Question 48160

Apr 28, 2016

${\text{1.25 g L}}^{- 1}$

Explanation:

The idea here is that you need to use the ideal gas law equation and the molar mass of nitrogen gas, ${\text{N}}_{2}$, 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 $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$, or $\text{273.15 K}$.

The ideal gas law equation looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, 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 $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - 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

${M}_{M} = {\text{28.0134 g mol}}^{- 1}$

This means that every mole of nitrogen gas has a mass of $\text{28.0134 g}$.

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

$n = \frac{m}{M} _ M$

Plug this into the ideal gas law equation to get

$P V = \frac{m}{M} _ M \cdot R T$

Rearrange to get

$P V \cdot {M}_{M} = m \cdot R T$

This will be equivalent to

$P \cdot {M}_{M} = \frac{m}{V} \cdot R T$

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

$\frac{m}{V} = \frac{P \cdot {M}_{M}}{R T}$

and therefore

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\rho = \frac{P \cdot {M}_{M}}{R T}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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)|)))

$\textcolor{w h i t e}{a}$
ALTERNATIVE APPROACH

As you know, one mole of any ideal gas kept at $\text{1 atm}$ and ${0}^{\circ} \text{C}$ occupies $\text{22.4 L}$ -- 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 $\text{28.0134 g}$. 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

$\rho = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{1.25 g L}}^{- 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$