Brownian motion. In this motion any particular molecule will follow a zig - zag path due to be collision with the other molecule or with the walls of the container.

One mole of an ideal gas at S.T.P occupies a volume of \(22.4 \times 10^{-3} \mathrm{~m}^{3}\)
∴ Number of moles \(/ \mathrm{m}^{3}=\mathrm{n}=\frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}}\)
\(\therefore n=2.69 \times 10^{25} \mathrm{moles} / \mathrm{m}^{3}\)
In terms of number of molecules and radius the mean free path at S.T.P can be written as
\(\lambda =\frac{1}{4 \pi \sqrt{2} r^{2} n}  \)
\(\text {Radius } r =\frac{3 \times 10^{-10}}{2}=1.5 \times 10^{-6} \mathrm{~m}\)
∴ Mean free path, \(\quad \lambda=\frac{1}{4 \times 3.14 \times 1.414 \times\left(1.5 \times 10^{-10}\right)^{2} \times 2.69 \times 10^{25}}\)
\(\lambda=0.0931 \times 10^{-6}=9.31 \times 10^{-8} \mathrm{~m}\)
Mean free path, \(\lambda \approx 9 \times 10^{-8} \mathrm{~m}\)

RMS velocity \(\mathrm{C} =\sqrt{\frac{3 R T}{M}} \)
\(\mathrm{T}=\mathrm{T}_{1} =273 \mathrm{~K}  \)
\(\therefore C =\sqrt{\frac{3 R \times 273}{M}}\)
When the RMS velocity is tripled
\(3 \mathrm{C}=\sqrt{\frac{3 R T}{M}}\)
Dividing equation is (2) by (1) we get
\(\frac{3 C}{C} =\sqrt{\frac{3 R T / M}{3 R \times 273 / M}}  \)
\(3 =\sqrt{\frac{T}{273}}  \)
\(\therefore 9 =\frac{T}{273}  \)
\(\therefore T =273 \times 9=2457 \mathrm{~K}\)
\(\therefore\) Temperature \(\mathrm{T}_{2}=2457 \mathrm{~K}, \mathrm{~T}_{1}=273 \mathrm{~K}\)

Molar mass of nitrogen m = 0.028 kg/mol
Temperature T = 20 + 273 = 293 K
Universal gas constant \( R=8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{k}\)
RMS speed of nitrogen (N₂),
\(v_{r m s} =\sqrt{\frac{3 R T}{m}}  \)
\(\left(\mathrm{~N}_{2}\right), v_{m s} =\sqrt{\frac{3 \times 8.31 \times 273}{0.028}}  \)
\(\left(\mathrm{~N}_{2}\right), v_{r m s} =\sqrt{2610 \times 10^{2}}=511 \mathrm{~m} / \mathrm{s}\)
Molar mass of Oxygen = 0.032 kg / mol
RMS speed of oxygen (O₂),
\(v_{m s}=\sqrt{\frac{3 R T}{m}}\)
RMS speed of (O₂) \(v_{m s} =\sqrt{\frac{3 \times 8.31 \times 293}{0.032}} \)
\(=\sqrt{2280 \times 10^{2}}\)
RMS speed of \(\left(\mathrm{O}_{2}\right), \quad v_{m s}=478 \mathrm{~m} / \mathrm{s}\)

The minimum number of independent coordinates needed to specify the position and configuration of a thermodynamic system in space is called degree of freedom of the system.

The escape speed of gases on the surface of Moon is much less than the root mean square speeds of gases due to low gravity. Due to this all the gases escape from the surface of the Moon.

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

The hydrogen atom has one proton and one electron. The mass of electron is negligible compared to the mass of proton.
Mass of one proton = 1.67\(\times\)10⁻²⁷kg.
One hydrogen molecule = 2 hydrogen
atoms = 2\(\times\)1.67\(\times\)10⁻²⁷kg.
The average speed
\(\overset { - }{ v } =\sqrt { \frac { 8KT }{ \pi m } } =1.60\sqrt { \frac { KT }{ m } } =\)
\(=1.60\sqrt { \frac { \left( { 1.38\times 10 }^{ -23 } \right) \times \left( 300 \right) }{ 2\left( 1.67\times { 10 }^{ -27 } \right) } } =1.78\times { 10 }^{ 3 }{ ms }^{ -1 }\)
(Boltzmann Constant k = 1.38\(\times\)10⁻²³ J K^-1)
The rms speed \({ v }_{ rms }=\sqrt { \frac { 3KT }{ m } } =1.73\sqrt { \frac { kT }{ m } } \)
\(=1.73\sqrt { \frac { \left( { 1.38\times 10 }^{ -23 } \right) \times \left( 300 \right) }{ 2\left( 1.67\times { 10 }^{ -27 } \right) } } =1.9\times { 10 }^{ 3 }{ ms }^{ -1 }\)
Most probable speed \({ v }_{ mp }=\sqrt { \frac { 2KT }{ m } } =1.41\sqrt { \frac { kT }{ m } } \)
\(=1.41\sqrt { \frac { \left( { 1.38\times 10 }^{ -23 } \right) \times \left( 300 \right) }{ 2\left( 1.67\times { 10 }^{ -27 } \right) } } =1.57\times { 10 }^{ 3 }{ ms }^{ -1 }\)
Note that v_rms > \(\overset { - }{ V } \) > v_mp

