2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Hint: Use the ideal gas law. $\endgroup$ – Todd Minehardt Aug 21 '15 at 15:20
  • 1
    $\begingroup$ And a useful number to know is the molar volume of an ideal gas. $\endgroup$ – Jon Custer Aug 21 '15 at 15:50
  • 1
    $\begingroup$ Your edit seems correct to me. Feel free to answer your own question so that it is useful for future users. $\endgroup$ – bon Aug 21 '15 at 17:55
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.