# Question 651b6

May 16, 2016

$\text{651.0 torr}$

#### Explanation:

The idea here is that you can use the ideal gas law equation to find the pressure in box $\text{B}$ before mixing the two gases in that box, and Boyle's Law to calculate the pressure of methane if placed alone in box $\text{B}$.

At that point, you can use Dalton's Law of Partial Pressures to find the total pressure in box $\text{B}$.

So, you know that you're dealing with two boxes of different volumes that contain different numbers of moles of gas.

So, the ideal gas law equation looks like this

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

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

If you take ${V}_{A}$ and to be the volume of box $\text{A}$, ${P}_{A}$ the pressure in box $\text{A}$, and $T$ to be the temperature of both boxes, you can say that

P_A * V_A = n_A * RT" " " "color(orange)(("*")) -> for box $\text{A}$

Now, you know that

${V}_{B} = 2 \times {V}_{A} \to$ the volume of box $\text{B}$ is twice that of box $\text{A}$

and

${n}_{B} = 2 \times {n}_{A} \to$ box $\text{B}$ contains twice as many moles of gas as box $\text{A}$

You can thus use the ideal gas law equation to write

${P}_{B} \cdot {V}_{B} = {n}_{B} \cdot R T \to$ for box $\text{B}$

But this is equivalent to

${P}_{B} \cdot \left(2 {V}_{A}\right) = \left(2 {n}_{A}\right) \cdot R T$

As you can see, you can use equation $\textcolor{\mathmr{and} a n \ge}{\left(\text{*}\right)}$ to conclude that since

${P}_{B} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} {V}_{A} = \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} {n}_{A} \cdot R T$

$\left\{\begin{matrix}{P}_{B} \cdot {V}_{A} = {n}_{A} \cdot R T \\ {P}_{A} \cdot {V}_{A} = {N}_{A} \cdot R T\end{matrix}\right.$

the pressure in box $\text{B}$ before the two gases are mixed will be equal to the pressure in box $\text{A}$

${P}_{B} = {P}_{A} = \text{434.0 torr}$

At this point, you can use Boyle's Law to find the pressure exerted by the gas in box $\text{A}$ if placed alone in box $\text{B}$, let's say ${P}_{\text{A in box B}}$.

As you know, volume and pressure have an inverse relationship when temperature and number of moles are kept constant.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{1} {V}_{1} = {P}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

${P}_{1}$, ${V}_{1}$ - the pressure and volume of the gas at an initial state
${P}_{2}$, ${V}_{2}$ - the pressure and volume of the gas at a final state

${P}_{A} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{A}}}} = {P}_{\text{A in box B}} \cdot \left(2 \textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{A}}}}\right)$

This will get you

P_("A in box B") = P_A/2 = "434.0 torr"/2 = "217.0 torr"

So, the moles of methane will exert a pressure of $\text{217.0 torr}$ when placed alone in box $\text{B}$.

According to Dalton's Law of Partial Pressures, the total pressure of a gaseous mixture will be equal to the sum of the partial pressures of each constituent of the mixture if left alone in the same volume at the same temperature as the mixture.

You will thus have

P_"total" = P_B + P_("A in box B")

This will get you

P_"total" = "434.0 torr" + "217.0 torr" = color(green)(|bar(ul(color(white)(a/a)"651.0 torr"color(white)(a/a)|)))

$\textcolor{w h i t e}{}$
ALTERNATIVE APPROACH

Here's an alternative method to use to find the total pressure of the mixture.

According to Dalton's Law of Partial Pressures, the partial pressure of a gas that's part of a gaseous mixture depends on its mole fraction and on the total pressure of the mixture.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{\text{gas i" = chi_"gas i" xx P_"total}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${P}_{\text{gas i}}$ - the partial pressure of gas $i$ in the mixture
${\chi}_{\text{gas i}}$ - the mole fraction of gas $i$
${P}_{\text{total}}$ - the total pressure of the mixture

Now, we've established that the gas in box $\text{B}$ exerts a pressure of $\text{434.0 torr}$. When you mix the two gases, the mole fraction of gas $\text{B}$ becomes

${\chi}_{B} = \text{number of moles of helium"/"total number of moles of gas}$

${\chi}_{B} = \left(8 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/((8 + 4)color(red)(cancel(color(black)("moles}}}}\right) = \frac{2}{3}$

This means that you have

${P}_{B} = {\chi}_{B} \times {P}_{\text{total}}$

and therefore

${P}_{\text{total}} = {P}_{B} / {\chi}_{B}$

This once again gets you

P_"total" = "434.0 torr"/(2/3) = color(green)(|bar(ul(color(white)(a/a)"651.0 torr"color(white)(a/a)|)))#