[Donaldson Tan]

this is a heat capacity problem from my textbook. i got stuck doing it.

show that for a gas which obeys the van der waals' equation fo state (p + a/v²)(v - b) = RT, Cv is a function of temperature only whereas Cp is a function of both pressure and temperature.

i wanted to use:
(dT/dv)_s = -(T/Cv)(dp/dT)_v
(dT/dp)_s = -(T/Cp)(dv/dT)_p

but i got stuck (dT/dv)_s and (dT/dp)_s because i don't know how constant entropy will make them vary. we cant use PV^Cp/Cv = constant because this is not a perfect gas.

i been studying from evening till dawn.. going to sleep now.. hope someone will enlighten me on this issue.

[Juan R.]

It would be very difficult to obtain this reply from usual chemical thermodynamics literature. I believe

As said the usual physical chemistry literature is very outdated  Some physicists textbook are also, but this is more strange. Geodeome, what is the textbook?

There is an emphasis on chemical literature on the equations of state. The equation of state is not fundamental. Please forgot many of theoretical chemical literature  

There are three levels of information in macroscopic thermal canonical science.

- The first level: fundamental level. The fundamental equation for potentials, especially for entropy S, is the basic equation. All thermodynamic information is contained therein. In fact, the ratios for any process (equilibrium is also a process because velocity zero is also a velocity) are guided by first order differentials of entropy in the macroscopic limit.

- Second level: Formed by thermal and caloric equations of state. The fact of that are not fundamental is easily see since there is no procedure for obtaining the fundamental equation from them (this indicated less information about the system). You have just the thermal equation of state.

- Third level is composed by many coefficients: compresion, etc. and expressions for Cv and Cp. The information is still lower.

You need the first level or the second more additional expressions.

In the first level one can see that the thermodynamic potential U is

U = U(ideal) + INT( T² (part (p/T) / part T ) ) dV

INT signifies integration between infinite and V. The partial is for constant V.

If you prefer to use the expresion for Cv (it arises directly from above U. The inverse is not possible).

Cv = Cv(ideal) + INT( T (part² p / part T² ) ) dV.

After use the relation between Cv and Cp.

Structure of equation for Cv shows that for any gas with p = f(T) with f a linear expresion Cv = Cv(ideal).

Therefore

Cv for a Van der Waals gas does not depend of temperature

This is not so strange. Cv is the change in kinetic energy per unit change in temperature. since that "VDW molecules" interact only with energy that depend on density (N/V), the value of Cv is unaffected by intermolecular forces in the VDW model.

[Donaldson Tan]

juan: my thermodynamics lecturer is a physicist, not a chemist. LOL.

[Juan R.]

Ok Geodome, I said

"usual physical chemistry literature is very outdated. Some physicists textbook are also, but this is more strange."

This (of course my) thinking also applies to lecturers, PhDs, proffesors, etc. not only to textbooks, but I believe that continue being true that usual chemical thermodynamics (i.e. papers, textbooks, monographs, and scientists) is really outdated. I was talking in an average sense not focusing in a special case.

It is like if i say that a men are, in general, more high that women. It is true (statistics is that, at least in Spain) and still there are Spansih women more high than me.


What is the textbook? I am intrigued  .

[Donaldson Tan]

Thermodynamics for Chemical Engineers  
K.E. Bett, J.S. Rowlinson, G. Saville

it's written by engineers from my department

[Juan R.]

Thanks Geodome. I don't know that textbook. I will atempt to look it!