The mass of a hydrogen atom is 3.32 times 10^ | vedclass

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Question

(C) Given: Mass of atom $m = 3.32 	imes 10^{-27} \, kg$,speed $v = 10^3 \, m/s$,number of atoms per second $n = 10^{23} \, s^{-1}$,area $A = 2 \, cm^

Vedclass · Accepted Answer

$2.347 	imes 10^3 \, N/m^2$

Vedclass · Answer

(C) Given: Mass of atom $m = 3.32 	imes 10^{-27} \, kg$,speed $v = 10^3 \, m/s$,number of atoms per second $n = 10^{23} \, s^{-1}$,area $A = 2 \, cm^2 = 2 	imes 10^{-4} \, m^2$,angle with normal $	heta = 45^{\circ}$.Since the collision is elastic,the magnitude of momentum before and after collision is $p = mv = (3.32 	imes 10^{-27}) 	imes (10^3) = 3.32 	imes 10^{-24} \, kg \cdot m/s$.The change in momentum along the normal to the wall is $\Delta p = |p_{final, normal} - p_{initial, normal}| = |(-mv \cos 	heta) - (mv \cos 	heta)| = 2mv \cos 	heta = 2p \cos 45^{\circ} = 2p 	imes \frac{1}{\sqrt{2}} = \sqrt{2}p$.The force exerted on the wall is $F = n 	imes \Delta p = n 	imes \sqrt{2}p$.The pressure $P = \frac{F}{A} = \frac{n \sqrt{2} p}{A} = \frac{10^{23} 	imes \sqrt{2} 	imes 3.32 	imes 10^{-24}}{2 	imes 10^{-4}}$.$P = \frac{1.414 	imes 3.32 	imes 10^{-1}}{2 	imes 10^{-4}} = \frac{4.69448 	imes 10^{-1}}{2 	imes 10^{-4}} = 2.347 	imes 10^3 \, N/m^2$.

The mass of a hydrogen atom is $3.32 \times 10^{-27} \, kg$. If $10^{23}$ atoms per second strike a wall of area $2 \, cm^2$ at an angle of $45^{\circ}$ with the normal to the wall and reflect elastically with a speed of $10^3 \, m/s$,the pressure exerted on the wall will be: $(N/m^2)$

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