Phase transition between two phases with different Cv

[Baibhab Bose]

Homework Statement

A many-body system undergoes a phase transition between two phases A and B at a
temperature ܶTc . The temperature-dependent specific heat at constant volume Cv of the two
phases are given by Cv(A)=aT^3+bT and Cv(B)=cT^3.Assuming negligible volume change of
the system, and no latent heat generated in the phase transition, ܶTc is...?

Relevant Equations

Clausius Clapeyron equation maybe.

I actually can't figure out what kind of phase transition it is and how to proceed through..!

 

[Lord Jestocost]

What does it mean for the order of a phase transition if there is no latent heat generated at the transition temperature T_c?

 

[Baibhab Bose]

What does it mean for the order of a phase transition if there is no latent heat generated at the transition temperature T_c?

Since 2nd order phase transition is accompanied by no heat change and 1st order transition does, that indicates this is a 2nd order transition!

 

[Lord Jestocost]

The next question is: Is there a thermodynamic function which shows a characteristic behavior at the transition temperature $T_c$ and which can be related to the heat capacities in both phases?

 

[Baibhab Bose]

In second order phase transition, Gibbs free energy remains constant.
dG=dU-Tds-SdT+PdV+VdP=0
TdS=0 (since no heat change)
SdT=0 (process at same temp)
PdV=0('negligible volume change)
so that leaves us
dG=dU+VdP
but if we write dU=CvdT then again dT=0.
So what to do?

 

[Lord Jestocost]

When dealing with a second order phase transition, the entropy is continuous at $T_c$ (it has a kink as the heat capacities of both phases differ from each other). See the figures on page 20 in
[PDF]
Phase Transitions and Phase Diagrams - Virginia

The temperature dependence of the entropy in both phases can be calculated as the temperature dependence of the heat capacities ($C_{V,A}(T)=aT^3+bT$ and $C_{V,B}(T)=cT^3$) in both phases are given:

$S_A (T) = \int_0^T \frac {C_{V,A}(T')} {T'} \, dT'$
$S_B (T) = \int_0^T \frac {C_{V,B}(T')} {T'} \, dT'$

(assuming $S_A(0)=0$ and $S_B(0)=0$).

As the entropy is continuous at $T_c$, one has $S_A(T_c) = S_B(T_c)$ . This equation can now be used to determine $T_c$.

 

[Baibhab Bose]

Okay, now I have learned how to use these conditions on phase transitions mathematically to approach a problem. And I have got the answer. thank you so much!

 

[Baibhab Bose]

$S_A (T) = \int_0^T \frac {C_{V,A}(T')} {T'} \, dT'$
$S_B (T) = \int_0^T \frac {C_{V,B}(T')} {T'} \, dT'$

I have one confusion here, you have defined the entropy at Tc by integrating Cv(T)/T from 0 to Tc. I don't understand why should T=0 be taken here as the lower limit. The transitions are not taking place around T=0, and T=0 is not of any special relevance here, then why?

 

[Lord Jestocost]

When dealing with a second order phase transition, the absolute values of the entropy of both phases are equal to each other at the phase transition point $T_c$. To calculate the absolute entropy of a material on base of experimental heat capacity data, one needs a reference point which is taken as the entropy $S (T)$ at $T=0K$. Under constant volume conditions, one thus gets for the absolute entropy at a certain temperature $T$:

$S(T)-S(0) = \int_0^T \frac {C_{V}(T')} {T'} \,dT'$