# Evaporative cooling power

by Angelos K
Tags: cooling, evaporative
 P: 39 I wish to understand evaporative cooling. Apparently the principle consists in pumping the vapour above a bath, hence extracting it's most energetic corpuscles and achieving cooling. The aim is to derive the cooling power. My textbook states: $$\frac{dQ}{dt} = L\frac{dn}{dt} \propto p$$ L being the latent heat per mol and p the gas pressure. I don't see why the proportionality holds. Sorry if I miss something simple, I'm battling a light cold ;-) [From here on one can use the gas pressure formula $$p = p_0 \exp(-\frac{L}{RT})$$ to demonstrate, how with linearly decreasing temperature the cooling rate decreases exponentially] Thanks in advance for any help!
 P: 39 Thanks! So my textbook seems to make several oversimplifying assumptions, right? What I don't get is how he gets to the relation: $$L\frac{dn}{dt} \propto p$$
 P: 39 The gas pressure formula [exponential increase with temperature] is derived in the book from the Clausius-Clapeyron equation, by neglecting the volume of the fluid phase compared to that of the gaseous one AND making an ideal gas assumption for the gas. I suppose the ideal gas behaviour to be a really rough approximation, even for helium, since the temperatures reached are around 1K [This wikipedia article discusses the Clausius-Clapeyron equation and its less general form $$\frac{dp}{dt} = \frac{Lp}{R{T}^2}$$ used here]