Respuesta :
Answer:
Partial pressure of methane: 1.18 atm
Partial pressure of ethane: 1.45 atm
Partial pressure of propane: 2.35 atm
Explanation:
Let the total moles of gases in a container be n.
Total pressure of the gases in a container =P = 5.0 atm
Temperature of the gases in a container =T = 23°C = 296.15 K
Volume of the container = V = 10.0 L
[tex]PV=nRT[/tex] (Ideal gas equation)
[tex]n=\frac{PV}{RT}=\frac{5.0 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.0564 mol[/tex]
Moles of methane gas =[tex]n_1=\frac{8.00 g}{16.04 g/mol}=0.4878 mol[/tex]
Moles of ethane gas =[tex]n_2=\frac{18.00 g}{30.07 g/mol}=0.5986 mol[/tex]
Moles of propane gas =[tex]n_3=?[/tex]
[tex]n=n_1+n_2+n_3[/tex]
[tex]n_3=n-n_1-n_2=2.0564 mol-0.4878 mol-0.5986 mol= 0.9700 mol[/tex]
Partial pressure of all the gases can be calculated by using Raoult's law:
[tex]p_i=P\times \chi_i[/tex]
[tex]p_i[/tex] = partial pressure of 'i' component.
[tex]\chi_1[/tex] = mole fraction of 'i' component in mixture
P = total pressure of the mixture
Partial pressure of methane:
[tex]p_1=P\times \chi_1=P\times \frac{n_1}{n_1+n+2+n_3}=P\times \frac{n_1}{n}[/tex]
[tex]p_1=5.00 atm\times \frac{0.4878 mol}{2.0564 mol}=1.18 atm[/tex]
Partial pressure of ethane:
[tex]p_2=P\times \chi_2=P\times \frac{n_2}{n_1+n+2+n_3}=P\times \frac{n_2}{n}[/tex]
[tex]p_2=5.00 atm\times \frac{0.5986 mol}{2.0564 mol}=1.45 atm[/tex]
Partial pressure of propane:
[tex]p_3=P\times \chi_3=P\times \frac{n_3}{n_1+n+2+n_3}=P\times \frac{n_3}{n}[/tex]
[tex]p_3=5.00 atm\times \frac{0.9700 mol}{2.0564 mol}=2.35 atm[/tex]
