For kinetic interpretation of temperature, see text
Numerical:
We know, KE. of one molecule of a perfect gas at temperature T is given by
\(E=\frac { 2 }{ 3 } kT\)
\(\\ \therefore \) KE. of \(\mu _{ 1 }\) molecules of a perfect gas at temperature T₁, E₁ = \(\left( \frac { 3 }{ 2 } k{ T }_{ 1 } \right) { \mu }_{ 1 }\)
KE. of \(\mu _{ 2 }\) molecules of a perfect gas at temperature T2, E₂ = \(\left( \frac { 3 }{ 2 } k{ T }_{ 2 } \right) { \mu }_{ 2 }\)
When both the gases are mixed, then the total KE. of the mixture is
\(E=\frac { 3 }{ 2 } k({ \mu }_{ 1 }{ T }_{ 1 }+{ \mu }_{ 2 }{ T }_{ 2 })\)
After mixing, the temperature of the mixture is T, therefore KE. of the mixture is given by
\(E'=\frac { 3 }{ 2 } kT({ \mu }_{ 1 })+\frac { 3 }{ 2 } k{ T }({ \mu }_{ 2 })\)
\(\\ =\frac { 3 }{ 2 } kT({ \mu }_{ 1 }+{ \mu }_{ 2 })\)
Since there is no loss of energy
\(\therefore E'=E\)
\(\\ or\ \frac { 3 }{ 2 } kT({ \mu }_{ 1 }+{ \mu }_{ 2 })=\frac { 3 }{ 2 } k({ \mu }_{ 1 }{ T }_{ 1 }+{ \mu }_{ 2 }{ T }_{ 2 })\)
\(\\ \therefore \ { T=\frac { { \mu }_{ 1 }{ T }_{ 1 }+{ \mu }_{ 2 }{ T }_{ 2 } }{ { \mu }_{ 1 }+{ \mu }_{ 2 } } } \) 

The entire structure of the kinetic theory of gases is based on the following assumptions which were first stated by Classius.
1. A gas consists of a very large number of molecules (of the order of Avogadro's number, 10²³), which are perfect elastic spheres. They are identical in all respects for a given gas and are different for different gases.
2. The molecules of a gas are in a state of incessant random motion. They move in all directions with different speeds, (of the order of 500 m/s) and obey Newton's laws of motion.
3. The size of the gas molecules is very small as compared to the distance between them. If typical size of a molecule is 2 \(\dot { A } \) .average distance between the molecules is \(\ge \)20 A. Hence volume occupied by the molecules is negligible in comparison to the volume of the gas.
4. The molecules do not exert any force of attraction or repulsion on each other, except during collision.
5. The collisions of the molecules with themselves and with the walls of the vessel are perfectly elastic. As such the momentum and the kinetic energy of the molecules are conserved during collisions, though their velocities change.
6. There is no concentration of the molecules at any point inside the container i.e., molecular density is uniform throughout the gas.
7. A molecule moves along a straight line between two successive collisions and the average straight distance covered between two successive collisions is called the mean free path of the molecules
8. The collisions are almost instantaneous, i.e., the time of collision of two molecules is negligible as compared to time interval between two successive collisions.
Numerical:
We know that
\(P=\frac { 1 }{ 3 } \rho { v }^{ 2 }\)
\(\\ \therefore { v }_{ rms }=\sqrt { ({ v }^{ 2 }) } =\left( \frac { 3p }{ \rho } \right) \)
Given that P = 1,0 x 10⁵  newton / meter² and
\(\rho =1.98\ kg/meter^{ 3 }\)
\(\\ \therefore \ { v }_{ rms }=\sqrt { \left[ \left\{ \frac { 3\times (1.0\times { 10 }^{ 5 } }{ 1.98 } \right\} \right] } =389\ meter/sec\)
From kinetic theory of gases, the root mean square speed is directly proportional to the square root of absolute temperature
\({ v }_{ rms }\propto \sqrt { T } \)
\(\\ \therefore \ \frac { { (v }_{ rms })_{ 30 } }{ { (v }_{ rms })_{ 0 } } =\sqrt { \left[ \left( \frac { 273+30 }{ 273+0 } \right) \right] } =\sqrt { \left[ \left( \frac { 303 }{ 273 } \right) \right] } =1.053\)
\(\\ or\ { (v }_{ rms })_{ 30 }={ (v }_{ rms })_{ 0 }\times 1.053\)
\(\\ =389\times 1.053=410\ meter/sec\)

