# A flask contains three times as many moles as H_2 as it does O_2 gas. If hydrogen and oxygen are the only present gases, what is the total pressure flask if the partial pressure due to oxygen is P?

Mar 12, 2016

${P}_{\text{total}} = 4 \times P$

#### Explanation:

According to Dalton's Law of Partial Pressures, the total pressure of a gaseous mixture will be equal to the individual partial pressure of each component of the mixture.

Simply put, to get the total pressure of a gaseous mixture, you add up the partial pressure you'd get for each component of that mixture if it occupied the volume of the mixture alone.

Mathematically, this can written as

color(blue)(|bar(ul(color(white)(a/a)P_"total" = sum_i P_icolor(white)(a/a)|)))" ", where

${P}_{\text{total}}$ - the total pressure of the gaseous mixture
${P}_{i}$ - the partial pressure of component $i$ of said mixture

In your case, the mixture is said to contain hydrogen gas, ${\text{H}}_{2}$, and oxygen gas, ${\text{O}}_{2}$, in a $3 : 1$ mole ratio.

This means that if you take ${n}_{{O}_{2}}$ to be the number of moles of oxygen gas present in the mixture, you can say that

n_(H_2) = 3 xx n_(O_2)" "color(purple)("(*)")

Now, let's assume that the mixture is being held in a container that has a volume of $V$ liters. If you take $P$ to be the partial pressure of oxygen gas, you can use the ideal gas law equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}}$

to express $V$ in terms of $P$ and ${n}_{{O}_{2}}$. You will thus have

P * V = n_(O_2) * RT implies V = n_(O_2)/P * RT" "color(red)("(*)")

Now, you can write the same thing for ${P}^{'}$, the partial pressure of hydrogen gas in the same volume $V$

${P}^{'} \cdot V = {n}_{{H}_{2}} \cdot R T \implies V = {n}_{{H}_{2}} / {P}^{'} \cdot R T$

Use equation $\textcolor{red}{\text{(*)}}$ to write

${n}_{{O}_{2}} / P \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}} = {n}_{{H}_{2}} / {P}^{'} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}$

Rearrange to find ${P}^{'}$, the partial pressure of hydrogen gas

${P}^{'} = {n}_{{H}_{2}} / {n}_{{O}_{2}} \cdot P$

Use equation $\textcolor{p u r p \le}{\text{(*)}}$ to write

${P}^{'} = \frac{3 \times \textcolor{red}{\cancel{\textcolor{b l a c k}{{n}_{{O}_{2}}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{n}_{{O}_{2}}}}}} \cdot P$

${P}^{'} = 3 \times P$

Finally, plug this into the equation that describes Dalton's Law of Partial Pressures to get the total pressure of the mixture, ${P}_{\text{total}}$

${P}_{\text{total}} = {P}^{'} + P$

${P}_{\text{total}} = 3 \times P + P = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 4 \times P \textcolor{w h i t e}{\frac{a}{a}} |}}}$