Answer :
Answer:
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 [tex]m_{O_2}[/tex] = 35 kg
The mass of the nitrogen [tex]m_{N_2}[/tex] = 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 [tex]O_2[/tex] = 70% = 0.70
Vapor percentage of [tex]O_2[/tex] = 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:
[tex]N_{total} = 1.0938+1.4286[/tex]
[tex]N_{total} = 2.5224 \ kmol[/tex]
So, the total no of moles in the whole system is:
[tex]N_f + N_g = 2.5224 --- (1)[/tex]
The total number of moles for oxygen in the system is
[tex]0.7 \ N_f + 0.34 \ N_g = 1.0938 --- (2)[/tex]
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)
[tex]N_f + N_g = 2.5224[/tex]
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:
[tex]m_{fO_2} = 0.7 \times 0.6564 \times 32[/tex]
[tex]m_{fO_2} = 14.703 \ kg[/tex]
The mass of oxygen in liquid phase = 14.703 kg
[tex]m_{g_O_2} = 0.34 \times 1.866 \times 32[/tex]
[tex]m_{g_O_2} = 20.302 \ kg[/tex]
The mass of oxygen in the vapor phase = 20.302 kg
