# Liquid in a vessel in a vacuum chamber

A vessel holding some liquid is placed in a vacuum chamber that is constantly pumped. Gas leaks from the vessel into the vacuum through a small hole, radius 4.15 micro m, but the pressure inside the vessel remains 38.6 kPa. After a day mass of the vessel has dropped from 100 g to 99.5 g. Assuming the container has at all times been at 70 K, suggest (with reasoning) the identity of the liquid.

See the attached file.

Â© BrainMass Inc. brainmass.com December 24, 2021, 7:22 pm ad1c9bdddfhttps://brainmass.com/physics/boltzmann-distribution/liquid-vessel-vacuum-chamber-183098

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please refer to the attachment.

The vessel with a pin hole contains the liquefied gas. Above the liquid in the vessel is the gas. The gas molecules are in a state of random motion, in the process colliding with each other as well as the sides of the container. Pressure on the sides of the container is the result of this bombardment of the sides by the gas molecules. Let the average speed with which a molecule collides with the sides of the container be v. Assuming a perfectly elastic collision, the molecule bounces back with the same speed resulting in a change of momentum = mv - (-mv) = 2mv (m is the mass of the molecule). Let there be n collisions per sec of the molecules with a certain area of the sides. Hence, change of momentum per sec or rate of change of momentum = 2mvn. As per Newton's second law of motion, rate of change of momentum = force. Hence, 2mvn gives the force acting on the given area of the sides.

The molecules which "collide" with the hole, escape from the vessel into the vacuum chamber. Area of the hole = Î r2 = 3.14 x (4.15 x 10-6)2 = 54.11 x 10-12 m2

Pressure inside the vessel = 38.6 x 103 Pa

Total force acting on the hole from within the vessel = Pressure x Area = 38.6 x 103 x 54.11 x 10-12 = 2088 x 10-9 N. Equating this to 2mvn we get :

2mvn = 2088 x 10-9 where n is the number of molecules escaping per sec ......(1)

Loss of gas due to effusion in 24 hours= 100 - 99.5 = 0.5g or 0.5 x 10-3 kg

Loss of gas per sec = 0.5 x 10-3/(24 x 60 x 60) = 5.7 x 10-9 kg/sec .......(2)

If n = number of molecules escaped per sec and m = mass of each molecule, then loss of gas per sec = mn ........(3)

From (2) and (3) we get : mn = 5.7 x 10-9 kg/sec .......(4)

Substituting for mn in (1) we get : 2v (5.7 x 10-9) = 2088 x 10-9

Or v = 183 m/s

As per Maxwellian speed distribution, average speed of gas molecules is given by :

________

Cave = âˆš(8kT/Î m) where k = Boltzmann constant, T = Temperature

Substituting Cave = 183 m/s, T = 70 K and k = 1.38 x 10-23 J/K, we get :

___________________

183 = âˆš(8x1.38x10-23x70/3.14m)

__________

Or 183 = âˆš246x10-23/m

Squaring both sides and solving for m, we get : m = 73.5 x 10-27 kg

Mass of a hydrogen atom = 1.6726 x 10-27 kg

Hence, molecular mass of the gas = 73.5 x 10-27/1.6726 x 10-27 = 44

As we know, CO2 has a molecular mass of 44. Hence, the liquid in the vessel is liquefied CO2.

Â© BrainMass Inc. brainmass.com December 24, 2021, 7:22 pm ad1c9bdddf>https://brainmass.com/physics/boltzmann-distribution/liquid-vessel-vacuum-chamber-183098