The mass of a hydrogen molecule is 3.32 times | vedclass

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(D) The change in momentum of one molecule colliding with the wall is calculated by considering the component of momentum perpendicular to the wall. T

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$2.35 	imes 10^3 \ N/m^2$

Vedclass · Answer

(D) The change in momentum of one molecule colliding with the wall is calculated by considering the component of momentum perpendicular to the wall. The initial momentum component normal to the wall is $p_n = mv \cos(45^\circ)$. Since the collision is elastic,the final momentum component normal to the wall is $-mv \cos(45^\circ)$.Change in momentum per molecule $\Delta p = mv \cos(45^\circ) - (-mv \cos(45^\circ)) = 2mv \cos(45^\circ)$.Given $m = 3.32 	imes 10^{-27} \ kg$,$v = 10^3 \ m/s$,and $\cos(45^\circ) = 1/\sqrt{2}$,the change in momentum per molecule is $\Delta p = 2 	imes (3.32 	imes 10^{-27}) 	imes 10^3 	imes (1/\sqrt{2}) = \sqrt{2} 	imes 3.32 	imes 10^{-24} \ kg \cdot m/s$.The force exerted on the wall is the total change in momentum per second: $F = n 	imes \Delta p$,where $n = 10^{23} \ s^{-1}$.$F = 10^{23} 	imes \sqrt{2} 	imes 3.32 	imes 10^{-24} = 3.32 	imes \sqrt{2} 	imes 10^{-1} \ N$.Pressure $P = F / A$,where $A = 2 \ cm^2 = 2 	imes 10^{-4} \ m^2$.$P = (3.32 	imes 1.414 	imes 0.1) / (2 	imes 10^{-4}) = (4.694 	imes 10^{-1}) / (2 	imes 10^{-4}) \approx 2.35 	imes 10^3 \ N/m^2$.

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

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