Question

What is the partial pressure of $C{O_2}$ in a container that holds $5$ moles of $C{O_2}$, $3$ moles of ${N_2}$, and $1$ mole ${H_2}$and has a total pressure of $1.05$ atm?

Answer 1

The amount of pressure which is exerted by every gas individually within a gaseous mixture is said to be the gas’s partial pressure. Since we are given number of moles of the gases along with the value of pressure, we will try to find a law that explains the relationship between mole fractions, partial pressure and total pressure.

Complete answer:
Let us first understand the values given to us;
There is a mixture of $C{O_2}$, ${N_2}$ and ${H_2}$ inside a container.
Number of moles of $C{O_2}$, ${N_2}$ and ${H_2}$ are $5,\;3$ and $1$ respectively.
The total pressure exerted by the gases onto the container is given as $1.05\;atm$.
From the given data we know that we can find the mole fraction of any constituent of the gas. This is because we are given the ‘number of moles’ present in each individual gas within the mixture of gases.
Note that the law which explains the relationship among partial pressures and number of moles is Dalton’s law of partial pressure.
Dalton’s law establishes the formula;
$\dfrac{{{P_x}}}{{{P_{total}}}} = \dfrac{{{n_x}}}{{{n_{total}}}} = \dfrac{{{V_x}}}{{{V_{total}}}}$
Here the part of the equation we can use is: $\dfrac{{{P_x}}}{{{P_{total}}}} = \dfrac{{{n_x}}}{{{n_{total}}}}$, since we have ${P_{total}} =$ the total pressure exerted, ${n_x} =$ number of moles of individual gases, ${n_{total}} =$ total number of moles in the gas. From here $\dfrac{{{n_x}}}{{{n_{total}}}}$ is called as mole fraction.
From the question we have;
Total pressure ${P_{total}} = 1.05\;atm$
$C{O_2}$’s mole count ${n_{C{O_2}}} = 5$
Total mole count is sum of each number of moles so ${n_{total}} = 9$
Now the law becomes;
$\Rightarrow \dfrac{{{P_{C{O_2}}}}}{{{P_{total}}}} = \dfrac{{{n_{C{O_2}}}}}{{{n_{total}}}}$
Now to find the partial pressure of $C{O_2}$, denoted here as ${P_{C{O_2}}}$ we can substitute all the given values in the formula of Dalton’s Law;
$\Rightarrow {P_{C{O_2}}} = \dfrac{{{n_{C{O_2}}}}}{{{n_{total}}}} \times {P_{total}}$
Putting in the values;
$\Rightarrow {P_{C{O_2}}} = \dfrac{5}{9} \times 1.05\;atm$
Simplifying we get the value of partial pressure of $C{O_2}$ that is;
$\Rightarrow {P_{C{O_2}}} = 0.58\;atm$
Therefore the partial pressure exerted by $C{O_2}$ within the contain of mixture of $C{O_2},\;{N_2}$ and ${H_2}$ is equal to $0.58\;atm$.

Note:
Remember that we are able to make equations to find the partial pressures only because we have assumed the ideal gas law to be true. In the ideal gas law the assumption states that every gas has behaves in the same way and it is independent of the intermolecular attraction and they will have the following equation at constant temperature and volume;
$P = n(\dfrac{{RT}}{V})$ ; where $P =$ pressure of gas, $V =$ constant volume, $T =$ given temperature, $R =$ universally accepted gas constant and $n =$ number of moles.