Applications
Now let us discuss the meaning of Eq. (45.7) and see why it answers the questions which we proposed in our last chapter. We considered the following problem: in kinetic theory it is obvious that an increase in temperature leads to an increase in pressure, because of the bombardments of the atoms on a piston. For the same physical reason, when we let the piston move back, heat is taken out of the gas and, in order to keep the temperature constant, heat will have to be put back in. The gas cools when it expands, and the pressure rises when it is heated. There must be some connection between these two phenomena, and this connection is given explicitly in Eq. (45.7). If we hold the volume fixed and increase the temperature, the pressure rises at a rate $(\ddpl{P}{T})_V$. Related to that fact is this: if we increase the volume, the gas will cool unless we pour some heat in to maintain the temperature constant, and $(\ddpl{U}{V})_T$ tells us the amount of heat needed to maintain the temperature. Equation (45.7) expresses the fundamental interrelationship between these two effects. That is what we promised we would find when we came to the laws of thermodynamics. Without knowing the internal mechanism of the gas, and knowing only that we cannot make perpetual motion of the second type, we can deduce the relationship between the amount of heat needed to maintain a constant temperature when the gas expands, and the pressure change when the gas is heated at constant volume! Now that we have the result we wanted for a gas, let us consider the rubber band. When we stretch a rubber band, we find that its temperature rises, and when we heat a rubber band, we find that it pulls itself in. What is the equation that gives the same relation for a rubber band as Eq. (45.3) gives for gas? For a rubber band the situation will be something like this: when heat $\Delta Q$ is put in, the internal energy is changed by $\Delta U$ and some work is done. The only difference will be that the work done by the rubber band is $-F\,\Delta L$ instead of $P\,\Delta V$, where $F$ is the force on the band, and $L$ is the length of the band. The force $F$ is a function of temperature and of length of the band. Replacing $P\,\Delta V$ in Eq. (45.3) by $-F\,\Delta L$, we get \begin{equation} \label{Eq:I:45:8} \Delta U = \Delta Q + F\,\Delta L. \end{equation} Comparing Eqs. (45.3) and (45.8), we see that the rubber band equation is obtained by a mere substitution of one letter for another. Furthermore, if we substitute $L$ for $V$, and $-F$ for $P$, all of our discussion of the Carnot cycle applies to the rubber band. We can immediately deduce, for instance, that the heat $\Delta Q$ needed to change the length by $\Delta L$ is given by the analog to Eq. (45.5): $\Delta Q = -T(\ddpl{F}{T})_L\,\Delta L$. This equation tells us that if we keep the length of a rubber band fixed and heat the band, we can calculate how much the force will increase in terms of the heat needed to keep the temperature constant when the band is relaxed a little bit. So we see that the same equation applies to both gas and a rubber band. In fact, if one can write $\Delta U = \Delta Q + A\,\Delta B$, where $A$ and $B$ represent different quantities, force and length, pressure and volume, etc., one can apply the results obtained for a gas by substituting $A$ and $B$ for $-P$ and $V$. For example, consider the electric potential difference, or “voltage,” $E$ in a battery and the charge $\Delta Z$ that moves through the battery. We know that the work done in a reversible electric cell, like a storage battery, is $E\,\Delta Z$. (Since we include no $P\,\Delta V$ term in the work, we require that our battery maintain a constant volume.) Let us see what thermodynamics can tell us about the performance of a battery. If we substitute $E$ for $P$ and $Z$ for $V$ in Eq. (45.6), we obtain \begin{equation} \label{Eq:I:45:9} \frac{\Delta U}{\Delta Z} = T\biggl(\ddp{E}{T}\biggr)_Z - E. \end{equation} Equation (45.9) says that the internal energy $U$ is changed when a charge $\Delta Z$ moves through the cell. Why is $\Delta U/\Delta Z$ not simply the voltage $E$ of the battery? (The answer is that a real battery gets warm when charge moves through the cell. The internal energy of the battery is changed, first, because the battery did some work on the outside circuit, and second, because the battery is heated.) The remarkable thing is that the second part can again be expressed in terms of the way in which the battery voltage changes with temperature. Incidentally, when the charge moves through the cell, chemical reactions occur, and Eq. (45.9) suggests a nifty way of measuring the amount of energy required to produce a chemical reaction. All we need to do is construct a cell that works on the reaction, measure the voltage, and measure how much the voltage changes with temperature when we draw no charge from the battery! Now we have assumed that the volume