The document derives the algebraic model of the ideal gas equation by approximating the random motion of gas molecules inside a cube box. It considers the force exerted on the walls of the box by a single molecule with mass and velocity. Using Newton's second law, it calculates the pressure exerted by a single molecule and then extrapolates this to the total pressure from all molecules. Through further rearrangement, it relates the pressure-volume relationship to the kinetic energy and temperature of the gas.
2. Imagine the
molecules of a gas
bouncing around
randomly inside a
cube box (x^3)
Now let us
approximate this
motion so that 1/3
of the molecules are
moving:
• From side to side
• Up and down
• Back to front
1/3
1/3
1/3
x
x
x
3. Now consider one of
these molecules with
mass ‘m’ and velocity
‘c’.
What is the force that
this molecule exerts on
the grey wall?
m
c
4. From Newton’s second law …
F = ma
= m.dc/dt
= d(mc)/dt
= dp/dt where ‘p’ = momentum
dp = mc – (-mc)
= 2mc
dt = dd/dc
= 2x/c
m
c
m
-c
5. So…
Dp = 2mc
Dt = 2x/c
F = dp/dt = 2mc = mc^2/x
2x/c
The pressure ‘p’ that the molecule exerts on the
wall…
P = F/A = mc^2/x = mc^2/x^3 = mc^2/V
x^2
6. The pressure of all the molecules (N) on THIS wall
…
P = N/3 x mc^2/V
(This is the same pressure which the other
molecules exert on the other walls).
Rearranging …
Pv = 1/3 x Nmc^2
7. Further rearranging …
3/2 x pv = N x ½mc^2
3/2 x pv = Total kinetic energy of the gas
Since pv = nRT …
3/2 x (nR/N) x T = ½mc^2
= 3/2 x (nR/N) x T
= 3/2 x (R/Na) x T
= 3/2 x kT
