Question

A 30-liter volume of gas at 25°C contains 12 g of methane, 1 g of nitrogen, and 15 g of carbon dioxide. Calculate (a) the moles of each gas present, (b) the partial pressure exerted by each gas, (c) the total pressure exerted by the mixture, and (d) the percentage by volume of each gas in the mixture. You may assume ideal gas behavior

Answer 1

Step-by-step 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:

`n_1=(12 g)/(16 g/mol)=0.75 mol`

Moles of nitrogen gas:

`n_2=(1 g)/(28 g/mol)=0.0357 mol`

Moles of carbon dioxide gas:

`n_3=(15 g)/(44 g/mol)=0.3409 mol`

b) Partial pressure exerted by each gas.

The total pressure of the gases can be calculated by using an ideal gas equation:

`PV=nRT`

`n=n_1+n_2+n_3=1.1266 mol`

`P=(1.1266 mol* 0.0821 atm L/mol K* 298.15 K)/(30 L)`

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

`p^o_(CH_4)=P* (n_1)/(n_1+n_2+n_3)=p* {n_1}{n}`

`p^o_(CH_4)=0.92 atm* (0.75 mol)/(1.1266 mol)=0.61 atm`

Partial pressure of nitrogen gas

`p^o_(N_2)=P* (n_2)/(n_1+n_2+n_3)=p* {n_2}{n}`

`p^o_(N_2)=0.92 atm* (0.0357 mol)/(1.1266 mol)=0.029 atm`

Partial pressure of carbon dioxide gas

`p^o_(CO_2)=P* (n_3)/(n_1+n_2+n_3)=p* {n_3}{n}`

`p^o_(CO_2)=0.92 atm* (0.3409 mol)/(1.1266 mol)=0.27 atm`

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:

`V_1=(n_1* RT)/(P)=(0.75 mol* 0.0821 atm L/mol K * 298.15 K)/(0.92 atm)=19.95 L`

Volume percentage of methane :

`(V_1)/(V)* 100=(19.95 L)/(30 L)* 100=66.5\%`

Volume of the nitrogen gas:

`V_2=(n_2* RT)/(P)=(0.0357 mol* 0.0821 atm L/mol K * 298.15 K)/(0.92 atm)=0.95 L`

Volume percentage of nitrogen:

`(V_2)/(V)* 100=(0.95 L)/(30 L)* 100=3.16\%`

Volume of the carbon dioxide gas:

`V_3=(n_3* RT)/(P)=(0.3409 mol* 0.0821 atm L/mol K * 298.15 K)/(0.92 atm)=9.06 L`

Volume percentage of carbon dioxide:

`(V_1)/(V)* 100=(9.06 L)/(30 L)* 100=30.2\%`