# Question 44e55

Jul 14, 2016

$\text{0.246 moles}$

#### Explanation:

As it's written, the problem is missing information about the temperature of the gas, so I will assume that you're working at room temperature, i.e. ${20}^{\circ} \text{C}$.

Now, the ideal gas law equation looks like this

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

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

Notice the units used for the universal gas constant

• atmospheres for pressure $\to \left[\text{atm}\right]$
• liters for volume $\to \left[\text{L}\right]$
• Kelvin for temperature $\to \left[\text{K}\right]$

You can convert the pressure from mmHg to atm by using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 atm " = " 760 mmHg}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The next thing to do is calculate the volume of the box.

You know that for a rectangular prism, the volume is given by

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} V = \text{length" xx "width" xx "height} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Notice that the dimensions of the box are given to you in centimeters. The thing to remember here is that you have

color(purple)(|bar(ul(color(white)(a/a)color(black)("1 dm " = " 10 cm")color(white)(a/a)|)))" " and " "color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L " = " 1 dm"^3")color(white)(a/a)|)))

The volume of the box will thus be equal to

V = (5.0 color(red)(cancel(color(black)("cm"))) xx "1 dm"/(10color(red)(cancel(color(black)("cm"))))) xx (25.0 color(red)(cancel(color(black)("cm"))) * "1 dm"/(10color(red)(cancel(color(black)("cm"))))) xx (50.0 color(red)(cancel(color(black)("cm"))) * "1 dm"/(10color(red)(cancel(color(black)("cm")))))#

$V = 5.0 \cdot 25.0 \cdot 50.0 \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{10} {\text{ dm}}^{3}$

$V = {\text{6.25 dm}}^{3}$

Expressed in liters, the volume will be

$V = \text{6.25 L}$

Finally, convert the temperature from degrees Celsius to Kelvin by using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{T \left[\text{K"] = t[""^@"C}\right] + 273.15} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and rearrange the ideal gas law equation to solve for $n$

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

Plug in your values to find

$n = \left(\frac{720.3}{760} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 6.25color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 20)color(red)(cancel(color(black)("K}}}}\right)$

$n = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{0.246 moles}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to three sig figs, but keep in mind that you only have two sig figs for the width of the box.