Problems on Thermodynamics?

Question

1) Horizontal cylinder (open at the one end) revolves around a vertical axis passing through its open end, on a constant angular speed s. If the surrounding air is an ideal gas of molecular weight M and pressure p, what is the pressure inside the cylinder?

2) A container of volume V contains ideal gas under initial pressure p0 and is connected to a rotational pump of actual volume ΔV. If during pumping the gas temperature remains constant, what is the pressure inside the container after N rotations of the pump.

How do you solve the above? Thanks.

Answer 1

1) Question 1 seems especially difficult since you have a (centripetal) acceleration field that changes very significantly with radius, unlike the atmospheric pressure equation which assumes a constant g over the <100 km altitude range in which it is usually applied. You can say this much:
Assuming this is a static problem (no continuous air flow in and out of the cylinder) and that we can reduce the open end to a small hole in the center, the we know that the pressure at the hole and along the centerline inside is at ambient pressure, or gauge pressure = 0. The centripetal acceleration on the wall of the cylinder is s^2 / radius R. If this were a constant field we could use the atmospheric pressure equation which assumes constant gravitational acceleration g and is an exponential function of altitude that involves g, the universal gas constant R, the temperature T and the molecular weight M. (The latter three being found in the universal gas law, PV=nRT.) But I can't go any further since the acceleration cannot reasonably be assumed constant.
2) The relevant part of the universal gas law, with R*T and V constant, is P1/n1 = P2/n2. I assume you're pumping out of the container, otherwise an item of information (outside pressure) is missing and getitng the answer would be a pretty simple exercise if you had it because each pump stroke would deliver the same amount of gas. However, pumping out of the container, each pump rotation removes deltaV/V times the current value of n. So the final quantity of gas n2 = n1 * (deltaV/V)^N, and the final gas pressure P2 = P1 * (deltaV/V)^N.

Answer 2

First initiate with P1V1 = P2V2 for this reason, P1V1 = 100KN/M (sq) x 0.0.5 m (cube) And, P2V2 = P2 X 0.001 m (cube). for this reason you have, 100KN/M (sq) x 0.0.5 m (cube) = P2 X 0.001 m (cube). P2 = 100KN/M (sq) x 0.0.5 m (cube) / 0.001 m (cube) as quickly as you're carried out looking P2, then you definately will use a similar records and plug them into the equation decrease than. P1V1/T1 = P2V2/T2 you already be attentive to, P1V1, and T1 = 127 ranges celsius. From the 1st equation above, you have already found P2, and you already be attentive to V2, so all we could desire to locate is T2. i desire this facilitates. :)