Respuesta :
Explanation:
Mass of methane gas = 12 g
Mass of nitrogen gas = 1 g
Mass of carbon dioxide = 15 g
Volume of the container ,V= 30 L
Temperature of the gases,T= 25°C = 298.15 K=
a) Moles of methane gas:
[tex]n_1=\frac{12 g}{16 g/mol}=0.75 mol[/tex]
Moles of nitrogen gas:
[tex]n_2=\frac{1 g}{28 g/mol}=0.0357 mol[/tex]
Moles of carbon dioxide gas:
[tex]n_3=\frac{15 g}{44 g/mol}=0.3409 mol[/tex]
b) Partial pressure exerted by each gas.
The total pressure of the gases can be calculated by using an ideal gas equation:
[tex]PV=nRT[/tex]
[tex]n=n_1+n_2+n_3=1.1266 mol[/tex]
[tex]P=\frac{1.1266 mol\times 0.0821 atm L/mol K\times 298.15 K}{30 L}[/tex]
P = 0.92 atm = Total pressure of the mixture
Partial pressure of all the gases can be determined by using Dalton's law of partial pressure
Partial pressure of methane gas
[tex]p^o_{CH_4}=P\times \frac{n_1}{n_1+n_2+n_3}=p\times {n_1}{n}[/tex]
[tex]p^o_{CH_4}=0.92 atm\times \frac{0.75 mol}{1.1266 mol}=0.61 atm[/tex]
Partial pressure of nitrogen gas
[tex]p^o_{N_2}=P\times \frac{n_2}{n_1+n_2+n_3}=p\times {n_2}{n}[/tex]
[tex]p^o_{N_2}=0.92 atm\times \frac{0.0357 mol}{1.1266 mol}=0.029 atm[/tex]
Partial pressure of carbon dioxide gas
[tex]p^o_{CO_2}=P\times \frac{n_3}{n_1+n_2+n_3}=p\times {n_3}{n}[/tex]
[tex]p^o_{CO_2}=0.92 atm\times \frac{0.3409 mol}{1.1266 mol}=0.27 atm[/tex]
c) The total pressure exerted by the mixture is 0.92 atm.
d) Percentage by volume of each gas in the mixture
Volume of the methane gas:
[tex]V_1=\frac{n_1\times RT}{P}=\frac{0.75 mol\times 0.0821 atm L/mol K \times 298.15 K}{0.92 atm}=19.95 L[/tex]
Volume percentage of methane :
[tex]\frac{V_1}{V}\times 100=\frac{19.95 L}{30 L}\times 100=66.5\%[/tex]
Volume of the nitrogen gas:
[tex]V_2=\frac{n_2\times RT}{P}=\frac{0.0357 mol\times 0.0821 atm L/mol K \times 298.15 K}{0.92 atm}=0.95 L[/tex]
Volume percentage of nitrogen:
[tex]\frac{V_2}{V}\times 100=\frac{0.95 L}{30 L}\times 100=3.16\%[/tex]
Volume of the carbon dioxide gas:
[tex]V_3=\frac{n_3\times RT}{P}=\frac{0.3409 mol\times 0.0821 atm L/mol K \times 298.15 K}{0.92 atm}=9.06 L[/tex]
Volume percentage of carbon dioxide:
[tex]\frac{V_1}{V}\times 100=\frac{9.06 L}{30 L}\times 100=30.2\%[/tex]