of the battery can be maintained constant, since we have omitted the $P\,\Delta V$ term when we set the work done by the battery equal to $E\,\Delta Z$. It turns out that it is technically quite difficult to keep the volume constant. It is much easier to keep the cell at constant atmospheric pressure. For that reason, the chemists do not like any of the equations we have written above: they prefer equations which describe performance under constant pressure. We chose at the beginning of this chapter to use $V$ and $T$ as independent variables. The chemists prefer $P$ and $T$, and we will now consider how the results we have obtained so far can be transformed into the chemists’ system of variables. Remember that in the following treatment confusion can easily set in because we are shifting gears from $T$ and $V$ to $T$ and $P$. We started in Eq. (45.3) with $\Delta U = \Delta Q - P\,\Delta V$; $P\,\Delta V$ may be replaced by $E\,\Delta Z$ or $A\,\Delta B$. If we could somehow replace the last term, $P\,\Delta V$, by $V\,\Delta P$, then we would have interchanged $V$ and $P$, and the chemists would be happy. Well, a clever man noticed that the differential of the product $PV$ is $d(PV) = P\,dV + V\,dP$, and if he added this equality to Eq. (45.3), he obtained \begin{alignat*}{2} \Delta(PV) &\,= P\,\Delta V &&+ V\,\Delta P\\[.65ex] \Delta U &\,= \Delta Q &&- P\,\Delta V\\ \hline \Delta(U + PV) &\,= \Delta Q &&+ V\,\Delta P \end{alignat*} In order that the result look like Eq. (45.3), we define $U + PV$ to be something new, called the enthalpy, $H$, and we write $\Delta H = \Delta Q + V\,\Delta P$. Now we are ready to transform our results into chemists’ language with the following rules: $U \to H$, $P \to -V$, $V \to P$. For example, the fundamental relationship that chemists would use instead of Eq. (45.7) is \begin{equation*} \biggl(\ddp{H}{P}\biggr)_T = -T\biggl(\ddp{V}{T}\biggr)_P + V. \end{equation*} It should now be clear how one transforms to the chemists’ variables $T$ and $P$. We now go back to our original variables: for the remainder of this chapter, $T$ and $V$ are the independent variables. Now let us apply the results we have obtained to a number of physical situations. Consider first the ideal gas. From kinetic theory we know that the internal energy of a gas depends only on the motion of the molecules and the number of molecules. The internal energy depends on $T$, but not on $V$. If we change $V$, but keep $T$ constant, $U$ is not changed. Therefore $(\ddpl{U}{V})_T = 0$, and Eq. (45.7) tells us that for an ideal gas \begin{equation} \label{Eq:I:45:10} T\biggl(\ddp{P}{T}\biggr)_V - P = 0. \end{equation} Equation (45.10) is a differential equation that can tell us something about $P$. We take account of the partial derivatives in the following way: Since the partial derivative is at constant $V$, we will replace the partial derivative by an ordinary derivative and write explicitly, to remind us, “constant $V$.” Equation (45.10) then becomes \begin{equation} \label{Eq:I:45:11} T\,\frac{\Delta P}{\Delta T} - P = 0;\quad \text{const $V$}, \end{equation} which we can integrate to get \begin{alignat}{2} \ln P &\;= \ln T + \text{const};\quad &&\text{const $V$},\notag\\[1ex] \label{Eq:I:45:12} P &\;= \text{const}\times T;\quad &&\text{const $V$}. \end{alignat} We know that for an ideal gas the pressure per mole is equal to \begin{equation} \label{Eq:I:45:13} P = \frac{RT}{V}, \end{equation} which is consistent with (45.12), since $V$ and $R$ are constants. Why did we bother to go through this calculation if we already knew the results? Because we have been using two independent definitions of temperature! At one stage we assumed that the kinetic energy of the molecules was proportional to the temperature, an assumption that defines one scale of temperature which we will call the ideal gas scale. The $T$ in Eq. (45.13) is based on the gas scale. We also call temperatures measured on the gas scale kinetic temperatures. Later, we defined the temperature in a second way which was completely independent of any substance. From arguments based on the Second Law we defined what we might call the “grand thermodynamic absolute temperature” $T$, the $T$ that appears in Eq. (45.12). What we proved here is that the pressure of an ideal gas (defined as one for which the internal energy does not depend on the volume) is proportional to the grand thermodynamic absolute temperature. We also know that the pressure is proportional to the temperature measured on the gas scale. Therefore we can deduce that the kinetic temperature is proportional to the “grand thermodynamic absolute temperature.” That means, of course, that if we were sensible we could make two scales agree. In this instance, at least, the two scales have been chosen so that they coincide; the proportionality constant has been chosen to be $1$. Most of the time man chooses trouble for himself, but in this case he made them equal!