Question

Suppose that a 1.00 cm^3 box in the shape of a cube is perfectly evacuated, except for a single particle of mass 1.05 10^-3 g. The particle is initially moving perpendicular to one of the walls of the box at a speed of 460 m/s. Assume that the collisions of the particle with the walls are elastic.

(a) Find the mass density inside the box.
(b) Find the average pressure on the walls perpendicular to the particle's path.
(c) Find the average pressure on the other walls.
(d) Find the temperature inside the box.

Answer 1

a)
Density equals mass per unit volume
ρ = m/V
The volume of cube with edge length L is L². So
ρ = m/L³ = 1.05×10⁻³ g / (1.0 cm³) = 1.05×10⁻³ g∙cm⁻³


b)
Consider one of the two walls, between the particle is moving. Since the collision is elastic and the particle is moving perpendicular to the wall, it recoils on the line of incidence, with same speed just in opposite direction.
So the change in momentum of the particle due to collision is
Δp = m∙v - m∙(-v) = 2∙m∙v
Assuming there is delay due to collision, the time elapsed between two collisions is same as the particle needs to travel twice along the cube:
Δt = 2∙L/v

The force exerted by particle on the wall equals its change in momentum per unit time:
F = Δp/Δt = 2∙m∙v / (2∙L/v) = m∙v²/L

The pressure equals force per unit area of the wall:
P = F/A = F/L² = m∙v²/L³
= 1.05×10⁻⁶ kg ∙ (460m∙s⁻¹) ² / (1.0×10⁻² m)³
= 22,218Pa

c)
Since there are no collisions with these walls, there is no force acting on these walls. Hence pressure is zero:
P = 0 Pa

d)
Assuming the particle behaves as ideal gas. Ideal gas law in molecular for,m states that:
P∙V = N∙k∙T
(N number of particles in the volume V , k boltzmann constant, Thermodynamic temperature)

Therefore
T = P∙V / (N∙k)
= (m∙v²/L³ ) ∙ L³ /(N∙k)
= m ∙ v² ∙ / (N∙k)
= 1.05×10⁻⁶ kg ∙ (460 m∙s⁻¹) ² / (1 ∙ 1.38×10⁻²³ J∙K⁻¹)
=1.61×10^22 K