 It states that the volume of a gn mass of a gas is inversely proportional to it pressure provided the temperature remains constant.
\(v\times \frac { 1 }{ P } (or)v=\frac { k }{ p } \) (or) pv = constant
Its value depends on
(i) mass of the gas
(ii) its temperature and
(iii) the units in which P and v are measured.
P₁ & V₁ - initial values of pressure and volume
P₂ & v₂ - Final values of pressure and volume
then accumulate the Boyle's law P₁ V₁ = P₂ v₂
Graph between P vs. V and P vs.\(\frac{1}{v}\) for a g_n mass a gas a constant temperature T are shown below.

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.

(i) Temperature of the gas
(ii) Pressure of the gas
(iii) Diameter of the gas molecules

We get \(P V=\frac{2}{3} U\) For a fixed pressure, the volume of the gas is proportional to internal energy of the gas or average kinetic energy of the gas and the average kinetic energy is directly proportional to absolute temperature. It implies that.
\(\mathrm{V} \propto \mathrm{T} \text { or }  \)
\(\frac{V}{T}=\text { constant }\)
This is Charles' law

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

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

\(\gamma=\frac{27}{17}\)
Number of moles of helium
\(\mathrm{n}=\frac{8}{4}=2\)
Number of moles of oxygen
\(n^{\prime}=\frac{16}{32}=\frac{1}{2}\)
For mono atomic Helium gas
\(\mathrm{f} =3 \)
\(\mathrm{C}_{\mathrm{V}} =\frac{f}{2} R \)
\(=\frac{3}{2} R \)
For diatomic oxygen gas
f = 5
\(\mathrm{C}_{\mathrm{V}} =\frac{f}{2} R \)
\(=\frac{5}{2} R \)
\(\mathrm{C}_{\mathrm{V}} \text { mixture } =\frac{n c_{v}+n^{\prime} C_{v}^{\prime}}{n+n^{\prime}} \)
\(=\frac{2 \times \frac{3}{2} R+\frac{1}{2} \times \frac{5}{2} R}{2+\frac{1}{2}} \)
\(C_{V} =\frac{3 R+\frac{5}{4} R}{\frac{5}{2}} \)
\(=\frac{17 R}{10} \)
\(\gamma =\frac{C_{p}}{C_{v}} \)
\(=1+\frac{R}{C_{V}} \)
\(=1+\frac{R}{\frac{17 R} {10}}\)
\(=1+\frac{10}{17} \)
\(=\frac{27}{17} \)

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

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.

To understand the microscopic origin of temperature in the same way.
Rewrite the equations
\(\mathrm{P}  =\frac{1}{3} n m \overline{v^{2}} \text { or } P \frac{1}{3} \frac{N}{V} m \overline{v^{2}} \quad \text { as }\left[n=\frac{N}{V}\right] \\ \)
\(\mathrm{P}  =\frac{1}{3} n m \overline{v^{2}} \text { or } P \frac{1}{3} \frac{N}{V} m \overline{v^{2}} \\ \)
\(\mathrm{PV}  =\frac{1}{3} N m \overline{v^{2}}\)   ....(1)
Comparing the equation (1) with ideal gas equation PV = Nkt
\(\mathrm{NkT} =\frac{1}{3} N m \overline {v^{2} } \)
\(\mathrm{kT} =\frac{1}{3} m v\overline v^{2}\)   ....(2)
Multiply the above equation by 3 / 2 on both sides,
\(\frac{3}{2} k T=\frac{1}{2} m \overline v^{2}\)
R.H.S of the equation is called average kinetic energy of a single molecule \((\overline{KE})\)
The average kinetic energy per molecule
\(\overline{K E}=\frac{3}{2} k T  \)
\(\frac{3}{2} k T =\frac{1}{2} \overline{m v^{2}}\)
Implies that the temperature of a gas is a measure of the average translational kinetic energy per molecule of the gas.

(i) All the molecules of a gas are identical, elastic spheres.
(ii) The molecules of different gases are different.
(iii) The number of molecules in a gas is very large and the average separation between them is larger than size of the gas molecules.
(iv) The molecules of a gas are in a state of continuous random motion.
(v) The molecules collide with one another and also with the walls of the container.
(vi) These collisions are perfectly elastic so that there is no loss of kinetic energy during collisions.
(vii) Between two successive collisions, a molecule moves with uniform velocity.
(viii) The molecules do not exert any force of attraction or repulsion on each other except during collision. The molecules do not possess any potential energy and the energy is wholly kinetic.
(ix) The collisions are instantaneous. The time spent by a molecule in each collision is very small compared to the time elapsed between two consecutive collisions.
(x) These molecules obey Newton's laws of motion even though they move randomly.