91Ó°ÊÓ

• Diameter of hydrogen molecule is$1.0×{10}^{-8}cm$
• Temperature of argon atoms$T=4000\text{ }K$
• Diameter of argon atom$3.0×{10}^{-8}$
• Density of argon atom$\mathrm{ÒÏ}=4.0×{10}^{19}\frac{atoms}{c{m}^3}$
• Argon atoms are stationary.

The expression for the RMS speed is given by,

$V_{rms}=\sqrt{\frac{3RT}{M}}$

Here$V_{rms}$ is the RMS speed, $R$is the universal gas constant,$T$ is the temperature and$M$ is the mass.

$7×{10}^{3}m/s$The speed of hydrogen molecule after escape:

The root-mean-square speed is given by

$v_{rms}=\sqrt{\frac{3RT}{M}}$

Where,

$M$is the molar mass of hydrogen, $M=2.02×{10}^{-3}kg/mo\text{l}$.

$T$is the temperature of argon atoms, equal to$4000\text{ }K$

$R$is the gas constant, and its value is given by$R=8.31J/mol.K$

Substituting these values intheabove equation,

$\begin{array}{c}{v}_{rms}=\sqrt{\frac{3\left(8.31\frac{\text{J}}{\text{mol.K}}\right)\left(4000\text{K}\right)}{2.02×{10}^{−3}\text{kg}/\text{mol}}}\\ =7×{10}^{3}m/s\end{array}$

Therefore the speed of hydrogen molecule is .

The closest distance between centers of ${\text{H}}_2$and $\text{Ar}$:

When the surfaces of the spheres that represent an ${\text{H}}_2$molecule and an Ar atom are touching, the distance between their centers is the sum of their radii.

$\begin{array}{c}d=r_1+r_2\\ =0.5×{10}^{-8}cm+1.5×{10}^{-8}cm\\ =2.0×{10}^{-8}cm\end{array}$

Therefore the closest possible centre is $2.0×{10}^{-8}cm$.

Initial number of collisions per second experienced by$H_2$atom:

The argon atoms are at rest, so in time$t$ the hydrogen atom collides with all the argon atoms in a cylinder of radius$d$ and length$vt$ , where$v$ is its speed, i.e,

$\text{No.ofcollisions}=\frac{\mathrm{Ï€}{d}^{2}vtN}{V}$

Where, $\mathrm{ÒÏ}=\text{N}/\text{V}$is the density or concentration of argon atoms.

$\mathrm{ÒÏ}=4.0×{10}^{19}atoms/c{m}^3$

The number of collisions per unit time is

$\text{No.ofcollisionspertime}=\frac{\mathrm{Ï€}{d}^{2}vN}{V}$

Substituting all values in the above equation, we get

$\text{No.ofcollisionspertime}=\mathrm{π}(2.0×{10}^{-10}m)(7×{10}^{3}m/s)(4×{10}^{25})$

$\text{No.ofcollisionspertime}=3.5×{10}^{10}collisions/second$

Therefore the initial number of collision experienced by the hydrogen molecule is$3.5×{10}^{10}collisions/second$ .