Condition for condensation from adiabatic expansion

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 3K views
jek8214
Messages
1
Reaction score
0

Homework Statement


Screen%20Shot%202017-04-15%20at%2012.31.43_zpsaqy8mauj.png

I'm stuck on part (c) of this question.

Homework Equations


$$T\frac{d}{dT}\bigg(\frac{L}{T}\bigg) \equiv \frac{dL}{dT} - \frac{L}{T}.$$
Clausius-Clapeyron equation:
$$ \frac{dp}{dT} = \frac{L}{T\Delta V} \approx \frac{L}{TV_{vap}}.$$

The Attempt at a Solution


My approach has been to find the equations for the adiabatic expansion in the p-T plane, given by
$$\ln\bigg(\frac{p}{p_0}\bigg) = \frac{C_{p_{vap}}}{R}\ln T$$
and the equation for the liquid-vapour phase coexistence curve, given by
$$\ln\bigg(\frac{p}{p_0}\bigg) = -\frac{L_0}{RT} + \frac{\Delta C_p}{R}\ln T.$$
Then for condensation to occur, the curves need to intersect (I can worry about the inequality later). This gives $$\frac{L_0}{T} + C_{p_{liq}}\ln T = 0.$$

Try as I might, I can't seem to turn this into the answer they want. I also don't see how any condition on intersection can be formed by considering the gradients. Is there something important I've missed/a slip in my algebra?
 
Physics news on Phys.org
I would start (c) by doing the following:
$$dS=\left(\frac{\partial S}{\partial T}\right)_PdT+\left(\frac{\partial S}{\partial P}\right)_TdP$$
So, at constant entropy,
$$dS=\left(\frac{\partial S}{\partial T}\right)_PdT+\left(\frac{\partial S}{\partial P}\right)_TdP=0$$
I would also use: $$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$