(i) All the molecules do not have the same vel.
(ii) The molecule which possess large velocity are able to overcome the molecular attraction and escape the liquid surface, which is known as evaporation.

\({ v }_{ rms }\times \frac { 1 }{ \sqrt { M } } \) so H₂ will leak more rapidly. Because of its smaller molecular mass.

(i) The temperature of monoatomic gas will rise by a large value.
(ii) In case of the monoatomic gas, the heat supplied is used entirely to increase the translational K.E of the molecules.
(iii) In case of the diatomic gas. The heat supplies is used to increase the translational, rotational and some times even the vibrational K.E of the molecules.
(iv) It is only the translational K.E which, increases the temperature.

Acceleration to kinectic interpretation of temperature
\(E=\frac { 3 }{ 2 } { k }_{ B }T\ (or)\ T=\frac { 2 }{ 3 } .\frac { E }{ { K }_{ B } } \)
Absolute temperature x average K.E of molecules
\(\therefore\) The temperature = 0 K, average K.E = 0.

The rms speed of the molecule of a gas is defined as the square root of the mean of the squared velocities of the molecule of a gas.
No, rms speed is different from the average speed.
\({ v }_{ rms }=\sqrt { \frac { { V }_{ 1 }^{ 2 }+{ V }_{ 2 }^{ 2 }+{ V }_{ 3 }^{ 2 } }{ 3 } } \)
\(\bar { V } \) = Average speed = \(\frac { { V }_{ 1 }+{ V }_{ 2 }+{ V }_{ 3 } }{ 3 } \)

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.

The average K.E of a gas molecule depends only on the absolute temperature of the gas and is directly proportional to it.

The Brownian motion increases
(i) with the decrease in single of the suspended particle
(ii) with the increases in temperature of the fluid
(iii) with the decrease indensity of the fluid
(iv) with the decrease in viscosity of the fluid

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 \(\mathrm{T}_{2}=2457 \mathrm{~K}, \mathrm{~T}_{1}=273 \mathrm{~K}\)

The average distance travelled by the molecule between two successive collisions is called mean free path
Mean free path \(\lambda=\frac{k T}{\sqrt{2} \pi d^{2} P}\)
Where K - Boltzmann's constant
T - Temperature
d - diameter of the molecule
P - Pressure

Law of equipartition energy states that the average kinetic energy of system of molecules in thermal equilibrium at temperature T is uniformly distributed to all degrees of freedom (x or y or z) directions of motion so that each degree of freedom will get \(\frac{1}{2}\)kT of energy.

\(\text {Temperature } \quad \mathrm{T}=80+273=353 \mathrm{~K} \)
\(\text {Pressure } P=5 \times 10^{-10} \mathrm{~N} / \mathrm{m}^{2} \)
\(\mathrm{k}_{\mathrm{B}}=\frac{R}{N} \quad \therefore R=K_{B}-N \)
\(P=n R T \)
\(=\mathrm{nk}_{\mathrm{B}} \mathrm{NT} \)
\(\text { Number of molecules } \mathrm{N}=\frac{P}{k_{B} T} \)
\(=\frac{5 \times 10^{-10}}{353 \times 1.3 .8 \times 10^{-23}} \)
\(=\frac{5 \times 10^{-10+23}}{487} \)
\(=0.0102 \times 10^{13} \)
\(\text {Number of molecules }=1.02 \times 10^{11}\)

This law states that at constant temperature and pressure, equal volumes of all gases contain same number of molecules. For two different gases at the same temperature and pressure, according to kinetic theory of gases. We get
\(\mathrm{P}=\frac{1}{3} n m \overline{v^{2}} \ or \ \mathrm{P}=\frac{1}{3} \frac{N}{V} m \overline{v^{2}}\)
\(\mathrm{P}=\frac{1}{3} \frac{N_{1}}{V} m_{1} v_{1}^{2} \)
\(=\frac{1}{3} \frac{N_{2}}{V} m_{2} v_{2}^{2}\)   ......(1)
where \(\overline{v_{1}^{2}} \ and \ \overline{v_{2}^{2}}\) are the mean square speed for two gases and \(\mathrm{N}_{1} \ and \ \mathrm{N}_{2}\) are the number of gas molecules in two different gases.
At the same temperature, average kinetic energy per molecule is the same for two gases.
\(\frac{1}{2} m_{1} \overline{v_{1}^{2}}=\frac{1}{2} m_{2} \overline{v_{2}^{2}}\)   ........(2)
Dividing the equation (1) by (2) we get
N₁ = N₂

We get \(P V=\frac{2}{3} U\) But the internal energy of an ideal gas is equal to N times the average kinetic energy \((\epsilon)\) of each molecule.
\(\mathrm{U}=\mathrm{N} \in\)
For a fixed temperature, the average translational kinetic energy \(\in\) will remain constant It implies that
\(\mathrm{PV} =\frac{2}{3} \mathrm{~N} \in  \)
\(\text {Thus } \mathrm{PV} =\text { constant }\)
Therefore, pressure of a given gas is inversely proportional to its volume provided the temperature remains constant. This is Boyle's law.

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