According to kinetic theory of gases,
\(p=\frac { 1 }{ 3 } { P }^{ { C }^{ 2 } }=\frac { 1 }{ 3 } \frac { M }{ V } { C }^{ 2 }\)
\(=\frac { 1 }{ 3 } \frac { M }{ V } KT\)   
\( T\propto PV\)      \((\because { C }^{ 2 }=KT.\)When k is constant)
Now as T is directly proportional to the product of P and V. If P and V are constant, then T is also constant.

(i) Average speed,
\({ v }_{ av }=\frac { 0+2+3+4+4+4+5+5+6+9 }{ 10 } \)
\(=\frac { 42 }{ 10 } =4.2\quad ms^{ -1 }\)
(ii) R.M.S. Speed,
\({ v }_{ rms }=\left[ \frac { { (0) }^{ 2 }+{ (2) }^{ 2+ }{ (3) }^{ 2 }+{ (4) }^{ 2 }+{ (4) }^{ 2 }+{ (4) }^{ 2 }+{ (5) }^{ 2 }+{ (5) }^{ 2 }+{ (6) }^{ 2 }+{ (9) }^{ 2 } }{ 10 } \right] ^{ 1/2 }\)
\(=\left[ \frac { 228 }{ 10 } \right] ^{ 1/2 }=4.77\quad ms^{ -1 }\)

Here \(\mu =3,{ 1 }=300K\) and for an ideal monoatomic gas
\({ C }_{ v }=\frac { 5 }{ 2 } R\)
As volume of gas is doubled (V₂ = 2V₁) at constant pressure, hence according to Charle's law
\({ T }_{ 2 }=\frac { { T }_{ 1 }{ V }_{ 2 } }{ { V }_{ 1 } } =\frac { 300\times 2V_{ 1 } }{ { V }_{ 1 } } =600\)
\(\therefore \) Gain in internal energy U₂ - U₁=\(\mu \) CV (T₂ - T₁) = 3\(\times \frac { 5 }{ 2 } R\times \)(600 - 300)
= 2250R = 2250 x 8.31J
= 1.87 x 10⁴J

Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed ideal) obeys the gas law. Since V and T are common to the two gases, we have P₁V = µ₁ RT and P₂V = µ2 RT, i.e. (P₁ /P₂ ) = (µ₁ / µ₂ ). Here 1 and 2 refer to neon and oxygen respectively. Since (P₁ /P₂ ) = (3/2) (given), (µ₁ / µ₂ ) = 3/2.
(i) By definition µ₁ = (N₁ /NA ) and µ₂ = (N₂ /NA ) where N₁ and N₂ are the number of molecules of 1 and 2, and NA is the Avogadro’s number. Therefore, (N₁ /N₂ ) = (µ₁ / µ₂ ) = 3/2.
(ii) We can also write µ₁ = (m₁ /M₁ ) and µ₂ = (m₂ /M₂ ) where m1 and m2 are the masses of 1 and 2; and M₁ and M₂ are their molecular masses. (Both m1 and M₁ ; as well as m₂ and M₂ should be expressed in the same units). If ρ₁ and ρ2 are the mass densities of 1 and 2 respectively, we have \( \frac{\rho_1}{\rho_2}=\frac{m_1 / V}{m_2 / V}=\frac{m_1}{m_2}=\frac{\mu_1}{\mu_2} \times\left(\frac{M_1}{M_2}\right) \)
\(=\frac{3}{2} \times \frac{20.2}{32.0}=0.947 \)

