Kinetic Theory

According to KTG, the gas exerts pressure because its molecule :

Suffer change in momentum when impinge on the walls of container.

$V_{rms}=\sqrt[]{\frac{3RT}{M}}$

& $v_P=\sqrt[]{\frac{2RT}{M}}$

$v_{rms}=\sqrt[]{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}_P$

$\mathrm{T}\mathrm{i}=-{50}^{?}\mathrm{C}$

$\begin{array}{rr}& =223\mathrm{K}\\ & {\mathrm{V}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\propto \sqrt[]{\mathrm{T}}\end{array}$

As $\begin{array}{rr}& =223\mathrm{K}\\ & {\mathrm{V}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\propto \sqrt[]{\mathrm{T}}\end{array}$  increased by 3 times

So ${\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\mathrm{f}}=4{\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\text{initial}}$

${\mathrm{T}}_{\mathrm{f}}=16{\mathrm{T}}_{\mathrm{i}}$

$=3568\mathrm{K}$

${\mathrm{T}}_{\mathrm{f}}=(3568-273{)}^{?}\mathrm{C}$

$={3295}^{?}\mathrm{C}$

Root mean square speed of gas molecules

${\nu }_{\text{rms}}=\sqrt[]{\frac{3RT}{M}}$

Pressure exerted by ideal Gas

$\mathrm{P}=\frac{1}{3}\rho v_{\mathrm{r}\mathrm{m}\mathrm{s}}^2$

$\mathrm{P}=\frac{1}{3}\mathrm{m}\mathrm{n}{\nu }^2$

$\rho =\mathrm{m}\mathrm{n},{\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}^2={\nu }^{-2}$

Average kinetic energy of a molecular

$\mathrm{K}\mathrm{E}=\frac{3}{2}\mathrm{K}\mathrm{T}$

Total internal energy of 1 mole of a diatomic gas

$\mathrm{U}=\frac{\mathrm{f}}{2}\mu \mathrm{R}\mathrm{T}$

$\mathrm{U}=\frac{5}{2}\mathrm{R}\mathrm{T}$ (For 1 mole diatomic gas)

$\stackrel{?}{\tau }=\stackrel{?}{\mathrm{P}}*\stackrel{?}{\mathrm{E}}$

$\left|\stackrel{?}{\tau }\right|=\mathrm{P}\mathrm{E}\mathrm{s}\mathrm{i}\mathrm{n}?\theta \mathrm{}\Rightarrow 4=\mathrm{q}*2\mathrm{a}*\mathrm{E}\mathrm{s}\mathrm{i}\mathrm{n}?{30}^{?}$

$\mathrm{q}=\frac{4}{\left(2*{10}^{-2}\right)*2*{10}^5*\left(\frac{1}{2}\right)}$

$=2*{10}^{-3}\mathrm{C}$

$=2\mathrm{m}\mathrm{C}$

v_rms = √ (3RT/M) and v_av = √ (8RT/πM)

⇒ v_rms / v_av = √ [ (3RT/M) / (8RT/πM)] = √ (3π/8)

According to principle of equi-partition of Energy, the average energy per molecules associated with each degree of freedom is (1/2)kT.

Use the equation to find ' N ':
P = 2/3 (v) (N/V) (1/2 mu²)

Mean relaxation time T = λ/V? = λ / √ (8RT/πm)
∴ T ∝ 1/√T

Frequency of collisions = 1/T = Vavg/λ = 600/ (3 * 10? ) = 2 * 10? sec? ¹