Recall from the previous post that the First Law of Thermodynamics can be expressed
\[ \alpha+\beta = -dU \]
where \(U\) is the total energy of a thermodynamic system and \(\alpha,\beta\) are one-forms whose integrals along a transformation express the work done by the system on its environment, and the heat supplied by the system to its environment, in the course of the transformation. (In Fermi’s notation \(\alpha=dL\), \(\beta=-dQ\), where \(L,Q\) should be thought of as functions not on the state space but on the path space of the state space – their values depend on how you got there). If the state space is a 2-manifold parametrized by volume \(V\) and pressure \(p\), then \(\alpha=pdV\). The thermal capacity of the system is the derivative \(dQ/dT\), that is the marginal amount of heat absorbed for an increase in temperature. There are two versions of the heat capacity, \(C_V\) (heat capacity at constant volume) and \(C_p\) (heat capacity at constant pressure). For the situation we’re considering, \(\alpha=pdV = 0\) at constant volume, so
\[ C_V = \left(\frac{\partial U}{\partial T}\right)_V,\qquad C_p = \left(\frac{\partial U}{\partial T}\right)_p + p\left(\frac{\partial V}{\partial T}\right)_p \]
by standard calculations with partial derivatives.
In particular we can consider ideal gases, which obey the equation of state \(pV = kRT\).
Joule Axiom For an ideal gas, the total energy \(U\) is a function of the temperature \(T\) only.
This was experimentally verified by Joule (he allowed a fixed quantity of gas to expand into a vacuum, while the whole set-up was contained in a calorimeter with which it was initially in thermal equilibrium; he observed only a small change in the temperature of the water in the calorimeter) and can apparently also be seen as a consequence of the second law (we’ll get to that, I hope). Moreover, it is experimentally found that the specific heat at constant volume \(C_V\) is more or less independent of the temperature; thus \(U=C_VT + \text{const}\), where we can set the constant to zero by adopting an absolute scale of temperature. From the ideal gas law, \( p (\partial V/\partial T)_p = kR \) and so from the displayed equation above \(C_p=C_V + kR\).
“It can be shown by an application of the kinetic theory” (says Fermi) that \(C_V= {\frac32}k R\) for a monatomic gas and \(C_V={\frac52}kR\) for a diatomic gas. (I have no idea how this is done, nor whether “diatomic” includes a compound like \(HF\) or only a symmetrical molecule like \(O_2\). Something to follow up on later.) Anyhow, based on this one can calculate the constant that Fermi denotes \(K\) which is the ratio \(C_p/C_V\) of the heat capacities at constant volume and constant pressure.: \(\frac53\) for a monatomic gas and \(\frac75\) for a diatomic.
Now recall the notion of an adiabatic transformation: one during which the system exchanges no heat with its environment. That is, an adiabatic transformation is one on which \(\beta\) restricts to zero. This gives a differential equation \(C_VdT+pdV=0\) along an adiabatic. But from the equation of state, \(kRdT = pdV+Vdp\). Substituting this and manipulating a bit we get
\[ KpdV +Vdp = 0\]
where \(K\) is the ratio of heat capacities as above, and therefore (integrating) the equation of an adiabatic in the \(p,V\) plane as \(pV^K=\text{constant}\). Compare with the equation \(pV=\text{constant}\) for an isothermal transformation of an ideal gas.
Example: important example in atmospheric physics is the computation of the lapse rate, that is the rate of decrease of atmospheric temperature with altitude. One can consider a rising parcel of air as expanding adiabatically (since air is a poor conductor of heat). As it expands it does work on its surroundings and must therefore cool. Let \(h\) denote height. Then
\[ \frac{dp}{dh} = – \rho g = – \frac{gM}{R}\frac{p}{T}, \]
where \(M\) is the molecular mass (average value for air about 29) and the expression for the density \(\rho\) comes from the equation of state. On the other hand, from the equation for an adiabatic,
\[ \frac{dT}{dp} = \left(\frac{K-1}{K}\right)\frac{T}{p}. \]
Multiplying these and using the chain rule
\[ \frac{dT}{dh} = – \frac{K-1}{K} \frac{gM}{R}. \]
Taking the atmosphere to be mostly diatomic so that \(K={\frac75}\) this gives a value of about \(-9.8\) degrees per kilometer (according to Fermi). The actual value is somewhat less – the condensation of water vapor is an additional effect which releases some energy and has been neglected here.