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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?

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  • $\begingroup$ @Poutnik, I read wikipedia. But I'm still not clear as to why we CAN change from pressure to vapour pressure? $\endgroup$
    – Natasha J
    Mar 22 at 12:40
  • $\begingroup$ @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? $\endgroup$
    – Natasha J
    Mar 22 at 12:44
  • $\begingroup$ 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? $\endgroup$
    – Natasha J
    Mar 22 at 13:27
  • $\begingroup$ What about when they are not in equilibrium? Is there another equation for that? $\endgroup$
    – Natasha J
    Mar 22 at 13:33
  • $\begingroup$ There all bets are off. There is infinite number of different possible nonequilibrium states, with spatial and temporal variation of parameters. $\endgroup$
    – Poutnik
    Mar 22 at 13:56

1 Answer 1

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For a single component system, p is the equilibrium vapor pressure and the pressure within the liquid. For a system with more than one component, if the gas phase is ideal and the Poynting correction for the effect of pressure on the Gibbs free energy of the liquid is negligible, p is the partial pressure of the component in the vapor phase. The overall pressure in the liquid and vapor can, of course, be higher than this.

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  • $\begingroup$ The atmosphere is a second component and changes the Phase Rule. It may or may not be negligible. $\endgroup$
    – jimchmst
    Mar 22 at 20:15
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    $\begingroup$ Like I said, if the Poynting correction is negligible (which is typically the case if the gas phase gives close to ideal behavior), even with a second component present (say air), the partial pressure of the component in the vapor phase is equal to the equilibrium vapor pressure of the pure component at the system temperature. $\endgroup$ Mar 22 at 22:12

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