# The density of Nitrogen at 0 degrees Celsius and 1 atmosphere is most nearly equal to which of the following quantities?

The density of nitrogen at 0 degrees Celsius and 1 atmosphere is most nearly equal to which of the following quantities?

a) 0.001 g/L

b) 0.01 g/L

c) 0.1 g/L

d) 1 g/L

e>) 10 g/L

The answer is d), 1 gram per litre.

I think this is how you do it, but please correct me if I'm missing something. I didn't end up using the standard molar volume, so this might be incorrect.

$$pV = nRT$$

We know the pressure is 1 atm, the temperature is about 273 K and R = 0.08.

Solving for $\frac {n}{V}$, we find $\frac {n}{V} = \frac {1}{22}$. Now, this is in moles per litre, so by multiplying by the molar mass of nitrogen (aproximately 28 g per mol), we find that we have about 1 g per litre as the density.

• Hint: Use the ideal gas law. – Todd Minehardt Aug 21 '15 at 15:20
• And a useful number to know is the molar volume of an ideal gas. – Jon Custer Aug 21 '15 at 15:50
• Your edit seems correct to me. Feel free to answer your own question so that it is useful for future users. – bon Aug 21 '15 at 17:55

Using the ideal gas law:
$\mathrm{PV = nRT}$
$\mathrm{\frac{n}{V} = \frac{P}{RT}}$

Where:
$\mathrm{T = 273 K}$
$\mathrm{P = 1~atm}$
$\mathrm{R = 0.0821~\frac{L~atm}{mol~K}}$
$\mathrm{MW~\ce{N2} = 28.0\frac{g}{mol}}$

We get:
$\mathrm{\frac{n}{V} = \frac{1~atm}{0.0821~\frac{L~atm}{mol~K}*273~K}}$
$\mathrm{\frac{n}{V} = 0.0446~\frac{mol}{L}}$

And finally:
$\mathrm{\ce{N2}~density = 0.0446~\frac{mol}{L}*28\frac{g}{mol}=1.2~\frac{g}{L}}$