Since oxygen is a diatomic molecule with 5 degrees of freedom. Degrees of freedom of molecules in 2 moles of oxygen = f₁
\(=2 \mathrm{~N} \times 5=10 \mathrm{~N}\)
Since argon is a mono atomic molecule with 3 degrees of freedom. Degrees of freedom of molecules in 4 moles of argon = f₂
\(=4 \mathrm{~N} \times 3=12 \mathrm{~N}\)
∴ Total degrees of freedom of the mixture = f
\(=\mathrm{f}_{1}+\mathrm{f}_{2}=22 \mathrm{~N}\)
According to the principle of law of equipartition energy, energy associated with each degree of freedom of a molecule
\(=\frac{1}{2} k T\)
∴ Total energy of the system \(=\frac{1}{2} k T \times 22\)
But k = R
∴ Total energy of the system \(=\frac{1}{2} R T \times 22=11 \mathrm{RT}\)

Pressure arises due to momentum transfer to the wall of the container.

The average speed
\(\overset { - }{ v } =\frac { 2+3+4+5+5+5+6+6+7+9 }{ 10 } =5.2{ ms }^{ -1 }\)
To find the rms speed, first calculate the mean square speed \(\overset { - }{ { v }^{ 2 } } \)
\(\overset { - }{ { v }^{ 2 } } =\frac { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 }+{ 5 }^{ 2 }+{ 5 }^{ 2 }+6^{ 2 }+{ 6 }^{ 2 }+{ 7 }^{ 2 }+{ 9 }^{ 2 } }{ 10 } \)
= 30.6ms²s^-2
The rms speed
\({ v }_{ rms }\sqrt { \overset { - }{ { v }^{ 2 } } } =\sqrt { 30.6 } =5.53{ ms }^{ -1 }\)
The most probable speed is 5 m s^-1 because three of the particles have that speed.

Boltzmann's constant \(\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{JK}^{-1}\)
\(\mathrm{k}_{\mathrm{B}}=\frac{R}{N} \quad \therefore R=K_{B} N\)
Now
\(\mathrm{P}=n R T=n k_{B} N T\)
∴ The number of molecules in the room
\(=\mathrm{nN}=\frac{P V}{T k_{B}} ;  \)
\(\text {Temperature }=27+273 =300 \mathrm{~K}  \)
\( \leq \frac{1.013 \times 10^{5} \times 25}{300 \times 1.38 \times 10^{-23}}  \)
\(=6.117 \times 10^{26} \text { molecules }  \)
\(=6.1 \times 10^{26} \text { molecules }\)

In 1827, Robert Brown, a botanist reported that grains of pollen suspended in a liquid moves randomly from one place to other. The random (Zig - Zag path) motion of pollen suspended in a liquid is called Brownian motion. In fact we can observe the dust particle in water moving in random directions. This discovery puzzled scientists for long time. There were a lot of explanations for pollen or dust to move in random directions were found adequate. After a systematic study, Wiener and Gouy proposed that Brownian motion is to the bombardment of suspended particles by bombardment of suspended particles by molecules of the surrounding fluid. But during 19⁺⁺⁺ century people did not accept that every matter is made up of small atoms or molecules. In the year 1905, Einstein gave systematic theory of Brownian motion based on kinetic theory and he deduced the average size of molecules.
According to kinetic theory any particle suspended in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner as shown in Figure. But when we put our hand in water it causes no random motion because the mass of our hand is so large that the momentum transferred. by the molecular collision is not enough to move our hand.

Factors affecting Brownian Motion:
(i) Brownian motion increases with increasing temperature.
(ii) Brownian motion decreases with bigger particle size, high viscosity and density of the liquid (or) gas.

From (9.26) \(\lambda =\frac { 1 }{ \sqrt { 2 } \pi { nd }^{ 2 } } \)
We have to find the number density n By using ideal gas law
\(n=\frac { N }{ V } =\frac { P }{ KT } =\frac { { 101.3\times 10 }^{ 3 } }{ 1.381\times { 10 }^{ -23 }\times 300 } \)
= 2.449\(\times\)10²⁵ molecues/m³
\(\lambda =\frac { 1 }{ \sqrt { 2 } \times \pi \times 2.449\times { 10 }^{ 25 }\times \left( 1.2\times { 1 }0^{ -10 } \right) ^{ 2 } } \)
\(=\frac { 1 }{ 15.65\times 10^{ 5 } } \)
λ = 0.63\(\times\)10⁻⁶m

The area under the graph will give total member of gas molecules in the system
\(\mathrm{n} =\frac{P V}{R T} \quad \mathrm{R}=k \)
\(\therefore \quad \mathrm{n} =\frac{P V}{k T} \)

\(C_{p}-C_{v}=R\)
For diatomic gas (N₂) No of degrees of freedom = 5
\(\therefore S_{p}-S_{v}=R / 28\)

Pressure is independent of internal energy.

