The answer depends on the kind of material you are considering. For an ideal gas, Cpm = Cvm + R. If it is a molecular gas, increasing temperature enables vibrational degrees of freedom, so that Cvm increases. Hence Cpm/Cvm = 1 + R/Cvm decreases.

There may be an exception for hydrogen: for example, e-H2 (equilibrium hydrogen) has a peak of the heat capacity at low temperatures.

For liquids, the answer is not simple indeed.

Answering the why is not as simple as it might seem. Thus I just redirect you to the long explanation here:

DOI: 10.1103/PhysRevB.78.104201

Both Cp and Cv increases with T, but increase of Cv is more than Cp, so specific heat ratio decreases with T. If you are looking for physical explanation, I don't know.

The answer depends on the kind of material you are considering. For an ideal gas, Cpm = Cvm + R. If it is a molecular gas, increasing temperature enables vibrational degrees of freedom, so that Cvm increases. Hence Cpm/Cvm = 1 + R/Cvm decreases.

There may be an exception for hydrogen: for example, e-H2 (equilibrium hydrogen) has a peak of the heat capacity at low temperatures.

For liquids, the answer is not simple indeed.

i remember seeing expressions for Cp,Cv as functions of Temperature.....now first try proving mathematically D(gamma)/DT to be negative if possible....i too will try for a theoretical explantion meanwhile. :)

Dear Surya,

if you have seen an expression for Cv(T) – instead of Cv(V, T) – you must have looked at a theory of either incompressible systems or ideal gases. In the former case, Cp = Cv; in the latter case, please see my previous posting.

In the general case (real gas), the question is not quite clear. Are we asked for gamma(T) along an isobaric or an isochoric path? Consider the case of following an isobaric path at the critical pressure: At low T, the sample would be in the liquid state;

at high T in the gas state; both states have a finite gamma value. But at the critical point, gamma would be infinite. Consequently, along such a path gamma(T) would not decline monotonously.

according to heat capacity relation Cp-Cv= is the product of (-T) x square of partial derivative of volume w.r.t. temp keeping pressure constant and negative partial derivative of pressure w.r.t. volume by keeping temp constant In which third term is negative and second term is positive hence cp-cv is always positive.

Mr.Ranjith.....to prove that Cp-Cv>0 is not relevant in any manner to the present discussion....moreover i dont think we need such a complicated expression in terms of partial derivatives and all....i remember a basic derivation in my 12th class saying Cp>Cv.....anyways here the discussion is regarding the behavior of Gamma as a function of Temp.....

As more degrees of freedom are activated (including vibrational DOF, dissociation, ionization, chemical reactions etc), gamma decreases. Keep in mind also the physical significance of gamma as "thermal stiffness" or the ability of a gas to build up pressure following an increase in internal energy.

I do not know in general, but in cases of gases and plasma the question can be answered easily if one consider the contribution of internal states and chemical processes. I have attached a paper discussing this behavior.

I have forgotten the attachment

Andi Petculescu provided the answer. But why to ask such a question which is explained in the elementary kinetic gases. By the equipartion of energy principle each degree of freedom contributes R/2 to the the value of c_v, Hence monatomic atoms give 3R/2 for the three translational degrees of freedom. As the temperature rises the rotational, then vibrational and finally electronic degrees of freedom are activated. Finally as has been mentioned c_p=c_v+R and the rest follows

Specific heat at constant volume increases, as U. Deiters explained.