An equimolar mixture of CO2/CO is in a 1 liter tank at 0C. Initial pressure is 8
ID: 942777 • Letter: A
Question
An equimolar mixture of CO2/CO is in a 1 liter tank at 0C. Initial pressure is 800 psi(a). Calculate the number of moles initially in the tank. If 10% of the moles are lost, what is the final tank pressure (assume the compressibility factor does not change significantly)? However, you are not sure whether the assumption of compressibility factor is correct. In order to validate the assumption, please calculate the compressibility factor at this new pressure. Determine how far were you in assuming a constant compressibility factor.
Explanation / Answer
P = 800psi
1 psi = 6,894.76 pascals
1pascal = 9.869*10-6 atm
P = 800 psi = 54.436 atm
V = 1 liter
T = 0C = 0+ 273 = 273K
PV = nRT
n = PV/RT
= 54.436*1/0.0821*273
= 54.436/22.4 = 2.43 moles
10% of the moles lost = 2.43*10/100 = 0.243 moles lose
PV = nRT
P = nRT/V
= 0.243*0.0821*273/1 = 5.44563 atm
final pressure is 5.44563 atm