I know that the relationship between pressure and temperature for an ideal gas is linear. The relationship between vapor pressure and temperature for a liquid, however, is exponential. To make it linear we take the natural log and end up with: [tex] \ln P = -\frac{\Delta H_{vap}}{RT} + b [/tex]. How did we get [tex] -\frac{\Delta H_{vap}}{RT}[/tex] to be the slope? The y-axis is [tex] \ln P [/tex] and the x-axis is [tex] \frac{1000}{T} [/tex].(adsbygoogle = window.adsbygoogle || []).push({});

Thanks

**Physics Forums - The Fusion of Science and Community**

# Clausius-Clapeyron equation

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Clausius-Clapeyron equation

Loading...

**Physics Forums - The Fusion of Science and Community**