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PhysicsW.E.P.MiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

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

Options:

Answer:

Solution:

$v = 10^{4} m s^{- 1}$
$m = 10^{- 26}$
$n = 10^{22}$
$A = 1 m^{2}$

$Δp = 2 mnv$

$Δp = 2 \times 10^{22} \times 10^{- 26} \times 10^{4} = 2$

$P = \frac{F}{A} = 2 N m^{- 2}$

Subject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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