Correct answer is (C)

The mean free path (λ) of molecules in a gas is given by the formula:

$\lambda =\frac{kT}{\sqrt{2}\pi d^{2}P}$

where k is the Boltzmann constant, T is the temperature in Kelvin, d is the diameter of the gas molecules, and P is the pressure.

At STP (standard temperature and pressure), the temperature is $273\mathrm{K}$ and the pressure is $1\mathrm{atm}$. We are given that the mean free path at STP is $1500d$. Let's denote the mean free path at $373\mathrm{K}$ as λ':

${\lambda }^{\prime }=\frac{k(373\mathrm{K})}{\sqrt{2}\pi d^2(1\mathrm{atm})}$

To find the ratio of the mean free path at $373\mathrm{K}$ to that at STP, we can divide λ' by λ:

$\frac{{\lambda }^{\prime }}{\lambda }=\frac{\frac{k(373\mathrm{K})}{\sqrt{2}\pi d^2(1\mathrm{atm})}}{\frac{k(273\mathrm{K})}{\sqrt{2}\pi d^2(1\mathrm{atm})}}$

The Boltzmann constant, pressure, and molecular diameter cancel out:

$\frac{{\lambda }^{\prime }}{\lambda }=\frac{373\mathrm{K}}{273\mathrm{K}}$

Now, we can solve for λ':

${\lambda }^{\prime }=\lambda \cdot \frac{373\mathrm{K}}{273\mathrm{K}}$

Substituting the given value of λ as $1500d$:

${\lambda }^{\prime }=1500d\cdot \frac{373\mathrm{K}}{273\mathrm{K}}$

${\lambda }^{\prime }\approx 2049d$

Thus, the mean free path of the molecules at $373\mathrm{K}$ while maintaining the standard pressure is approximately $2049d$.

Correct answer is (B)

1. The collision frequency (f) is given by the formula :

$f=\sqrt{2}\pi d^{2}v{n}_v$

Where:

- $d$ is the diameter of the molecule,

- $v$ is the average velocity of the molecules, and

- $n_v$ is the number density (number of molecules per unit volume).

2. From this equation, we can see that the collision frequency (f) is directly proportional to the number density ($n_v$), because all the other variables ($d$ and $v$) are constant :

$f\propto n_v$

3. Therefore, the ratio of two different collision frequencies ($f_1$ and $f_2$) is equal to the ratio of their corresponding number densities ($n_{v1}$ and $n_{v2}$):

$\frac{f_1}{f_2}=\frac{n_{v1}}{n_{v2}}$

4. Given that the number density increased from $3\times {10}^{19}$ to $12\times {10}^{19}$, we can substitute these values into the equation to find the ratio of the collision frequencies:

$\frac{f_1}{f_2}=\frac{3\times {10}^{19}}{12\times {10}^{19}}$

5. Simplifying this equation gives:

$\frac{f_1}{f_2}=0.25$

So, the ratio of the collision frequency of air molecules before and after the increase in number is 0.25.

Correct answer is (B)

According to kinetic theory of gases,

A. The motion of the gas molecules freezes at 0 K.

B. The mean free path decreases on increasing the number density of the molecules as $\mu =\frac{1}{\sqrt{2}\pi n{d}^2}\Rightarrow \mu \propto \frac{1}{n}$.

C. The mean free path increases on increasing the volume. Now if temperature is increased by keeping the pressure constant the volume should increase that is mean free path increases.

D. K.E._avg per molecule per degree of freedom is $\frac{1}{2}{k}_{B}T$.

Correct answer is (D)

$I_{mean}=\frac{RT}{\sqrt{2}\pi d^2{N}_{A}P}$

$=\frac{1.38\times 300\times {10}^{-23}}{\sqrt{2}\times 3.14\times {(0.3\times {10}^{-9})}^2\times 1.01\times {10}^5}$

$=102\times {10}^{-9}$ m

$=102$ nm

Correct answer is (25)

$\because$ mean free path

$\lambda =\frac{1}{\sqrt{2}\pi d^{2}n}$

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{d_2^2{n}_2}{d_1^2{n}_1}$

$={\left(\frac{5}{10}\right)}^2=0.25=25\times {10}^{-2}$