Now we can write And: or: Taking S = S(U,V) we can also write: By comparing the two: Extremely important – definition of temperature!!! Also important, but perhaps slightly less

Heat capacity can be expressed in terms of entropy: (not in the book, but it is a very useful material) Considering that: and that: We can write: If V = const., then: Meaning that: Very useful!!!

Another example: the famous “throttling process” (a.k.a. “Joule-Thomson process” or “J-T effect”) This is a constant enthalpy (isoenthalpic) process – i.e., the enthalpy remains constant during the entire process of transferring the gas from the left part of the apparatus to the right part. It is easy to show that (next slide).

Let’s “balance the checkbook”: Initial energy of the gas is: U 1 Work done on the system when the left piston is moved in: p 1 V 1 Work done by the gas when the right piston is moved out: p 2 V 2 Final energy of the gas: U 2 Energy added to the system Energy given away by the system

So, we can write: The initial and the final enthalpy is the same – ergo, H = const. in the process. Taking T=T(H,p) : And because H=const.:

BAD NEWS…. we don’t know the derivative One way of determining this derivative is to use the legendary “Maxwell Relations”. But we don’t know them yet…. There is another recipe – that presented in the Book – it is not easy to “discover” that procedure without having much practice in solving problems in classical thermodynamics. To make it clear, we will go through it in a step-by-step manner. We begin with taking the enthalpy differential: And the First Law (let me drop the “bar” in dQ):

By combining the two, we have: Now divide the latter by T : OK, at this moment we can use the Euler Criterion:

Left side: is the derivative we need, note! Now the right side: To process the derivative, we will use the ordinary chain rule (on next slide)

Continued from the preceding slide: We now have both sides processed, and we can equate them: We can also use the standard volume expansion coefficient we have introduced a while ago:

To get: So, finally we have all needed components, and we can write The expression for the T change in the J-T process: Which is a well-known result you can find in every thermal physics texbook!!!

Last thing: let’s calculate the drop in T in an ideal gas. It is easy to find the volume expansion coefficient for an ideal gas: Hence: An ideal gas does not change its temperature In the J-T process! So, why all that publicity around an effect that is not? Well, because J-T experiment showed that in experiments with air, nitrogen, and other gases the temperature actually does change (in most cases, it decreases). J-T process major practical application is in liquifying gases.