Answer to Question #92288 in Molecular Physics | Thermodynamics for Shoyeb Hasan Shafin

Question

Precisely one million molecules strikes a surface with an area of 1.00×10^-2 m^2 in one second and bounce elastically. each of these molecules strike at an angle of 60° from normal, each has a mass of 3.30×10^-27 kg, and the average speed of these million molecules is 330 m/s. i) what is the magnitude of total momentum of the molecules before they strike the surface? How large is its component perpendicular to the surface? ii) what average pressure do these molecules exert on this surface during the one second they are striking it?

Answer 1

Millions means 10⁶

Let N is the total no of particles and given as 10⁶particles

v is the average velocity of particles striking the wall at 60 degree angle from normal to the wall

m is the mass of particle

1) Momentum is just the product of mass and velocity of particles

Thus total Momentum of particles through which particles strike the wall at 60 degree angle normal to wall and is given by

$P=Nmv$

$P=$10x3.30×10^-27x330

P=1.10 x 10^-18 kg m/s

And horizontal component of momentum or the component perpendicular to the wall is 1.10 x 10-18x $Cos60^0$ =0.55 x 10^-18kg m/s.

2) Pressure= Impulse force/ Area= Change of momentum per unit time/ Area

$Pressure=\frac{dP/dt}{A}$

The particle strike the wall and back with same velocity with same mass this the change of momentum is double the original momentum

Thus pressure=2x0.55 x 10^-18/(1x10^-2)

=1.1x10^-16 pa