Question

Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 ∘C, the total pressure in the container is 5.00 atm . Calculate the partial pressure of each gas in the container.

Answer 1

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

$PV=nRT$ (Ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.0 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.0564 mol$

Moles of methane gas =$n_1=\frac{8.00 g}{16.04 g/mol}=0.4878 mol$

Moles of ethane gas =$n_2=\frac{18.00 g}{30.07 g/mol}=0.5986 mol$

Moles of propane gas =$n_3=?$

$n=n_1+n_2+n_3$

$n_3=n-n_1-n_2=2.0564 mol-0.4878 mol-0.5986 mol= 0.9700 mol$

Partial pressure of all the gases can be calculated by using Raoult's law:

$p_i=P\times \chi_i$

$p_i$ = partial pressure of 'i' component.

$\chi_1$ = mole fraction of 'i' component in mixture

P = total pressure of the mixture

Partial pressure of methane:

$p_1=P\times \chi_1=P\times \frac{n_1}{n_1+n+2+n_3}=P\times \frac{n_1}{n}$

$p_1=5.00 atm\times \frac{0.4878 mol}{2.0564 mol}=1.18 atm$

Partial pressure of ethane:

$p_2=P\times \chi_2=P\times \frac{n_2}{n_1+n+2+n_3}=P\times \frac{n_2}{n}$

$p_2=5.00 atm\times \frac{0.5986 mol}{2.0564 mol}=1.45 atm$

Partial pressure of propane:

$p_3=P\times \chi_3=P\times \frac{n_3}{n_1+n+2+n_3}=P\times \frac{n_3}{n}$

$p_3=5.00 atm\times \frac{0.9700 mol}{2.0564 mol}=2.35 atm$