Deduce (a) Boyle’s law (b) Charles law (c) Graham’s law (d) Dalton’s law from kinetic gas equation.

Boyles’s law : According to kinetic gas equation
$\begin{array}{l}\mathrm{PV}=\frac{1}{3}{\mathrm{mnU}}_{\mathrm{rms}}^2\text{ (or) }\mathrm{PV}=\frac{2}{3}\left(\frac{1}{2}{\mathrm{mnU}}_{\mathrm{rms}}^2\right)\text{ (or) }\\ \mathrm{PV}=\frac{2}{3}{\mathrm{E}}_{\mathrm{K}}\dots .\text{ (1) }\end{array}$
But KE is directly proportional to the absolute temperature
${\mathrm{E}}_{\mathrm{K}}\propto \mathrm{T}\text{ (or) }{\mathrm{E}}_{\mathrm{K}}=\mathrm{K}.\mathrm{T} -----\left(2\right)$
From (1) and (2)
$\mathrm{PV}=\frac{2}{3}\mathrm{KT} -----\left(3\right)$
At constant temperature, PV=constant
This is Boyle’s law
Charles law :
From equation (3)
$\mathrm{PV}=\frac{2}{3}\mathrm{KT};\mathrm{V}=\frac{2}{3}\frac{\mathrm{KT}}{\mathrm{P}}$
At constant P, V = constant × T (or) V $\propto$ T
This is Charles law
Graham’s law :
According to kinetic gas equation
$\mathrm{PV}=\frac{1}{3}{\mathrm{mnU}}_{\mathrm{rms}}^2$
For one mole of gas n = N
$\therefore \mathrm{mN}=\mathrm{M}$
If another gas is present in the same vessel instead of first gas, the pressure of that gas according to kinetic gas equation.
$\begin{array}{l}\mathrm{PV}=\frac{1}{3}{\mathrm{MU}}_{\mathrm{rms}}^2\therefore {\mathrm{U}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{PV}}{\mathrm{M}}}\left(\therefore \mathrm{d}=\frac{\mathrm{M}}{\mathrm{V}}\right)\\ {\mathrm{U}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{P}}{\mathrm{d}}}\end{array}$
At constant pressure
${\mathrm{U}}_{\mathrm{rms}}=\frac{\text{ constant }}{\sqrt{\mathrm{d}}}\left(\text{ or }\right){\mathrm{U}}_{\mathrm{rms}}\propto \frac{1}{\sqrt{\mathrm{d}}}$
This is Graham’s law.
Dalton’s law:
According to kinetic gas equation
$\mathrm{PV}=\frac{1}{3}\mathrm{m}\cdot \mathrm{n}\cdot {\mathrm{U}}_{\mathrm{rms}}^2$
n₁ molecules of a gas is present in a vessel of volume ‘V’. If mass of each molecules is m₁ and U_rms is the RMS velocity. The pressure of the gas according to kinetic gas equation,
${\mathrm{P}}_1=\frac{1}{3}\cdot \frac{{\mathrm{m}}_1{\mathrm{n}}_1{\mathrm{U}}_{{\mathrm{rms}}_1}^2}{\mathrm{V}};{\mathrm{P}}_2=\frac{1}{3}\cdot \frac{{\mathrm{m}}_2{\mathrm{n}}_2{\mathrm{U}}_{{\mathrm{rms}}_2}^2}{\mathrm{V}}$
If the mixture of these two gases are taken in the same vessel of volume ‘V’, the total pressure of the mixture is
$\mathrm{P}=\frac{1}{3}\cdot \frac{{\mathrm{m}}_1{\mathrm{n}}_1{\mathrm{U}}_{{\mathrm{rms}}_1}^2}{\mathrm{V}}+\frac{1}{3}\cdot \frac{{\mathrm{m}}_2{\mathrm{n}}_2{\mathrm{U}}_{\mathrm{rms}}^2}{\mathrm{V}}$
$\mathrm{P}={\mathrm{P}}_1+{\mathrm{P}}_2$
This is Dalton’s law of partial pressures.