Question

Question #48160

Answer 1

`"1.25 g L"^(-1)`

Explanation:

The idea here is that you need to use the ideal gas law equation and the molar mass of nitrogen gas, `"N"_2`, to find a relationship between the mass of a sample of nitrogen gas and the volume it occupies at STP.

Now, STP conditions will most likely be given to you as a pressure of `"1 atm"` and a temperature of `0^@"C"`, or `"273.15 K"`.

The ideal gas law equation looks like this

`color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "`, 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("atm" * "L")/("mol" * "K")`
`T` - the absolute temperature of the gas

As you know, a compound's molar mass tells you the mass of one mole of that substance. IN your case, nitrogen gas has a molar mass of

`M_M = "28.0134 g mol"^(-1)`

This means that every mole of nitrogen gas has a mass of `"28.0134 g"`.

You can thus replace the number of moleso f nitrogen gas, `n`, in the ideal gas law equation by the ratio between a mass `m` and the molar mass of the gas

`n = m/M_M`

Plug this into the ideal gas law equation to get

`PV = m/M_M * RT`

Rearrange to get

`PV *M_M = m * RT`

This will be equivalent to

`P * M_M = m/V * RT`

But since density, `rho`, is defined as mass per unit of volume, you can say that you have

`m/V = (P * M_M)/(RT)`

and therefore

`color(purple)(|bar(ul(color(white)(a/a)color(black)(rho = (P * M_M)/(RT))color(white)(a/a)|)))`

Now all you have to do is use the STP values for pressure and temperature and the molar mass of nitrogen gas to get

`rho = (1 color(red)(cancel(color(black)("atm"))) * "28.0134 g" color(red)(cancel(color(black)("mol"^(-1)))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 273,15color(red)(cancel(color(black)("K")))) = color(green)(|bar(ul(color(white)(a/a)"1.25 g L"^(-1)color(white)(a/a)|)))`

`color(white)(a)`
ALTERNATIVE APPROACH

As you know, one mole of any ideal gas kept at `"1 atm"` and `0^@"C"` occupies `"22.4 L"` -- this is known as the molar volume of a gas at STP.

Moreover, you know that one mole of nitrogen gas has a mass of `"28.0134 g"`. Since density tells you mass per unit of volume, all you have to do now is determine the mass of one liter of nitrogen gas at STP

`1 color(red)(cancel(color(black)("L N"_2))) * overbrace((1 color(red)(cancel(color(black)("mole N"_2))))/(22.4color(red)(cancel(color(black)("L N"_2)))))^(color(purple)("molar volume of a gas at STP")) * overbrace("28.0134 g"/(1color(red)(cancel(color(black)("mole N"_2)))))^(color(blue)("molar mass of N"_2)) = "1.25 g"`

Since this is how many grams you get per liter of nitrogen gas at STP, it follows that its density will be

`rho = color(green)(|bar(ul(color(white)(a/a)"1.25 g L"^(-1)color(white)(a/a)|)))`