# If a gaseous mixture is made by combining 3.54 g of Ar and 2.59 g of Kr in an evacuated 2.50 L container at 25 degrees C, what are the partial pressures of each gas and what is the total pressure by the gaseous mixture?

Jul 1, 2016

Here's what I got.

#### Explanation:

The first thing to do here is use the molar masses of the two elements to determine how many moles of each you have present in your sample.

3.54 color(red)(cancel(color(black)("g"))) * "1 mole Ar"/(39.95color(red)(cancel(color(black)("g")))) = "0.08861 moles Ar"

2.59 color(red)(cancel(color(black)("g"))) * "1 mole Kr"/(83.80color(red)(cancel(color(black)("g")))) = "0.03091 moles Kr"

Now, the partial pressure of each gas in the mixture will be equivalent to their pressure if placed alone in the same container and kept at the same temperature.

Your tool of choice 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, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

So, you're placing the sample of argon alone in the container. Its pressure will be

$P V = n R T \implies P = \frac{n R T}{V}$

In this case, you have

P_"Ar" = (0.08861 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(2.50color(red)(cancel(color(black)("L"))))

P_"Ar" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.868 atm")color(white)(a/a)|)))

Now place the sample of krypton alone in the container. Its pressure will be

P_"Kr" = (0.03091 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(2.50color(red)(cancel(color(black)("L"))))

P_"Kr" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.303 atm")color(white)(a/a)|)))

I'll leave both answers rounded to three sig figs.

As you can see, the pressure of the gas depends exclusively on how many moles are present in the sample if and only if volume and temperature are kept constant.

According to Dalton's Law of Partial Pressures, the pressure of the mixture will be equal to the sum of the partial pressures of each constituent $i$ of the mixture.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{\text{mixture}} = {\sum}_{i} {P}_{i} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you will have

${P}_{\text{mixture" = P_"Ar" + P_"Kr}}$

P_"mixture" = "0.868 atm" + "0.303 atm" = color(green)(|bar(ul(color(white)(a/a)color(black)("1.171 atm")color(white)(a/a)|)))

I'll leave the answer rounded to three decimal places.