Air in a 0.3 m3 cylinder is initially at a pressure of 10 bar and a temperature of 330K. The cylinder is to be emptied by opening a valve and letting the pressure drop to that of the atmosphere. What will be the temperature and mass of gas in the cylinder if this is to be accomplished? A. In a manner that maintains the temperature of the gas at 330 K. B. In a well-insulated cylinder. For simplicity assume, in part (b), that the process oc- curs suficiently rapidly that there is no heat transfer between the cylinder walls and the gas. The gas is ideal, and Cp 29 (J/mol K).

(a) Temperature = 330 K, and mass = 0.321 kg

(b) T₂ = 171.56 K, mass = 0.32223 kg

Explanation:

For a constant temperature process we have

p₁v₁ = p₂v₂

Where p₁ = initial pressure = 10 bar = 1000000 Pa

p₂ = final pressure = 1 atm = 101325 Pa

v₁ initial volume = 0.3 m³

v₂ = final volume = unknown

From the relation we have v₂ = 2.96 m³

Therefore at constant temperature 2.93 m³ - 0.3 m³ or 2.66 m³ will be expelled from the container

Temperature = 330 K, and mass =

Also from the relation p1v1 = mRT1

We have, (1000000×0.3)/(8314×330) = 109..337 mole

For air mass

Mass = 3.171 kg

After opening we have

p2v2/(RT1) = n2 = 11.07 mol or 0.321 kg

or

(b) This is said to be adiabatic condition hence

Here

But cp = 29 (J/mol K).

and p₁v₁ = RT₁ therefore R = 1000000*0.3/330 = 909.1 J/mol·K

And For perfect gas γ = 1.4

Hence T₂ = 171.56 K

γ =cp/cv therefore cv=cp/γ = 29/1.4 = 20.714 (J/mol K). and R =cp-cv = 8.29 J/mol·K

Therefore p1v1/(RT1) = 109.66 moles and we have

p2v2/(R×T2) = 11.11 mole left

For air that is 0.32223 kg

the atomic mass is the bottom number on the element.

-steel jelly