Finding the density of nitrogen at 0 degrees Celsius and 1 atmosphere


> 
> The density of nitrogen at $\pu{0 ^\circ C}$ and $\pu{1 atm}$ is most nearly equal to which of the following quantities?  
> 
> (a) $\pu{0.001 g/L}$; (b) $\pu{0.01 g/L}$; (c) $\pu{0.1 g/L}$;
> (d) $\pu{1 g/L}$; (e) $\pu{10 g/L}$.
> 
> 
> 


The answer is (d), $\pu{1 g L-1}$.


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 $\pu{1 atm}$, the temperature is about $\pu{273 K}$ 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 (approximately $\pu{28 g mol-1}$), we find that we have about $\pu{1 g}$ per litre as the density.




Using the ideal gas law
\begin{align}
 pV &= nRT, &
 \frac nV &= \frac{p}{RT},
\end{align}
where
\begin{align}
 T &= \pu{273 K},\\
 p &= \pu{1 atm},\\
 R &= \pu{0.0821 L atm mol-1 K-1},\\
 M(\ce{N2}) &= \pu{28.0 g mol-1},
\end{align}  

we get
\begin{align}
 \frac{n}{V} &= 
 \frac{\pu{1 atm}}{\pu{0.0821 L atm//mol K} \times\pu{273 K}},\\
 \frac{n}{V} &= \pu{0.0446 mol//L},
\end{align}  

and finally  

$$
\rho(\ce{N2}) 
 = \pu{0.0446 mol//L} \times \pu{28 g//mol} 
 = \pu{1.2 g//L}.
$$  

Therefore answer (d) is correct.


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