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 : The partial pressure of 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.

and,

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.

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 =

T = Temperature of gas =

Putting values in above equation, we get:

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

and,

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

According to the Raoult's law,

where,

= partial pressure of gas

= total pressure of gas  = 1.12 atm

= mole fraction of gas

and,

Thus, the partial pressure of is, 0.281 atm and 0.839 atm respectively and the total pressure in vessel is, 1.12 atm.