(1)We have seen that the results of the Joule expansion (valid for ideal gases) demonstrated experimentally that for an ideal gas,
It would be advantageous to be able to calculate this quantity from an equation of state or other pVT data. There is an equation which we will prove later, but which we introduce now because it is so useful, called the "thermodynamic equation of state, which will allow us to do this. It allows us to calculate the derivative in Equation (1) from an equation of state.
The equation is,
This equation will be proved easily once we have the second law of thermodynamics. For now we will just accept it conditionally until it can be proved. Notice that the right- hand side contains nothing but pVT data. We can see that the equation is at least plausible by checking that it does give zero for an ideal gas.(2)
For an ideal gas,
so
Then,
We can also check to see what our thermodynamic equation of state would give for a van der Waals gas. For the van der Waals gas we find,
so
and
We know that a is small and n2/V 2 will be small except at very high pressures (densities).
(The above result can be understood based on what is going on in the gas. When a gas expands at constant temperature it absorbs heat from the surroundings and does work on the surroundings. If the gas is ideal the heat and work exactly balance so that there is no change in the internal energy of the gas. In a van der Waals gas - and real gases - the expansion must also overcome the intermolecular forces so part of the heat absorbed from the surroundings goes to overcoming the intermolecular forces. The a term in the van der Waals equation of state accounts for intermolecular forces. If you calculate the work in expanding a van der Waals gas you will see that that the part of the work that is proportional to a is positive so that this work was done on the system - it raised the internal energy of the system.)In most cases
is still pretty small, even for a van der Waals gas.
There is a companion to Equation (2),
This equation can be derived (without the second law) from Equation (2) so that if Equation (2) is correct, so is Equation (3).(3)
We will leave it to the reader to show that Equation (3) gives zero for an ideal gas. Applying this equation to the van der Waals gas is a little more involved and not particularly enlightening.
Relationship Between Cp and CV
Cp and CV are related to each other and their difference can be calculated from an equation of state. We wish to prove that
Let's begin with the definitions of Cp and H,(8d)
The second term in (4c) is in an acceptable form, but the first term is not. (The wrong variable is being held constant.) To deal with the first term regard U as U = U(T,V). Then,(4a, b, c)
Now divide Equation (5) by dT and hold p constant. (Your calculus teacher won't like this, but you can prove that the result is correct and that this procedure will always work.) We obtain,(5)
Now we substitute Equation (6) for the appropriate term in Equation (4c) to get,(6)
But(7,a b)
Substituting Equation (2) in for (∂U/∂T)V in Equation (7b) gives
Which is the desired result. We will let the reader show, using Euler's chain relation and the definitions of α and κ , that this relation can be rewritten as(8a, b, c, d)
The second term on the right of Equation (9) is necessarily positive because κ is always positive. α can be negative (water near 0oC), but it appears here as the square. Thus Cp > CV. For solids and liquids the second term on the right of Equation (9) is usually small. For gases it can be large. For an ideal gas we found earlier that α = 1/T and κ = 1/p so that(9)
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