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I am confused about what the Clausius-Clapeyron equation is saying basically. I learned it from Finn's Thermal Physics textbook. Here is the illustration they used to derive it.

enter image description here

Matter of two different phases can only coexist when they have the same specific gibbs function.

At A,

$$g_1(T,P) = g_2(T,P)$$

And at B,

$$g_1(T+dT,P+dP)=g_2(T+dT,P+dP)$$

I'm not sure why this is necessarily the case, but Finn merely states this.

From the previous equation, via a Taylor expansion and the thermodynamics identity for G,

$$v = \left(\frac{\partial g}{\partial P}\right)_T$$

$$s = -\left(\frac{\partial g}{\partial T}\right)_P$$

Then, a way to rewrite the equation $$g_1(T+dT,P+dP)=g_2(T+dT,P+dP)$$ is

$$\frac{dP}{dT} = \frac{\Delta S}{\Delta V} $$

Then, with $L = T \Delta S$,

$$\implies \frac{dP}{dT} = \frac{L}{T \Delta V}$$

The mathematics of arriving from the equation in terms of specific gibbs functions and to the final result is fine with me, but I don't actually know what this represents. To me, it is one of two things:

  1. It is either the rate of change of pressure with temperature such that a substance remains on a phase boundary and does not transition

  2. Or it informs how the volume and entropy change from a phase transition based on its signature.

But, to be honest I really have no idea what this is. I understand the mathematics but not the underlying principles here.

For instance, for these two graphs:

enter image description here

Would following along the lines where it says $\frac{dP}{dT} > 0$ or $\frac{dP}{dT} < 0$ be the lines one would have to follow to not change phase? Or, for instance, is the $\frac{dP}{dT} > 0$ in (a) showing the effects of volume and entropy (that the volume and entropy would increase) for melting would account for a $dP/dT$ that is positive here?

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    $\begingroup$ The derivation is only concerned with points along the boundary. So the relation just talks about the boundary line, right? $\endgroup$ – Aaron Stevens Apr 12 at 12:21
  • $\begingroup$ Did you really want to focus on the SL equilibrium, or are you more interested is LV equilibrium? $\endgroup$ – Chet Miller Apr 12 at 14:32
  • $\begingroup$ @AaronStevens I believe so. $\endgroup$ – sangstar Apr 12 at 20:19
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Some would call the equation that you have derived the Clapeyron equation and if one of the phases is a gas they would call it the Clausius-Clapeyron equation.

I am confused about what the Clausius-Clapeyron equation is saying basically.

The equation is telling you about the slope of the pressure against temperature graph when two phases coexist with one another.

If $L$ and $\Delta V$ do not change by very much over a certain temperature range then integrating your equation gives $P_2-P_1 = \dfrac {L}{\Delta V}\ln{\dfrac{T_2}{T_1}}$ so if you have knowledge of one point on the graph, $(P_1,T_1)$, the (Clausius-)Clapeyron equation enables you to find other points $(P_2,T_2)$.
In other words the prssure and the temperature are not independent of one another.

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  • $\begingroup$ Is it the required behavior of pressure with temperature required for two phases to coexist? $\endgroup$ – sangstar Apr 14 at 16:59
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I'm not sure why this is necessarily the case, but Finn merely states this.

For a mixture of two phases in equilibrium, when mass of the phase 1 is $m_1$ and that of phase 2 is $m_2$, the total Gibbs free energy $$G(T,P)=m_1g_1(T,P)+m_2g_2(T,P)$$ must be at its minimum. Using $m_1+m_2=\text{constant}$, and the fact that for an infinitesimal transfer of mass from one phase to another, the change in $G(T,P)$ must be zero i.e. $$dG(T,P)= dm_1~g_1(T,P)+dm_2~ g_2(T,P)=0,$$ one obtains (since $dm_1=-dm_2$) $$g_1(T,P)=g_2(T,P).$$

  1. It is either the rate of change of pressure with temperature such that a substance remains on a phase boundary and does not transition
  2. Or it informs how the volume and entropy change from a phase transition based on its signature.

If $\frac{dP}{dT}>0$, it implies that the with the increase in pressure the melting point increases and if $\frac{dP}{dT}<0$, with the increase in pressure, the melting point decreases.

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  • $\begingroup$ I still don't understand. Why does $\frac{dP}{dT} > 0$ imply an increasing pressure implies increase in melting point? What does increasing pressure constitute on a $PT$ diagram? Going up or down along an isotherm? $\endgroup$ – sangstar May 3 at 10:02
  • $\begingroup$ If I have a value for $\frac{dP}{dT}$, how do I what Wikipedia does to find the pressure needed to melt ice at $-7$ Celsius? The Wikipedia link I'm referring to is at en.wikipedia.org/wiki/… $\endgroup$ – sangstar May 3 at 10:11

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