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# Connecting Thermal Derivatives with Experiment

## Prerequisites

For this section, students should have:

- A working ability to perform total differentials.
- Basic understanding of what a partial differential represents in a physical process.
- The skills to predict how heat will transfer in a simple thermodynamic process.
- Understanding of the concepts of heat capacity and specific heat.

## In-class Content

### Lecture: The Thermodynamic Identity

##### Lecture notes from Dr. Roundy's 2014 course website:

The **internal energy** is clearly a state function, and thus its differential must be an exact differential. $$dU = \text{ ?}$$ $$= đQ - đW$$ $$ = đQ - pdV \text{ only when change is quasistatic}$$

This $−pdV$ term can be a bit confusing at first. You are accustomed to work being $Fdx$. With a little thought, you can recognize pp as the force per unit volume, and the ratio of $dV$ and $dx$ as the area. The minus sign comes from the fact that a positive pressure pushes outwards. What is this $đQ$? As it turns out, we can define a state function $S$ called entropy and so long as a process is done reversibly $$đQ = TdS \text{ only when change is quasistatic}$$ so we find out that $$dU = TdS - pdV$$

The fact that the $T$ in this equation is actually the physical temperature measured by our thermometers was originally an observation based on experiment. At this point, the entropy $S$ is just some weird heat-related state function.

If you decide to get a thermodynamics tattoo, my recommendation would be to choose the thermodynamic identity

$$dU = TdS - pdV$$

It is far and away the most fundamental and essential equation, and one which you will need to come back to again and again. It contains hidden within it (if you remember the First Law) the thermodynamic definition of entropy.

### Activity: Name the Experiment I

Link to Name the Experiment I Activity

**Activity Highlights**

- This small group activity is designed to help students connect thermodynamic derivatives with experiments that measure those quantities.
- Students design an experiment to measure a given partial derivative.
- The compare-and-contrast wrap-up discussion highlights the significance of which variable is held constant in a partial derivative.

### Lecture: Heat Capacity (10 minutes)

- This lecture should bring in points that were previously covered in a discussion of the Dulong and Petit Rule.
- It should be noted that although the heat capacity depends on many different factors, the values should all be reasonably close to the value given by Dulong and Petit

##### Lecture notes from Dr. Roundy's 2014 course website:

As we learned last week, **heat capacity** is amount of energy required to raise the temperature of an object by a small amount. $$C \sim \frac{đ Q}{\partial T}$$ $$đ Q = C dT \text{ At constant what?}$$ If we hold the volume constant, then we can see from the first law that $$dU = đQ - pdV$$ since $dV=0$ for a constant-volume process, $\newcommand\myderiv[3]{\left(\frac{\partial #1}{\partial #2}\right)_{#3}}$ $$C_V = \myderiv{U}{T}{V}$$ But we didn't measure $C_V$ on Monday, since we didn't hold the volume of the water constant. Instead we measured $C_p$, but what is that? To distinguish between different sorts of heat capacities, we need to specify the sort of path used. So, for instance, we could write $$đQ=Tds$$ $$đQ=C_αdT+?dα$$ $$TdS=C_αdT+?dα$$ $$dS= \frac{C_\alpha}{T} dT + \frac{?}{T}d\alpha$$ $$C_\alpha = T \myderiv{S}{T}{\alpha}$$

This may look like an overly-tricky derivative, so let's go through the first law and check that we got it right in a few cases. I'll do the $C_V$ case. We already know that $$dU=đQ-pdV$$ $$C_V= \myderiv{U}{T}{V}$$ $$= \myderiv{U}{S}{V} \myderiv{S}{T}{V}$$ $$= T \myderiv{S}{T}{V}$$ where the second step just uses the ordinary chain rule.

## Homework

- (FirstLaw) PRACTICE
*What goes here?*You heat an insulated piston with a resistor. You run 5 A through the resistor at 10 V for a total of 10 seconds. The pressure is fixed at 1 Pa (which is 1 N/m$^2$). If the system expands by 0.1 m$^3$, what is the change in internal energy of the system?

You heat an insulated piston with a resistor. You run 5 A through the resistor at 5 V for a total of 10 seconds. The pressure is fixed at 1 Pa (which is 1 N/m$^2$). If the system expands by 0.5 m$^3$, what is the change in internal energy of the system?

Consider and insulated cylinder full of an ideal gas, whose internal energy is given by \[U=\frac{3}{2} N k_B T\] What happens to the temperature of the gas when I compress the insulated piston? Why?

- (BottleInBottle)
*What goes here?*The internal energy of helium gas at temperature $T$ is to a very good approximation given by \begin{align} U &= \frac32 Nk_BT \end{align} Consider a very irreversible process in which a small bottle of helium is placed inside a large bottle, which otherwise contains vacuum. The inner bottle contains a slow leak, so that the helium leaks into the outer bottle. The inner bottle contains one tenth the volume of the outer bottle, which is insulated. What is the change in temperature when this process is complete? How much of the helium will remain in the small bottle? \begin{center} \includegraphics[width=3in]{../figs/bottle-in-bottle} \end{center}

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