If 10^{22} gas molecules,each of mass 10^{-26 | vedclass

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(D) Pressure is defined as the normal force exerted per unit area.Force is equal to the rate of change of momentum,i.e.,$F = \frac{\Delta p}{\Delta t}

Vedclass · Accepted Answer

None of these

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(D) Pressure is defined as the normal force exerted per unit area.Force is equal to the rate of change of momentum,i.e.,$F = \frac{\Delta p}{\Delta t}$.For an elastic collision with a surface,the change in momentum of a single molecule is $\Delta p = mv - (-mv) = 2mv$.Given:Number of molecules per second,$N = 10^{22} \ s^{-1}$Mass of each molecule,$m = 10^{-26} \ kg$Speed of each molecule,$v = 10^4 \ m/s$Area,$A = 1 \ m^2$Total force exerted on the surface is $F = N 	imes (2mv)$.$F = 10^{22} 	imes 2 	imes 10^{-26} 	imes 10^4 = 2 \ N$.Pressure $P = \frac{F}{A} = \frac{2 \ N}{1 \ m^2} = 2 \ N/m^2$.Since the calculated value $2 \ N/m^2$ does not match any of the given options,the correct choice is 'None of these'.

If $10^{22}$ gas molecules,each of mass $10^{-26} \ kg$,collide with a surface (perpendicular to it) elastically per second over an area of $1 \ m^2$ with a speed of $10^4 \ m/s$,the pressure exerted by the gas molecules will be of the order of:

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