An oxygen–nitrogen mixture consists of 35 kg of oxygen and 40 kg of nitrogen. This mixture is cooled to 84 K at 0.1 MPa pressure. Determine the mass of the oxygen in the liquid and gaseous phase.

The mass of oxygen in liquid phase = 14.703 kg

The mass of oxygen in the vapor phase = 20.302 kg

Explanation:

Given that:

The mass of the oxygen = 35 kg

The mass of the nitrogen = 40 kg

The cooling temperature of the mixture T = 84 K

The cooling pressure of the mixture P = 0.1 MPa

From the equilibrium diagram for te-phase mixture od oxygen-nitrogen as 0.1 MPa graph. The properties of liquid and vapor percentages are obtained.

i.e.

Liquid percentage of = 70% = 0.70

Vapor percentage of = 34% = 0.34

The molar mass (mm) of oxygen and nitrogen are 32 g/mol and 28 g/mol respectively

Thus, the number of moles of each component is:

number of moles of oxygen = 35/32

number of moles of oxygen =  1.0938 kmol

number of moles of nitrogen = 40/28

number of moles of nitrogen = 1.4286  kmol

Hence, the total no. of moles in the mixture is:

So, the total no of moles in the whole system is:

The total number of moles for oxygen in the system is

From equation (1), let N_f = 2.5224 - N_g, then replace the value of N_f into equation (2)

∴

0.7(2.5224 - N_g) + 0.34 N_g = 1.0938

1.76568 - 0.7 N_g + 0.34 N_g = 1.0938

1.76568 - 0.36 N_g = 1.0938

1.76568 - 1.0938 = 0.36 N_g

0.67188  = 0.36 N_g

N_g = 0.67188/0.36

N_g = 1.866

From equation (1)

N_f + 1.866 = 2.5224

N_f = 2.5224 - 1.866

N_f = 0.6564

Thus, the mass of oxygen in the liquid and vapor phases is:

The mass of oxygen in liquid phase = 14.703 kg

The mass of oxygen in the vapor phase = 20.302 kg

answer:

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explanation: