Question

The mass of a hydrogen molecule is$3.32\times {10}^{-27}\mathrm{kg}$. If ${10}^{23}$ hydrogen molecules strike, per second, a fixed wall of area $2{\mathrm{cm}}^2$ at an angle of $45°$ to the normal, and rebound elastically with a speed of ${10}^3\mathrm{m}/\mathrm{s}$, then the pressure on the wall is nearly:

Answer 1

Step 1 : Given data:

$\mathrm{u}={10}^3\mathrm{m}/\mathrm{s}\mathrm{m}=3.32\times {10}^{-27}\mathrm{kg}\mathrm{n}={10}^{23}{\mathrm{H}}_2$molecules per sec.

Step 2: Apply due to total change in momentum

Change in momentum along $\mathrm{y}-$axis of one molecule ${\mathrm{H}}_2$

$\Delta \mathrm{P}=mu\cos 45°-$$\left(-mu\cos 45°\right)$

$\Rightarrow \Delta \mathrm{P}=2\mathrm{mucos}45°$

Total Change in momentum along $\mathrm{y}-$axis of all ${\mathrm{H}}_2$molecule in one second

$\Delta {\mathrm{P}}_{\mathrm{T}}=2\mathrm{nmucos}45°$

Area of the wall $\mathrm{A}=2{\mathrm{cm}}^2$

To convert centimeter into meter

A=$2\times {10}^{-4}{\mathrm{m}}^2$

Step 3 : To calculate net pressure:

Net pressure as $\mathrm{P}=\frac{\Delta {\mathrm{P}}_{\mathrm{T}}}{\mathrm{A}}$

Substitute these values in the net pressure equation as

$=\frac{2\times {10}^{23}\times 3.32\times {10}^{-27}\times {10}^3\times 0.707}{2\times {10}^{-4}}$

Net pressure $=2.35\times {10}^3\mathrm{N}/{\mathrm{m}}^2$

Therefore the net pressure $=2.35\times {10}^3\mathrm{N}/{\mathrm{m}}^2$