suppose that a 1.00 cubic cm box in the shape of a cube is perfectly evacuated,
ID: 1895876 • Letter: S
Question
suppose that a 1.00 cubic cm box in the shape of a cube is perfectly evacuated, except for a single particle of mass 1.13 x 10-3g. The particle is initially moving perpendicular to one of the walls of the box at a speed of 355 m/s. Assume that the collisions of the particle with the walls are elastic. find the mass density inside the box. find the average pressure on the walls perpendicular to the particle's path. Find the average pressure on the outer walls. Find the temperature inside the box.Explanation / Answer
the mass density is a dirac delta function. Let the direction in which particle is moving ve x direction. Also, let it be at y = z = 0
Then, mass density inside the cube should be of the form: (x,y,z) = a*(x-x(t))(y)(z).
Where x(t) is the current location of the particle. We need to find a. For it note that total mass in side the cube is m. Where m is the mass of the particle. So, the integral pf over all x, y, z gives m.
Thus, a = m. So, (x,y,z) = m*(x-x(t))(y)(z).
the average mass density is total mass in the cube divided by the Volume of cube
Average = 1.13*10-3 g/cm3
The length of one side (d) of the cube is cube root of Volume, so d = 1 cm = 0.01 m
the speed of the particle is 355 m/s.
The particle collides with the wall and reverses its velocity but speed remains same, then it collides with the opposite wall again reversing its direction of motion and so on. Per collision the particle transfers a momentum = 2*m*355 to a wall. Or
p = 8 * 10-4 kg m/s
Also, number of collisions of the particle in 1 second is given by the distance travelled by the particle in one second divided by the side of the cube.
So, n = 355/0.01 = 35500 collision/s
Since there are 2 walls, only half of these collisions are with a single wall.
So, total force exerted on a wall perpendicular to its direction of motion is
F = np/2 = 35500 * 8 * 10-4/2 = 14.24 N
Now, area of one wall = 1 cm2 = 0.0001 m2, thus average pressure on the wall perpendicular to the direction of motion of the particle = F/Area = 14*104 Pa
Since, the particle does not collide with any of the other walls, so pressure on other walls = 0
Considering it to be ideal gas systen, the total energy is = 3/2 KBT
Since, the only contribution to total energy is from Kinetic energy, and K.E = 1/2mv2 then
(1/2)*1.13*10-6 *3552 = 3/2 * 1.38 * 10-23 * T
Thus, T = 3.4398128 × 1021 K
Note that temperature is very large not because the particle has large velocity or that it is just a single particle. But, this is so because the mass of the particle is very large compared to the molecules and atoms that we generally consider.
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