We know that
\(\lambda =\frac { kT }{ \sqrt { 2 } \pi { d }^{ 2 }\rho } =\frac { m }{ \sqrt { 2 } \pi { d }^{ 2 }\rho } =\frac { 1 }{ \sqrt { 2 } \pi n{ d }^{ 2 } } \)
\(\lambda \) depends upon:
(i) diameter (d) of the molecule, smaller the 'd', larger is the mean free path \(\lambda \)
(ii) \(\lambda \propto T\) i.e., higher the temperature larger is the \(\lambda \)
(iii) \(\lambda \propto \frac { 1 }{ p } \)i.e., smaller the pressure larger is the \(\lambda \)
(iv) \(\lambda \propto \frac { 1 }{ p } \) i.e., smaller the density (p), larger will be the \(\lambda \)
(v) \(\lambda \propto \frac { 1 }{ p } \) i.e., smaller the number of molecules per unit volume of the gas, larger is the \(\lambda \)

We know that P = \(\frac { 1 }{ 3 } \frac { M }{ V } { C }^{ 2 }\)
In the given problem, M is constant
\(\therefore \frac { { p }_{ 1 } }{ { p }_{ 2 } } =\frac { { C }_{ 1 }^{ 2 } }{ { V }_{ 1 } } \times \frac { { V }_{ 2 } }{ { C }_{ 2 }^{ 2 } } =\frac { { V }_{ 2 } }{ { V }_{ 1 } } \left[ \frac { { C }_{ 1 } }{ { C }_{ 2 } } \right] ^{ 2 }\)
or   \(\frac { { p }_{ 1 } }{ { p }_{ 2 } } =\frac { 2 }{ 1 } \times \frac { 16 }{ 1 } =32\)

Here, \(\rho\) = 0.09 kg m^-3 
At S.T.P., Pressure P = 1.01 x 10⁵ pa.
According to kinetic theory of gases.
\(p=\frac { 1 }{ 3 } \rho { C }^{ 2 }\quad or\quad C=\sqrt { \frac { 3p }{ \rho } } \)
\(=\sqrt { \frac { 3\times 1.01\times { 10 }^{ 5 } }{ 0.09 } } =1837.5\quad { ms }^{ -1 }\)
Volume occupied by one mole of hydrogen at S.T.P. = 22.4 liters = 22.4 x 10^-3 m³
\(\therefore \) Mass of hydrogen, M = volume x density
= 22.4 x 10^-3 x 0.09
= 2.016 x 10^-3kg
Average K.E/mole =\(\frac { 1 }{ 2 } \)MC2
=\(\frac { 1 }{ 2 } \) x (2.016 x 10^-3) x (1837.5)²
= 3403.4 J

(i) The moon  has small gravitational; force and hence the escape velocity is small .As the moon is in tyhe proximity of the earth as seen from the sun, the moon has the same amount of heat per unit area  as that of the earth , The air molecules have l;arge range of speeds.
Even though the rms speed of the air molecules is smaller than the escape velocity on the moon, a significant number of  molecules have speed greater than escape velocity and they escape.
Now, rest of the molecules arrange the speed distribution for the equilibrium temperature. Again, a significant number of molecules escape as their speeds exceed escape sppeed. Hence, over a long time the moon has lost most of its atmosphere.
(ii) As the molecules move higher , their potential energy increases and hence kinetic energy decreases and hence temperature reduces.
At greater height, more volume is available and gas expands and hencde some cooling takes place.

\(Given,\ Diameter\ molecule,d=2\overset { 0 }{ A }=2\times { 10 }^{ -10 }m\)
\(\\ At\quad STP,\ one\ mole\ of\ gas\ (or\ 22.4\ L)\ of\ gas\ have\)
\(\\ { N }_{ A }=6.023\times { 10 }^{ 23 }molecules\)
\(\\ \therefore Number\ density\ of\ nitrogen\ molecules\)
\(\\ n=\frac { { N }_{ A } }{ 22.4\quad L } =\frac { 6.023\times { 10 }^{ 23 }{ m }^{ -3 } }{ 22.4\times { 10 }^{ -3 }{ m }^{ 3 } } =2.69\times { 10 }^{ 25 }{ m }^{ -3 }\)
\(\\ \therefore Mean\ free\ path\ of\ nitrogen\ at\ STP\ condition,\)
\(\\ \lambda =\frac { 1 }{ \sqrt { 2\pi n{ d }^{ 2 }\ } } \)
\(\\ \lambda =\frac { 1 }{ 1.414\times 3.142\times (2.69\times { 10 }^{ 25 })\times (2\times { 10 }^{ -10 })^{ 2 } }\)
\( \\ =2.1\times { 10 }^{ -7 }m\)