Average K.E of each degree of freedom
\(=\frac{1}{2} k T=\frac{1}{2} N T[k=N]\)
N - no. of moles
T - temperature

As Temperature of room A is less than that of room B evidently Room A has more air in it.

(i) We know from postulates of kinetic theory that the molecules of a gas are in random motion and they collide with each other.
(ii) Between two successive collisions, a molecule moves along a straight path with uniform velocity.
(iii) This path is called mean free path. Consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume.
(iv) Assume that only one molecule is in motion,and all others are at rest.

(v) If a molecule moves with average speed v in a time t, the distance travelled is vt.
(vi) In this time t, consider the molecule to move in an imaginary cylinder of volume nd2vr.
(vii) It collides with any molecule. whose center is within this cylinder. Therefore, the number of collisions is equal to the number of molecules in the volume of the imaginary cylinder.
(viii) It is equal to \(\pi\)d²vtn. The total path length divided by the number of collisions in time t is the mean free path.
Mean free pat, \(\lambda =\frac{distance \ travelled}{Number \ of \ collisions}\)
\(\lambda =\frac { vt }{ n{ \pi d }^{ 2 }vt } =\frac { 1 }{ n{ \pi d }^{ 2 } } \)   ...(1)
(ix) Though we have assumed that only one molecule is moving at a time and other molecules are at rest, in actual practice all the molecules are in random motion.
(x) So the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations (you will learn in higher classes) the correct expression for mean free path .
\(\therefore \ \lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2 } } \)   ...(2)
(xi) The equation (1) implies that the mean free path is inversely proportional to number density.
(xii) When the number density increases the molecular collisions increases and it decreases the distance travelled by the molecule before collisions:
Case1: Rearranging the equation (2) using 'm' (mass of the molecule)
\(\therefore \ \lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2} mm } \)
But mn = mass per unit volume = p (density of the gas)
\(\therefore \ \lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2 } p} \)
Also we know that PV = NkT
P =\(\frac{N}{V}\)KT= nKT
\(\therefore n =\frac{P}{KT}\)
Substituting n = \(\frac{P}{KT}\) in equation, we get
\(\lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2 }P } \)

In general our interest our interest is to find how many gas molecules have the range of speed from v to v + dv. This is given by Maxwell's speed distribution function.
\({ N }_{ v }=4\pi N{ \left( \frac { m }{ 2\pi KT } \right) }^{ \frac { 3 }{ 2 } }{ v }^{ 2 }{ e }^{ \frac { { mv }^{ 2 } }{ 2KT } }\)   ....(1)
The above expression is graphically shown as follows

From the figure it is clear that, for a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The rms speed, average speed and most probable speed are indicated in the figure. It can be seen that the rms speed is greatest among the three. To Know the number of molecules in the range of speed between \(50 \mathrm{~m} \mathrm{~s}^{-1} \ and \ 60 \mathrm{~m}\mathrm{s}^{-1}\), we need to integrate \(\int_{50}^{60} 4 \pi N\left(\frac{m}{2 \pi k T}\right)^{\frac{3}{2}} v^{2} e^{\frac{m v^{2}}{2 k T}} d v=\mathrm{N}\left(50\right.\ to \ \left.60 \mathrm{~ms}^{-1}\right)\). In general the number of molecules within the range of speed v and v + dv is given by
\(\int_{v}^{v+d v} 4 \pi N\left(\frac{m}{2 \pi k T}\right)^{\frac{3}{2}} v^{2} e^{\frac{m v^{2}}{2 k T}} d v=N(v \text { to } v+d v) \)
The exact integration is beyond the scope of the book. But we can infer the behaviour of gas molecules from the graph.
(i) The area under the graph will give the total number of gas molecules in the system.
(ii) Figure shows the speed distribution graph for two different temperatures. As temperature increases, the peak of the curve is shifted to the right. It implies that the average speed of each molecule will increase. But the area under each graph is same since it represents the total number of gas molecules.