Question

Derive the ideal gas law from a particle moving with a velocity v_x a rectangular box with dimensions L_x, L_yand L_z.

a). Draw a picture of the system

b). Drive the gas law.

Answer 1

Part a

Lz

Part b

Consider the x direction, the particle travels in a straight line along the x direction with a velocity Vx and collodes with the wall. The time between collision is,

$t=\frac{2L_x}{v_x}$

The momentum change of the particle when it collides with the container is,

$\Delta p=mv_x-(-mv_x)=2mv_x$

Force is defined as the rate of chage of momentum and can be given by,

$F=\frac{\Delta p}{t}=\frac{2mv_x*v_x}{2L_x}=\frac{mv_x^{2}}{L_x}$

The pressure exerted by this molecule on the wall of the container is,

$P=\frac{F}{A}=\frac{\frac{mv_x^{2}}{L_x}}{L_y*L_z}=\frac{mv_x^{2}}{L_x*L_y*L_z}=\frac{mv_x^{2}}{V}$

If N molecules are travelling in then x direction with velocities, Vx1, Vx2,... we have,

$P=\frac{m}{V}[v_{x1}^{2}+v_{x2}^{2}+v_{x3}^{2}+...+v_{xN}^{2}]$

This can be replaced by the mean square speed,

$P=\frac{Nm\overline{v_x^{2}}}{V}$

This is for the x direction only. A similar derrivation can be done for the y and z directions. As the pressure exerted on each wall of a conatiner is equal, the overall pressure is,

$P=\frac{Nm\overline{v^{2}}}{3V}$

where;

$\overline{v^{2}}=\overline{v_x^{2}}+\overline{v_y^{2}}+\overline{v_z^{2}}$

The kinetic energy is given by,

$E=\frac{m\overline{v^{2}}}{2}$

Therefore,

$2E=m\overline{v^{2}}$

We substitute this into our pressure equation,

$P=\frac{Nm\overline{v^{2}}}{3V}=\frac{2NE}{3V}$

we can replace the term 2/3*E by kT, where k is a constant,

$P=\frac{NkT}{V}$

The constant k(boltzmann constant) can be replaced by R/Na, where Na is the avogardro no. and r is the universal gas constant,

$P=\frac{NRT}{N_aV}$

Now, N/Na is no of moles (n),

$P=\frac{nRT}{V}$

which gives us,

$PV=nRT$