Answer:
(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
