# At NTP, what is the mass of "22.4 L" of nitrogen gas?

Aug 11, 2017

Well, I can set things up for you to calculate the mass... but weight is in newtons, $\text{N}$, or ${\text{kg"cdot"m/s}}^{2}$... and you have to decide what the chemical formula of nitrogen is... you have a 50% chance of guessing it right!

Normal Temperature and Pressure, NTP, is apparently ${20}^{\circ} \text{C}$ and $\text{1 atm}$. This should not be confused with STP, which is at ${0}^{\circ} \text{C}$, not ${20}^{\circ} \text{C}$, and not ${25}^{\circ} \text{C}$.

The "mass" version of the ideal gas law can be derived.

$P V = n R T$

$n = \frac{P V}{R T}$

To get the units from $\text{mols}$ to $\text{g}$, simply multiply by the molar mass $M$ in $\text{g/mol}$.

$n M \equiv m = \frac{P V M}{R T}$

In this case, we have:

• $P$, pressure, in $\text{atm}$, of the ideal gas within the container.
• $V$, volume, in $\text{L}$, of the ideal gas.
• $M$, molar mass, in $\text{g/mol}$, of the ideal gas.
• $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$, the universal gas constant.
• $T$, temperature, in $\text{K}$, within the container filled with ideal gas.

In case you couldn't tell, we're assuming ideal gases... So, the mass is:

color(red)(m) = (cancel"1 atm" cdot 22.4 cancel"L" cdot M" g/"cancel"mol")/(0.082057 cancel"L"cdotcancel"atm""/"cancel"mol"cdotcancel"K" cdot (20 + 273.15 cancel"K"))

= color(red)(???)

Now, it's up to you to decide what the molar mass is.

What is the molar mass of "nitrogen"? Is it $\text{14.007 g/mol}$, or $\text{28.014 g/mol}$? Why would you have to choose? (It matters!)