# Homework Help: Vapour Pressure of Water

1. Mar 18, 2012

### kudoushinichi88

1. The problem statement, all variables and given/known data
This is problem in Mandl's book on Stats Phys; The vapour pressure of water at 298.15 K is 23.75 mmHg. What is the vapour pressure of water at 273.16 K, given that the latent heat of evaporation of water at 298.15 and 273.16 K is 43,991 and 45,049 J/mol respectively?

2. Relevant equations
According to the Hints, we use this equation;

$$\frac{d}{dT}\ln P = \frac{1}{P} \frac{dP}{dT} = \frac{ML_{12}}{R T^2}$$

where M is the gram-molecular weight and L_12 is the latent heat of vaporization.

3. The attempt at a solution
Integrating, we get
$$P=A \exp\left({- \frac{M L_{12}}{RT}}\right)$$

where A is an arbitrary constant. but when i sub in the values, I get some ridiculous number. where did I go wrong??

Btw the answer is 4.58 mmHg while i got something of 10^6 in magnitude.