3 Marks Question

Question 13 Marks

Prove that the average kientic energy of molecule is equal to $\frac{3}{2} kT$

Answer

According to the kinetic equation of gases, $
\begin{array}{l}PV=\frac{1}{3} m Nc^2 \\\text {or } PV=\frac{2}{3} \times \frac{1}{2} m Nc^2\end{array}$
Hence, $P V=\frac{2}{3}$ K.E.
$ \left(\because \text { K.E. }=\frac{1}{2} m N c^2 \text { and mass }=mN\right) $
So,$\text { K.E. }=\frac{3}{2} \text { P.V. }$
According to the ideal gas equation. $PV = nRT$ Kinetic energy for $\text {n}$ molecules of a gas
$\text {K.E.}$ $=\frac{3}{2} n$ $RT$
$\therefore \quad$ $\text {K.E.}\propto$ $T$ Here $T$ is the absolute temperature.
$\therefore \quad$ $\text {K.E.}=\frac{3}{2} kT$ Here $k$ is Boltzmann constant.

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Question 23 Marks

What is Graham's law of gas diffusion ? Propound this law on the basis of molecular motion theory.

Answer

Graham's gas diffusion law : According to this law, the rate of diffusion of a gas at constant pressure is inversely proprotional to the square root of its density.
$\text {That is}$ $\quad D\propto \frac{1}{\sqrt{p}}$
Where $\rho$ is gas density.
On the basis of molecular kinetic theory of gas $
\begin{array}{l}
PV=\frac{1}{3} m N \overline{C^2} \\
\text { or } \overline{C^2}=\frac{3 PV}{m N}=\frac{3 PV}{\text { Mass }}=\frac{3 P}{\rho} \because \text { Density }=\frac{\text { Mass }}{\text { Volume }} \\
\text { or } \sqrt{C^2}=C_{rms}=\sqrt{\frac{3 P}{P}} \\
\text { or } C_{rms} \propto \frac{1}{\sqrt{\rho}} \text { while } P \text { is constant. }
\end{array}
$
Since the rate of diffusion depends on the root mean square velocity of the gas molecules is proportional to the sequence, therefore
Rate of diffusion(D) $\propto \frac{1}{\sqrt{\rho}}$.
This is Graham's law of diffusion.

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Question 33 Marks

According to the molecular kinetic theory of gases, what is the mean free path?

Answer

Mean free path : The distance that gas molecules travel before colliding with another molecule is called mean free path. The value of free path may be different in different collisions. We can understand in another way like this.
The avarage distance travelled by a molecule between two consecutive collisions is called mean free path. Let us represent this $\bar{\lambda}$
Mean free path $=\frac{\text { Total distance travelled }}{\text { Number of collisions }}$
If any molecule travel the distance $\lambda_1, \lambda_2, \lambda_3, \ldots \ldots \ldots \lambda_{ N }$ after $\text {t}$ time with $\text {N}$ different collisions then mean free path value will be
$\bar{\lambda}=\frac{\lambda_1+\lambda_2+\lambda_3 \ldots \ldots \ldots \lambda_{ N }}{ N }$
Average free path is inversely proportional to the density of the gas and proportional to the mass of the molecule
$\bar{\lambda} \propto \frac{1}{\rho}$
Free path depend on temperature and pressure that means
$\begin{array}{l}\quad \quad\quad\bar{\lambda} \propto T \\\text {and}\quad \bar{\lambda} \propto \frac{1}{\rho}\end{array}$

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