# Applying the Clausius-Clapeyron relation

An example of the Clausius-Clapeyron equation, according to Wikipedia, is:

I'm a bit confused with the process.

As far as I'm concerned, this slope is the rate of change of pressure with respect to temperature such that the material being used remains at its phase boundary. At the phase boundary, phase transitions occur.

So, to find the pressure needed to melt ice at $$-7$$ degrees Celsius, we'd need to find the phase boundary line for at the $$-7$$ degrees Celsius isotherm, as this is the required pressure per kelvin on the ice to melt it at that temperature.

Why then, to find melt ice at a temperature below $$0$$ degrees Celsius the author considered at just $$0$$ degrees Celsius? I can't imagine this equation with $$T = 273 \ K$$ can predict the pressure needed to melt ice at $$T = 266 \ K$$.

I think I'm just confused generally with this equation and thought process.. What is $$\Delta T$$ and $$T$$ here? I thought $$T$$ is the isotherm temperature for this phase boundary. What is $$\Delta P$$? I derived it using differentials $$d$$, not $$\Delta$$'s, and I'm always a bit cautious when treating $$d$$'s as $$\Delta$$'s and then treating the whole thing as a fraction.

• At 0 C, the corresponding pressure at the phase boundary is 1 atm. The equation tells you the change from this pressure required for the melting point to be -7 C. The 1 atm value is going to be negligible compare to the required pressure at -7 C, so it is neglected. – Chet Miller May 1 at 20:07