** **

Consider a cube of side L containing a molecule of a gas with momentum mv_{x}.
Each time it hits and recoils from a wall, it imparts an impulse, i = 2 * mv_{x}.
It reapeats this with a frequency, f = v_{x}/(2*L) so the force on the
wall, F = f * i = mv_{x}^{2}/L. For a number [N] of molecules,
the force is F = Nmv_{x}^{2}/L. Now since the molecule may move
in any one of three directions the average velocity [v^{2}] = 3 * v_{x}^{2}.
Thus, the pressure on any one wall [P] = F/A = 1/3 Nmv^{2}/V and PV =
1/3 Nmv^{2}.

Since the kinetic energy [Ek] = 1/2 Nmv^{2}, PV = 2/3 Ek and Ek = 3/2
PV.

And since PV = nRT = NkT, E = 3/2 NkT and T = 2/3 E / (Nk)

Note: R = 8.31 J/mol-K and k = R/6*10^{23} = 1.38 * 10-^{23}
J/molecule-K

**Isothermal Expansion**

For isothermal (constant temperature) processes, consider PV = NkT. Since T
is constant, P = NkT / V, and pressure is directly inverse to volume.

**Adiabatic Expansion**

For adiabatic (no heat flows in or out to the control volume) expansion,
temperature is free to vary. Temperature generally falls as volume rises and as
pressure falls. In particular, PV^{g} = K where g = cp/cv. For diatomic
gasses such as nitrogen and oxygen g = 1.4. Therefore P = P_{0}*(V_{0}
/ V)^{1.4} and pressure drops a little faster than it would if the gas
were free to absorb heat and maintain constant temperature.

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References: __Teaching Introductory Physics__, Swartz, 1998, AIP Press,
pages 236 and 239.