Question

A hydrogen molecule (diameter $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathit{c}\mathit{m}$), travelling at the rms speed, escapes from a $\mathbf{4000}\mathit{K}$ furnace into a chamber containing $\text{cold}$argon atoms (diameter) at a density of.

a) What is the speed of the hydrogen molecule?

b) If it collides with an argon atom, what is the closest their centres can be, considering each as spherical?

c) What is the initial number of collisions per second experienced by the hydrogen molecule? (Hint: Assume that the argon atoms are stationary. Then the mean free path of the hydrogen molecule is given by

Mean free path$\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }{\mathbf{d}}^{\mathbf{2}}\mathbf{N}\mathbf{/}\mathbf{V}}$

Short answer

a) The speed of the hydrogen molecule, after it escapes from a furnace into a chamber containing cold argon atoms, is$7\times {10}^{3}m/s$ .

b) The closest distance between the centers when the hydrogen molecule collides with the argon atom is

$2\times {10}^{-8}cm$.

c) Initial number of collision per second experienced by the hydrogen molecule is$3.5\times {10}^{10}collisions/second$

.

Step-by-step solution

Given data

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

Understanding the concept

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.

(a) Calculate the speed of the hydrogen molecule

$7\times {10}^{3}m/s$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, $M=2.02\times {10}^{-3}kg/mo\text{l}$.

$T$is the temperature of argon atoms, equal to$4000K$

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

Substituting these values intheabove equation,

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

Therefore the speed of hydrogen molecule is .

(b) Calculate the closest their centres can be if the hydrogen molecule collides with an argon atom

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.

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

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

(c) Calculate the initial number of collisions per second experienced by the hydrogen molecule

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,

$\text{No.ofcollisions}=\frac{\pi d^{2}vtN}{V}$

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

$\rho =4.0\times {10}^{19}atoms/c{m}^3$

The number of collisions per unit time is

$\text{No.ofcollisionspertime}=\frac{\pi d^{2}vN}{V}$

Substituting all values in the above equation, we get

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

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

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