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The mass of a hydrogen molecule is 3.32 \times 10^{-27} \mathrm{~kg}. If 10^{23} molecules are colliding per second on a stationary wall of an area 2 \mathrm{~cm}^{2} of an angle of 45^{\circ} to the normal to the wall and reflected elastically with a speed 10^{3} \mathrm{~m} / \mathrm{sec} the pressure exerted on the wall will be -

Option: 1

2.3 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}


Option: 2

3.6 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}


Option: 3

4.3 \times 10^{3} \mathrm{~N} / \mathrm{m}^{3}


Option: 4

None of these


Answers (1)

best_answer

figure:

Molecules moving along A O collides the wall of O and returns along O B where,

\angle A O N=\angle B O N=45^{\circ} \text { as the impact is elastic - }

\begin{aligned} \therefore\left|\vec{p}_{1}\right| & =\left|\overrightarrow{p_{2}}\right|=p=m v \\ & =3.32 \times 10^{-24} \mathrm{kgm} / \mathrm{sec} \end{aligned}

the change in momentum along the normal

\begin{aligned} (\Delta P)_{n}=\left|\vec{P}_{2 n}-\vec{P}_{1 n}\right|=2 P_{n} & =2 p \cos 45^{\circ} \\ & =\sqrt{2} p \end{aligned}

If f is the collision frequency the force applied on the wall

f=\left(\Delta P_{n}\right) \times f=\sqrt{2} p f

\therefore$ Pressure $P=\frac{F}{A}=\frac{\sqrt{2} P f}{A} on substituting

\begin{aligned} & A=2 \times 10^{-4} \mathrm{~m}^{2} \text { and } f=10^{23} \text { per sec } \\ & P=\frac{\sqrt{2} \times 3.32 \times 10^{-24} \times 10^{23}}{2 \times 10^{-4}}=2.3 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2} \end{aligned}

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shivangi.shekhar

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