Connecting Thermal Derivatives with Experiment

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.

Link to Name the Experiment I Activity

Activity Highlights

• 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

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.