Respuesta :
Answer:
Following are the solution to this question:
Explanation:
let,
[tex]\text{m= mass of the gas}\\\text{M= molar weight of the gas}\\ m_{N_2} = 1 \kg \\T_{N_2}= 298 \ K\\ M_{N_2} = 28 \ \frac{kg}{kmol}\\ P_{N_2}= 300 \ kPa\\m_{O_2} = 3 \ kg \\T_{O_2} = 298 \ K \\M_{O_2}=32 \ \frac{kg}{kmol}\\ P_{O_2} = 500 \ kPa \\T_m= 298 \ K[/tex]
Calculating the mole of nitrogen:
[tex]N_{N_2} = \frac{m_{N_2}}{M_{N_2}}[/tex]
[tex]=\frac{1}{28} \\\\= 0.0357 \ kmol[/tex]
Calculating the mole oxygen:
[tex]N_{O_2} = \frac{m_{O_2}}{M_{O_2}}\\\\=\frac{3}{32} \\\\= 0.09375 \ kmol[/tex]
Calculating the total mole:
[tex]N_m=N_{N_2}+N_{O_2}\\\\= 0.0357+0.09375\\\\= 0.1295 \ kmol[/tex]
Calculating the volume of nitrogen:
[tex]V_{N_2} =\frac{N_{N_2} \times R_{u}\times T_{N_2}}{P_{N_2}}\\\\= \frac{0.0357 \times 8.314 \times 298}{300}\\\\= 0.295 m^3[/tex]
Calculating the volume of oxygen:
[tex]V_{O_2} =\frac{N_{O_2} \times R_{u}\times T_{O_2}}{P_{O_2}}\\\\= \frac{0.09375 \times 8.314 \times 298}{500}\\\\= 0.465 m^3[/tex]
Calculating the total volume:
[tex]V_m=V_{N_2}+V_{O_2}\\\\=0.295+0.465 \\\\=0.76 \ m^3[/tex]
Calculating the mixture pressure:
[tex]P_m = \frac{N_{m} \times R_u \times T_m}{V_m} \\\\= \frac{0.1295 \times 8.314 \times 298}{0.76} \\\\= 422.2\ kPa[/tex]
