# Clausius-Clapeyron phase question

1. May 8, 2014

### unscientific

1. The problem statement, all variables and given/known data

Part (b): Find the temperature in which the pressure is twice the atmospheric pressure.

2. Relevant equations

3. The attempt at a solution

I've done every part except part (b).

Part (b)

$$\Delta v \approx v_{gas} = \frac{RT}{p m'}$$

Therefore the clausis-clapeyron equation reads:

$$\frac{dp}{dT} = \frac{L}{T \Delta v'} = \frac{L m' p}{RT^2}$$

$$\int \frac{1}{p} dp = \frac{Lm'}{R} \int \frac{1}{T^2} dT$$

$$ln p = -\frac{Lm'}{RT} + const.$$

$$p = p_0 exp(-\frac{Lm'}{RT})$$

Taking $p = 2p_0 = 10^5 Pa$, so $\frac{p}{p_0} = 2$. What's the point of giving us the atmospheric pressure then? All we need is the ratio of pressures.

This gives an answer of $T = 7000K$, which is insanely hot water. (So hot that the sun gets jealous).

2. May 8, 2014

### Staff: Mentor

You're right about not having had to be supplied with the numerical value of the atmospheric pressure, but your integration is not correct. You did not determine the constant of integration correctly. If p = p0, your equation predicts that the corresponding temperature is infinite, and not 373K.

3. May 9, 2014

### unscientific

That's right. So at $T = T_0$ and $P = P_0$:

$$ln P_0 = -\frac{Lm'}{RT_0} + const.$$
$$const. = ln (P_0) + \frac{Lm'}{RT_0}$$

Thus,

$$ln \left(\frac{P}{P_0}\right) = \frac{Lm'}{R}\left(\frac{1}{T_0} - \frac{1}{T}\right)$$

Where $T_0 = 300 K$ and $P_0 = 10^5 Pa$.

This gives a temperature of $313 K = 40 ^oC$, which seems reasonable.

Last edited: May 9, 2014