Re: Text omission = lost pedagogical opportunity

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Bob Sciamanda asks how we would conclude the ideal gas law from Boyle's,
Charles', and Gay-Lussac's laws:

> Given for a fixed quantity of ideal gas:
> 1) at constant T,    PV = a constant
> 2) at constant P,   V/T = a constant and
> 3) at constant V,   P/T = a constant    (Obviously, P is absolute
> pressure and T is absolute temperature - all quantities are positive),
> how does one deduce that PV = CT for all states, where C is independent
> of the variables P,V,T?
>
> There is more than one way.  How do you do it?  Above, I refer to a
> particular argument which serves a wider pedagogical purpose.

The method I have in mind is useful in that similar reasoning is used in
later courses when it comes time to introduce the students to the method
of separation of variables for solving homogeneous linear PDEs.

Now 1) implies PV = f(T) (1'), 2) implies V/T = g(P) (2'), and 3) implies
p/T = h(V) (3') for some functions f, g and h, each of only one argument.
Solving 2') for V and substituting into 1'), and solving 3') for P and
substituting into 1') results in PV = f(T) = PTg(P) = TVh(V).  Now divide
all four sides of this triple equality by T.  The result is:
PV/T = f(T)/T = Pg(P) = Vg(V).  Now each of the 2nd, 3rd, and 4th sides of
this triple equality each depend on a single variable, but in each case
that variable is a different one.  The only way for a function of one
variable to be identically equal to another function of a different
variable is for both of these functions to be the same constant function of
its own argument variable.  This is seen by varying one of the variables on
one side of the equation while holding the other variable constant.  The
side whose variable is held constant must *be* constant.  But this constant
is then equal to the 'supposedly' variable function of the other variable.
This function is this thus a constant function.  We repeat this argument
for the other variable on the other side of the equation and conclude that
both functions are the same constant function.

By applying this reasoning pairwise to the various sides of our equality:
f(T)/T = Pg(P) = Vg(V) we conclude that all three expression are the same
constant function C. Substitution of this constant back into our above
equation for PV/T gives PV/T = C.  Multiplying this equation by T gives
the usual form of PV = CT.                                             QED

David Bowman
dbowman@gtc.georgetown.ky.us