Question

Question #44e55

Answer 1

`"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^@"C"`.

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` - the universal gas constant, usually given as `0.0821("atm" * "L")/("mol" * "K")`
`T` - the absolute temperature of the gas

Notice the units used for the universal gas constant

• atmospheres for pressure `-> ["atm"]`
• liters for volume `-> ["L"]`
• Kelvin for 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 `n`

`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.