet bulb temperature and ambient temperature, $Q$ is the heat absorbed by the water as it evaporates, $n_{air}$ is the number of moles of air, and $c_P$ is the heat capacity at constant pressure of air.
Using $Q = n_{water} L,$ where $n_{water}$ is the number of moles of water that evaporate and $L$ is the latent heat of fusion of water,
$$\Delta T = \dfrac{n_{water}L}{n_{air}c_P}$$
The ideal gas law says
$$\dfrac{n_{water}}{n_{air}} = \dfrac{\Delta P_{vapor}}{P_{air}}$$
where $\Delta P_{vapor}$ is the change in the water vapor pressure as the water evaporates.
I have
$$\begin{align*} \Delta P_{vapor} & = .5\cdot 3.5\times 10^3 \;\mathrm{Pa} \\ L & = 40.8 \;\mathrm{kJ/mol} \\ c_P & = 30 \;\mathrm{J/(K\;mol)} \end{align*}$$
$c_P$ is calculated from $c_P = \dfrac72 R$ for a diatomic gas. The source on $L$, the latent heat of vaporization of water, is here.
Putting in these numbers, I find
$$\Delta T \approx 24 K$$
However, calculators like this one give $\Delta T \approx 7\;\mathrm{K}$.
My calculation leaves off some effects such as the heat needed to cool the vapor, but because the partial pressure of the vapor is small compared to that of air, this should be a small effect. So where is the factor of $\approx 3-4$ error coming from?
