Question

A gas cylinder contains 55% nitrogen, 20% oxygen, and 25% carbon dioxide by mass, at 760 mm pressure. Calculate the partial pressure of each gas.

Answer 1

According to Dalton's law of partial pressure the total pressure exerted by the mixture of ideal gases is equal to the sum of the partial pressure exerted by the individual gas in the container.
${{P}_{T}}={{P}_{1}}+{{P}_{2}}+{{P}_{3}}...+{{P}_{n}}$

Complete answer:
Let us assume that the mass of the gas is 100 g.
So, 100 g gas will contain 55 g nitrogen, 20 g oxygen, and 25 g carbon dioxide.
Now, we know that the atomic mass of nitrogen is 28 u, the atomic mass of oxygen is approximately 20 u, and the atomic mass of carbon is 12 u.
In the free state, nitrogen gas exists as dinitrogen $({{N}_{2}})$, so its molecular mass will be
${{M}_{{{N}_{2}}}}=2\times 14=28$
Similarly, in the free state, oxygen gas exists as dioxygen $({{O}_{2}})$, so its molecular mass will be
${{M}_{{{O}_{2}}}}=2\times 16=32$
And finally, the molecular mass of carbon dioxide $(C{{O}_{2}})$ gas will be
${{M}_{C{{O}_{2}}}}=12+2\times 16=44$
Now, the number of moles of the gas in the container can be calculated by the formula
$n=\dfrac{W}{M}$
Where W is the mass of gas present and M is the molecular mass of the gas.
So, the moles of nitrogen gas in the container will be
${{n}_{{{N}_{2}}}}=\dfrac{55}{28}$= 1.964 moles
The moles of oxygen gas in the container will be
${{n}_{{{O}_{2}}}}=\dfrac{20}{32}$= 0.625 moles
And the moles of the carbon dioxide gas in the container will be
${{n}_{C{{O}_{2}}}}=\dfrac{25}{44}$= 0.568 moles
The total number of moles in the container is
${{n}_{T}}={{n}_{{{N}_{2}}}}+{{n}_{{{O}_{2}}}}+{{n}_{C{{O}_{2}}}}$
${{n}_{T}}$= 1.964 + 0.625 + 0.568
${{n}_{T}}$ = 3.157 moles
The ratio of moles of a particular gas to the total number of moles in a container is known as the mole ratio of that gas. It is denoted by symbol X.
$X=\dfrac{n}{{{n}_{T}}}$
Now we know that the pressure exerted by each gas in a container filled with more than one gas is known as partial pressure.
For an ideal gas, the ratio of the partial pressure of a gas and the total partial pressure exerted by the system is equal to the mole ratio of the gas.
X =$\dfrac{n}{{{n}_{T}}}=\dfrac{P}{{{P}_{T}}}$
Or, we can say that the partial pressure of a gas in a system
$P=\dfrac{n}{{{n}_{T}}}\times {{P}_{T}}$
Now, it is given to us that the total pressure of the container ${{P}_{T}}$= 760 mm.
So, the partial pressure of the nitrogen gas will be
${{P}_{{{N}_{2}}}}=\dfrac{1.964}{3.157}\times 760$= 472.8 mm
The partial pressure of oxygen gas will be
${{P}_{{{O}_{2}}}}=\dfrac{0.625}{3.157}\times 760$= 150.5 mm
The partial pressure of carbon dioxide will be
${{P}_{C{{O}_{2}}}}=\dfrac{0.568}{3.157}\times 760$= 136.7 mm

Note:
It should be noted that while Dalton's law of partial pressure holds when the pressure is low, whereas it deviates significantly at high pressures. Also, it is assumed that the mixture of gases in the container is non-reactive or ideal.