# Question #44e55

##### 1 Answer

#### 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.

Now, the **ideal gas law** equation looks like this

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#

Here you have

#P# - the pressure of the gas

#V# - the volume it occupies

#n# - the number of moles of gas

#R# - theuniversal gas constant, usually given as#0.0821("atm" * "L")/("mol" * "K")#

#T# - theabsolute temperatureof the gas

Notice the units used for the universal gas constant

atmospheresfor pressure#-> ["atm"]# litersfor volume#-> ["L"]# Kelvinfor temperature#-> ["K"]#

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = " 760 mmHg")color(white)(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

#color(blue)(|bar(ul(color(white)(a/a)V = "length" xx "width" xx "height"color(white)(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 * 25.0 * 50.0 * 1/10 * 1/10 * 1/10" dm"^3#

#V = "6.25 dm"^3#

Expressed in *liters*, the volume will be

#V = "6.25 L"#

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#

and rearrange the ideal gas law equation to solve for

#PV = nRT implies n = (PV)/(RT)#

Plug in your values to find

#n = (720.3/760 color(red)(cancel(color(black)("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"))))#

#n = color(green)(|bar(ul(color(white)(a/a)color(black)("0.246 moles")color(white)(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.