Suppose that a 1.00 cm3 box in the shape of a cube is perfectly evacuated, except for a single particle of mass 1.13 x 10^-3 g. The particle is initially moving perpendicular to one of the walls of the box at a speed of 525 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.
a) d=m/v
d= (1.13*10^-3)/1
d= 1.13e-3 g/cm^3
c) 0 Pa
To solve this problem, we need to use the laws of physics, including the principles of kinetic theory and gas laws. Let's break down the problem step by step.
(a) Find the mass density inside the box:
The mass density (ρ) is defined as the mass (m) divided by the volume (V):
ρ = m/V
Given that the particle has a mass of 1.13 x 10^-3 g and the volume of the box is 1.00 cm^3, we can convert the units to kg and m^3:
ρ = (1.13 x 10^-3 g) / (1.00 cm^3)
= (1.13 x 10^-6 kg) / (1.00 x 10^-6 m^3)
= 1.13 kg/m^3
Therefore, the mass density inside the box is 1.13 kg/m^3.
(b) Find the average pressure on the walls perpendicular to the particle's path:
The average pressure (P) on the walls can be calculated using the equation:
P = 2 * ρ * v^2
where ρ is the mass density and v is the velocity of the particle perpendicular to the wall.
Given that the speed of the particle is 525 m/s, the equation becomes:
P = 2 * (1.13 kg/m^3) * (525 m/s)^2
= 1.13 kg/m^3 * 275,625 m^2/s^2
= 311,312.5 N/m^2
Therefore, the average pressure on the walls perpendicular to the particle's path is 311,312.5 N/m^2.
(c) Find the average pressure on the other walls:
Since the particle is moving perpendicular to one of the walls, its collision with the other walls is assumed to be perfectly elastic. In an elastic collision, momentum is conserved, but kinetic energy is also conserved. Therefore, the average pressure on the other walls will be the same as the average pressure on the walls perpendicular to the particle's path, which we calculated to be 311,312.5 N/m^2.
Therefore, the average pressure on the other walls is also 311,312.5 N/m^2.
(d) Find the temperature inside the box:
To find the temperature, we can use the ideal gas law equation:
P * V = n * R * T
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Since the box is perfectly evacuated except for a single particle, the number of moles, n, can be approximated as the mass of the particle divided by the molar mass of the particle. Assuming the particle is an ideal gas, we can use the molar mass of air, which is approximately 29 g/mol.
n = (1.13 x 10^-3 g) / (29 g/mol)
≈ 3.90 x 10^-5 mol
Given that the volume of the box is 1.00 cm^3, we can convert it to m^3:
V = (1.00 cm^3) / (1.00 x 10^6 cm^3/m^3)
= 1.00 x 10^-6 m^3
Now we can rearrange the ideal gas law equation to solve for temperature (T):
T = (P * V)/(n * R)
Substituting the values:
T = (311,312.5 N/m^2) * (1.00 x 10^-6 m^3) / (3.90 x 10^-5 mol * 8.314 J/(mol*K))
= 1.00 x 10^12 N m^2 / (3.90 x 10^-5 J K)
≈ 2.56 x 10^16 K
Therefore, the temperature inside the box is approximately 2.56 x 10^16 Kelvin.
Note: The temperature obtained is very high, which indicates that the assumption of an ideal gas for the particle may not be valid in this case.