Respuesta :
Explanation:
Moles of nitrogen gas = [tex]n_1=\frac{1.20 g}{28 g/mol}=0.0428 mol[/tex]
Moles of oxygen gas = [tex]n_2=\frac{0.77 g}{32 g/mol}=0.0240 mol[/tex]
Mole fraction of nitrogen gas=[tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]
[tex]\chi_1=\frac{0.0428 mol}{0.0428 mol+0.0240 mol}=0.6407\approx 0.64[/tex]
Mole fraction of oxygen gas=[tex]\chi_2=1-\chi_1=1-0.6407=0.3593\approx 0.36[/tex]
Total umber of moles in container :
n =[tex]n_1+n_2[/tex]= 0.0428 mol + 0.0240 mol = 0.0668 mol
Volume of the container = V = 1.65 L
Temperature of the container = T = 15°C = 288.15 K
Total pressure in the container = P
Using an ideal gas equation:
[tex]PV=nRT[/tex]
[tex]P=\frac{0.0668 mol\times 0.0821 atm L/mol k\times 288.15 K}{1.65 L}[/tex]
P = 0.9577 atm
Partial pressure of nitrogen gas = [tex]p^{o}_1[/tex]
Partial pressure of nitrogen gas = [tex]P^{o}_2[/tex]
Partial pressure of nitrogen gas and oxygen gas can be calculated by using Dalton's law of partial pressure:
[tex]p^{o}_i=p_{total}\times \chi_i[/tex]
[tex]p^{o}_1=P\times \chi_1=0.6135 atm\approx 0.61 atm[/tex]
[tex]p^{o}_2=P\times \chi_2=0.3441 atm\approx 0.34 atm[/tex]
