Question

. Air in a 0.3 m3 cylinder is initially at a pressure of 10 bar and a temperature of 330 K. The cylinder is to be emptied by opening a valve and letting the pressure drop to that of the atmosphere. What will be the final temperature and mass of gas in the cylinder if this is accomplished:

a) In a manner that maintains the temperature of the gas at 330 K?

b) In a well-insulated cylinder (Answer: Tfinal = 171 K).

Answer 1

Given data :

Cylinder volume V1 = 0.3 m3

Initially pressure P1 = 10 bar = 10*10^5 Pa = 10 ^6 Pa

Temperature T1 = 330 K

a) cylinder emptied by a valve and letting pressure drop

P2 = 1 atm = 101325 Pa

For this condition temperature remains same T2 = 330 K

This process is isothermal process for this process we have relation,

PV = constant

P1V1 = P2V2

V2 = P1V1/P2 = 10^6Pa*0.3m3/101325Pa = 2.96 m3

After letting pressure drop 1 atm then gas having volume V2 = 2.96 m3

Then we find volume of air expelled = 2.96 -0.3 = 2.66 m3

We have data for final condition,

P2 = 101325 Pa

V2 = 2.96 m3

T2 = 330 k

Then by ideal gas equation,

Initial moles of gas,  

P1V1= n1RT2

n1 = P1V1/RT1 = (10^6 Pa * 0.3m3)/(8.314Pa.m3/mol.k)*330k = 109.31 mol

Expelled air moles

n2 = P2V2/RT2 = 101325Pa*2.66m3/8.314*330 = 98.23 mol

Moles having cylinder after explled air,

n = n1 - n2 = 109.31 - 98.23 = 11.07 moles

Or becuase cylinder accomplish air then final volume of air Inside the cylinder V = 0.3 m3 remain same

n = 101325*0.3/8.314*330 = 11.07 mol

Molecular weight of air, M = 29 gm/mol

Mass of air inside the tank after getting pressure 1 atm,

m = n*M = 11.07mol*29 gm/mol = 321.12 gm = 0.321 kg

mass of air = 0.321 kg

Final temperature of air T2 = 330K.

b) well insulated cylinder it means this process is adiabatic process.

PV ^Y = constant

Y - specific heat ratio = Cp/Cv = 1.4 for air

P2V2^Y = P1V1^Y

V2 = V1*(P1/P2)^(1/1Y)

V2 = 1.539 m3

Final volume of air,

PV^1.4 = const

T2 = T1*(P2/P1)^(Y-1)/Y = 330*(101325/10^6)^(1.4-1)/1.4

T2 = 171.56 K

final temperatue of the gas when pressure drop to 1 atm, T2 = 171.56 K

amount of air inside the tank after having final condition

P2 = 101325 Pa

V2 = 1.539 m3

T2 = 171.56 K

Expelled air, V = V2 - V1 = 1.539 - 0.3 = 1.239 m3

Initial moles of air n1 = 109.31 moles

by ideal gas equation,

Moles of expelled air,

n2 = P2V2/RT2 = 101325Pa*1.239m3/(8.314Pa.m3/mol.k)*171.56 k = 88.01 mol

After getting pressure drop moles of air having accomplished cylinder,

n = n1 - n2 = 109.31 - 88.01 = 21.29 mol

Or n = 0.3*101325/8.314*171.56 = 21.29 mol

Mass of air left m = 21.29 mol*29 gm/mol = 617.52 gm = 0.617 kg

mass of air left m = 0.617 kg.

Final temperature of gas T2 = 171.56 K