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The Gibbs Phase Rule indicates that for a two phase, single component thermodynamic system we will have one independent intensive parameter. Given that the Degree of Freedom is $1$ means that fixing one intensive parameter would fix the entire state of the system.

Any property $x$ is just a function of one other property say $y$, i.e. $x=f(y)$. This gives for two phase system $P=f(T)$, but why isn't $v=f(T)$?

As we know $v$ can have a value anywhere between $v_f$ and $v_g$. Are there any assumptions for the Phase Rule, other than equilibrium, which would explain this?

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    $\begingroup$ Hi akshay, welcome to engineering.SE. I don't understand the question. If the volume is fixed for a two phase mixture, the temperature and pressure can of course be changed, but not independently of each other (in accordance with Gibb's phase rule). Can you elaborate on the situation that you have in mind where the phase rule is violated by editing your question? $\endgroup$ Jul 27 '15 at 12:26
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Specific Volume is a intensive property too.

Using Gibbs you get 1 degree of freedom for your single component, 2 phase mixture.

f = N - P + 2

f = 1 - 2 + 2 = 1

It is not yet fixed what you have to use as your degree of freedom. For your example, at a fixed specific volume v (which equals $\frac{1}{\rho}$) you can either change T or p. But keep in mind that T(p,v) and p(T,v). So a change in pressure results in a change in Temperature, too and vice versa.

Does that clear up your confusion about the rule?

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Look at the diagramms. If you set $v$ you have 'locked' you state. You see now that you cannot change $p,T$ independently. And regarding your comment for example you can see that $T(v)|_p$.

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  • $\begingroup$ Degree of freedom is 1 means that fixing 1 intensive property would fix state of system. Any property 'x' is just function of one other property say 'y', i.e. x=f(y). This gives for two phase system P=f(T), but why isnt v=f(T)? As we know v can have value anywhere between vf and vg. $\endgroup$ Jul 28 '15 at 11:42
  • $\begingroup$ @akshayrathore Updated my answer $\endgroup$
    – idkfa
    Jul 30 '15 at 8:55

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