Notation for finite and infinitesimal changes

After marking last year's exam, I realised that some students have a problem with the manipulation of finite and infinitesimal changes. I hope the following helps.

If we are dealing with functions of state, there are three ways they could enter:

• As totals, for instance in the following expressions for a classical ideal gas
  
  $$PV=nRT;\qquad E={3\over 2}nRT; \qquad S=nR\left(\ln V+{\textstyle{3\over 2}}\ln E\right)+S_0.$$
  
  

• As differentials, for instance in the fundamental thermodynamic relation,
  
  $${\rm d}E= T {\rm d}S -P {\rm d}V$$
  
  

• As finite changes, for instance for a process at constant volume with a heat capacity which is independent of temperature
  
  $$\Delta S\equiv S_2-S_1= C \ln{T_2\over T_1}$$
  
  

There are two basic rules:

• If one term in an equation involves an infinitesimal, all the other terms must also contain one (and only one!).
• If finite changes are involved, all terms will have reference to the initial and final states.

Introducing things like heat and work which aren't functions of state complicates things slightly; the first rule still holds (though we write ${}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}Q$ and ${}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}W$). However changes in functions of state can be related to amounts of heat and work, as in $\Delta E=Q+W$.

Another common mistake is to assume that if something involving a differential holds, the same relation will hold for finite changes. This is not generally true. For instance for an isothermal expansion of an ideal gas $E$ is constant, so

$$T{\rm d}S=P{\rm d}V \quad\hbox{and}\quad T\Delta S=\int_{V_1}^{V_2} P{\rm d}V \quad\hbox{but}\quad T\Delta S\ne P\Delta V.$$

Why not? Well $T$ is constant so $T{\rm d}S$ can be integrated to $T\Delta S$, but $P$ certainly isn't constant so $\int P{\rm d}V\ne P \int {\rm d}V$. What we can do here is substitute $P=nRT/V$, giving $T\Delta S=nR \ln(V_2/V_1)$.

Similarly,

$${\rm d}S= {{}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}Q^{\rm rev}\over T} \not\Rightarrow \Delta S={Q^{\rm rev}\over T}$$

unless T is constant. However if the heat capacity is constant, as is a reasonably good approximation for most solids and gases at STP, we can write ${}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}Q=C{\rm d}T$ and integrate to get $\Delta S= C \ln(T_2/ T_1)$.