\(Moles\ of\ helium\ \left( { \mu }_{ He } \right) =\frac { 16 }{ 4 } =4\)
\(\\ Moles\ of\ oxygen\ \left( { \mu }_{ { O }_{ 2 } } \right) =\frac { 16 }{ 32 } =\frac { 1 }{ 2 } \)
\(\\ As\ helium\ is\ monoatomic,\ so\ degree\ of\ freedom\ of\ helium,\ f=3,\ so\ { C }_{ { V }_{ He } }=\frac { f }{ 2 } R=\frac { 3 }{ 2 } R\)
\(\\ As\ oxygen\ is\ diatomic,\ so\ degree\ of\ freedom\ of\ oxygen,\ f=5,\ so\)
\(\\ { C }_{ { V }_{ { O }_{ 2 } } }=\frac { f }{ 2 } R=\frac { 5 }{ 2 } R\)
\(\\ \therefore { C }_{ V\ mixture }=\frac { { \mu }_{ He }{ C }_{ V_{ He } }+{ \mu }_{ { O }_{ 2 } }{ C }_{ { V }_{ { O }_{ 2 } } } }{ { \mu }_{ He }+{ \mu }_{ { O }_{ 2 } } } \)
\(\\ =\frac { 4\times \frac { 3 }{ 2 } R+\frac { 1 }{ 2 } \times \frac { 5 }{ 2 } R }{ 4+\frac { 1 }{ 2 } } =\frac { 29 }{ 18 } R\)
\(\gamma =\frac { { C }_{ p } }{ { C }_{ v } } [of\ mixture]\)
\(\\ { \gamma }_{ mixture }=1+\frac { R }{ { C }_{ { V }_{ mixture } } } =1+\frac { R }{ \frac { 29 }{ 18 } R } =1.62 as\ { C }_{ p }-{ C }_{ v }=R]\)

Given, for monoatomic gas, 
\({ \mu }_{ 1 }=1,{ C }_{ { v }_{ 1 } }=\frac { 3 }{ 2 } R\ and\ for\ a\)
\(\\ diatomic\ gas,\ { \mu }_{ 2 }=3\ and\ { C }_{ { v }_{ 2 } }=\frac { 5 }{ 2 } R\)
\(\\ \therefore Total\ heat\ energy\ required\ to\ raise\ the\ temperat5ure\ of\ mixture\ by\ \triangle T.\)
\(\\ \triangle U={ \mu }_{ 1 }{ C }_{ { v }_{ 1 } }\triangle T+{ \mu }_{ 2 }{ C }_{ { v }_{ 2 } }\triangle T\)
\(\\ \triangle U=1\times \frac { 3 }{ 2 } R\triangle T+3\times \frac { 5 }{ 2 } R\triangle T=9R\triangle T\ \ ---\ (i)\)
\(\\ Let\ { C }_{ { v }_{ m } }\ be\ the\ molar\ specific\ heat\ of\ the\ mixture\ at\ constant,\ volume\ and\ as\ totla\ number\ of\ moles\ of\ mixture.\)
\(\\ { \mu }_{ m }=1+3=4\)
\(\\ \therefore Heat\ energy\ required\)
\(\\ \triangle U={ \mu }_{ m }{ C }_{ { V }_{ m } }\triangle T\)
\(\\ \Rightarrow \triangle U=4{ C }_{ { V }_{ m } }\triangle T\ ---\ (ii)\)
\(\\ From\quad Eqs.(i)\ and\ (ii),\ we\ have\)
\(\\ 9R\triangle T=4{ C }_{ { V }_{ m } }\triangle T\ \Rightarrow \ \triangle { C }_{ { V }_{ m } }=\frac { 9 }{ 4 } R=2.25\ R\)