Question

A mixture of 6.00 g of O2 (g) and 9.00 g of CH4 (g) is placed in a 15.0 L vessel at 0o C. What is the partial pressure of each gas? What is the total pressure in the vessel?

Answer 1

Answer : The partial pressure of
`O_2\text{ and }CH_4` is, 0.281 atm and 0.839 atm respectively.

The total pressure in vessel is, 1.12 atm.

Explanation :

First we have to calculate the moles of oxygen and methane.

`\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}=(6.00g)/(32g/mol)=0.188mol`

and,

`\text{Moles of }CH_4=\frac{\text{Given mass }CH_4}{\text{Molar mass }CH_4}=(9.00g)/(16g/mol)=0.562mol`

Now we have to calculate the total moles of gas.

Total moles of gas = Moles of oxygen + Moles of methane

Total moles of gas = 0.188 + 0.562

Total moles of gas = 0.75 mol

Now we have to calculate the total pressure of gas by using ideal gas equation.

`PV=nRT`

where,

P = total pressure of gas = ?

V = total volume of gas = 15.0 L

n = total number of moles gas = 0.75 mole

R = Gas constant =
`0.0821L.atm/mol.K`

T = Temperature of gas =
`0^oC=273+0=273K`

Putting values in above equation, we get:

`P* 15.0L=0.75mole* (0.0821L.atm/mol.K)* 273K`

`P=1.12atm`

Now we have to calculate the mole fraction of oxygen and methane.

`\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }O_2+\text{Moles of }CH_4}=(0.188)/(0.188+0.562)=0.251`

and,

`\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }O_2+\text{Moles of }CH_4}=(0.562)/(0.188+0.562)=0.749`

Now we have to calculate the partial pressure of oxygen and methane.

According to the Raoult's law,

`p_i=X_i* p_T`

where,

`p_i` = partial pressure of gas

`p_T` = total pressure of gas = 1.12 atm

`X_i` = mole fraction of gas

`p_(O_2)=X_(O_2)* p_T`

`p_(O_2)=0.251* 1.12atm=0.281atm`

and,

`p_(CH_4)=X_(CH_4)* p_T`

`p_(CH_4)=0.749* 1.12atm=0.839atm`

Thus, the partial pressure of
`O_2\text{ and }CH_4` is, 0.281 atm and 0.839 atm respectively and the total pressure in vessel is, 1.12 atm.