# Question #4d7ef

Aug 9, 2016

${V}_{2} = \frac{{P}_{f} - {P}_{a}}{P} _ 2 \cdot {T}_{2} / {T}_{1} \cdot {V}_{1}$

#### Explanation:

The idea here is that the number of moles of hydrogen gas, ${\text{H}}_{2}$, are kept constant when going from the initial state of the gas to the final state of the gas.

Your starting point here will be the ideal gas law equation, which looks like this

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

Here you have

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant
$T$ - the absolute temperature of the gas

Now, you know that when the gas is first generated in a volume ${V}_{1}$ at a temperature ${V}_{1}$, the total pressure of the gas + air mixture goes from ${P}_{a}$ to ${P}_{f}$.

This means that the pressure of hydrogen gas is equal to

${P}_{{\text{H}}_{2}} = {P}_{f} - {P}_{a}$

In other words, the total pressure of the hydrogen gas + air mixture is given by the sum of the partial pressures of its two components, air and hydrogen gas, as given by Dalton's Law of Partial Pressures.

You can thus use the ideal gas law equation to write

${P}_{{\text{H}}_{2}} \cdot {V}_{1} = n \cdot R \cdot {T}_{1}$

which is equivalent to

$\left({P}_{f} - {P}_{a}\right) \cdot {V}_{1} = n \cdot R \cdot {T}_{1} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Now focus on the second state of the gas. You want the same number of moles, $n$, kept at a temperature ${T}_{2}$, which corresponds to partial pressure ${P}_{2}$, to occupy a volume ${V}_{2}$.

Once again, use the ideal gas law equation to write

${P}_{2} \cdot {V}_{2} = n \cdot R \cdot {T}_{2} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

All you have to do now is divide equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ by equation $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to get rid of the number of moles of hydrogen gas and of the universal gas constant

$\frac{\left({P}_{f} - {P}_{a}\right) \cdot {V}_{1}}{{P}_{2} \cdot {V}_{2}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{n}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R}}} \cdot {T}_{1}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{n}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R}}} \cdot {T}_{2}}$

Rearrange to isolate ${V}_{2}$ on one side of the equation

${P}_{2} \cdot {V}_{2} \cdot {T}_{1} = \left({P}_{f} - {P}_{a}\right) \cdot {V}_{1} \cdot {T}_{2}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{V}_{2} = \frac{{P}_{f} - {P}_{a}}{P} _ 2 \cdot {T}_{2} / {T}_{1} \cdot {V}_{1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$