Question

Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second?

Answer 1

(a) The mean speed of the molecules is 1780 m/s

(b) The number of molecules strike each square metre of the wall per second is 1.2 × 10²⁸.

Given:

Pressure = P = 1 atm = 10⁵ Pascals

Temperature = T  = 300 K

Mass = M = 2 g $=2\times10^{-3} \ kg$

Solution:

(a) Mean velocity is given by the formula:

$V_{\mathrm{avg}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$

On substituting known values, we get,

$V_{\mathrm{avg}}=\sqrt{\frac{8 \times 8.3 \times 300}{3.14 \times 2 \times 10^{-3}}}$

$\therefore V_{\mathrm{avg}}=1781.004 \approx 1780 \ \mathrm{m} / \mathrm{s}$

(b) Angle the molecule strike = 45°

The number of molecules which are striking per unit area is given by the formula:

$n=\frac{\text { Force }}{\sqrt{2} m v \times \text { Area }}$

Pressure is force per unit area.

$n=\frac{\text { Pressure }}{\sqrt{2} m V}$

On substituting the known values, we get,

$n=\frac{10^{5}}{\frac{\sqrt{2} \times 2 \times 10^{-3} \times 1780}{6 \times 10^{23}}}$

$n=\frac{3}{\sqrt{2} \times 1780} \times 10^{31}$

$n=1.19 \times 10^{-3} \times 10^{31}$

Thus, the number of molecules is:

$\therefore n=1.19 \times 10^{28} \approx 1.2 \times 10^{28}$

Answer 2

(a) The mean speed of the molecules is $1780 \frac{m}{s}$

(b) Number of molecules strike each square meter of the wall per second $1.2 \times 10^{28}$

Explanation:

Given Data

$P=10^{5} \mathrm{Pa}$

T = 300 K

Mass of Hydrogen ,  $\mathrm{M}=2 \times 10^{-3} \mathrm{kg}$

(a) we know average speed  

$v=\sqrt{\frac{8 R T}{\pi M}}$

$=\sqrt{\frac{8 \times 8.3 \times 300 \times 7}{2 \times 10^{-3} \times 22}}$

$=\sqrt{\frac{2400 \times 58.1}{44 \times 10^{-3}}}$

$=\sqrt{\frac{139440}{44 \times 10^{-3}}}$

$=\sqrt{3169090.909}$

$=1780 \frac{m}{s}$

Let's take a cubic volume of 1 $m^{3}$.

V=1 $m^{3}$

Momentum of 1 natural molecule before collision to the striking surface = $\mu sin45^\circ$

Momentum of 1 normal molecule to striking surface following collision = $- \mu sin45^\circ$

$\text { Shift in molecule impulse }=2 \mathrm{\mu} \sin 45^{\circ}=2 \mathrm{\mu} \frac{1}{\sqrt{2}}=\sqrt{2} \mathrm{\mu}$

$\text {Change in n molecules impulse} =2 \mu \sin 45^{\circ}=2 \mu \frac{1}{\sqrt{2}}=\sqrt{2} \mu$

Let the time taken to change the momentum be omitted.

$\text {Force per unit area by single molecule }=\sqrt{2} \frac{\mu}{\Delta t}$

$\text {Pressure observed from collision with molecules} =\sqrt{2} \frac{\mu}{\Delta t}=10^{5}$

 $n = \frac{\sqrt{2} \frac{\mu}{\Delta t}}{\sqrt{2} \frac{\mu}{\Delta t}}$

$n=\frac{10^{5}}{\sqrt{2} \mu}$

$6.0 \times 10^{23} \text { molecules }=2 \times 10^{-3} \mathrm{kg}$

$\text { 1 molecule }=\frac{2 \times 10^{-3}}{6 \times 10^{28}}=3.3 \times 10^{-27} \mathrm{kg}$

$n=\frac{10^{5}}{\sqrt{2} \times 3.3 \times 10^{-27} \times 1780}$

$n=1.2 \times 10^{28}$

Therefore, the mean speed of the molecules is $1780 \frac{m}{s}$  and the number of molecules strike each square meter of the wall per second is  $1.2 \times 10^{28}$