Question

1) A mixture of oxygen and ammonia at 273.15 K and 1.00 atm has a volume of 150.0 cm .This mixture is cooled to the temperature of liquid nitrogen at which ammonia freezes out

and the remaining gas is removed from the vessel. The vessel is allowed to warm to 273.15 K and 1 atm, and the volume is now 85.0 cm . Calculate the mole fraction of ammonia in the

original mixture.

2) (a) Use the van der Waals equation to calculate the volume occupied by one mole of Xe gas at 400 K and 20.0 atm pressure. The van der Waals constants are a = 4.19 atm L² mol^-2 and b= 0.051 L mol^-1 .

(b) Determine which is more dominant, the attractive forces or repulsive forces under these conditions. (You must provide an explanation to receive credit).

3) (a) Calculate the critical constants (Pc, Tc, Vc) for CO2 if the van der Waals constants are a = 22.8 dm⁶ bar mol^-2 and b = 0.146 dm³ mol^-1

(b) Calculate the reduced pressure and the reduced temperature when CO2 is found at 1.25 atm and 145C.

Answer 1

Question 1.

According to the ideal gas equation:

PV = nRT

Where 'P' is the pressure of the mixture of oxygen and ammonia = 1 atm

'V' is the volume of the mixture of oxygen and ammonia = 150 cm³ = 150 mL = 0.15 L

'n' is the total no. of moles of oxygen and ammonia

'R' is the universal gas constant = 0.08206 L.atm/mol.K

'T' is the absolute temperature = 273.15 K

i.e. 1 atm * 0.15 L = n * 0.08206 L.atm/mol.K * 273.15 K

Therefore, the total no. of moles of oxygen and ammonia = 0.006692 mol

After removing oxygen gas:

PV = nRT

Where 'P' is the pressure of the ammonia = 1 atm

'V' is the volume of ammonia = 85 cm³ = 85 mL = 0.085 L

'n' is the total no. of moles of ammonia

'R' is the universal gas constant = 0.08206 L.atm/mol.K

'T' is the absolute temperature = 273.15 K

i.e. 1 atm * 0.085 L = n * 0.08206 L.atm/mol.K * 273.15 K

Therefore, the total no. of moles of ammonia = 0.003792 mol

Therefore, the mole fraction of ammonia = moles of ammonia/total moles of oxygen and ammonia = 0.003792 mol/0.006692 mol

= 0.57