# Pressure in Clapeyron and clausius Clapeyron equation

The Clapeyron equation is $$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta S}{\Delta V}$$. Here, is the change in pressure 'dp' actually a change in external pressure that is being applied on the whole system?

Clausius Clapeyron equation:

$$\int_{p_1}^{p_2}\frac1p\mathrm{d}p=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\mathrm{d}T$$, where $$p_2$$ is vapour pressure at $$T_2$$ and $$p_1$$ is vapour pressure at $$T_1$$ respectively.

I am confused as to how can we get from pressure on the whole system to vapour pressure? Am I missing something? Or is Clausius Clapeyron equation derived in such a way from Clapeyron equation so as to relate the vapour pressures with corresponding temperatures?

• @Poutnik, I read wikipedia. But I'm still not clear as to why we CAN change from pressure to vapour pressure? Commented Mar 22 at 12:40
• @Poutnik. I read it again. It says that "The Clausius–Clapeyron relation can be used to find the relationship between pressure and temperature along phase boundaries". Along the phase boundaries where gas is in equilibrium with condensed phase vapour pressure is equal to external pressure. Hence we can put vapour pressure. Is this correct? Commented Mar 22 at 12:44
• So my understanding that CC equation is when gas-liquid equilibrium with vapour pressure being equal to external pressure is correct? When we are not in equilibrium C equation is more appropriate where dp is change in external pressure? Commented Mar 22 at 13:27
• What about when they are not in equilibrium? Is there another equation for that? Commented Mar 22 at 13:33
• There all bets are off. There is infinite number of different possible nonequilibrium states, with spatial and temporal variation of parameters. Commented Mar 22 at 13:56