If the vapour density for a gas is 20, then what is the volume of "20 g" of this gas at NTP?

Jun 27, 2017

I get about $\text{11.93 L}$, assuming ideality all the way through.

This is an old term for the ratio of the density to the density of ${\text{H}}_{2} \left(g\right)$...

${\rho}_{V} = {\rho}_{\text{gas}} / {\rho}_{{H}_{2}}$

Densities here are in $\text{g/L}$. From the ideal gas law,

$P M = \rho R T$,

where $M$ is molar mass in $\text{g/mol}$ and the remaining variables should be well-known...

• $P$ is pressure in $\text{atm}$ as long as $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$.

• $T$ is temperature in $\text{K}$.

Thus, $\rho = \frac{P M}{R T}$, and the ratio of the densities for ideal gases is the ratio of the molar masses:

rho_"gas"/rho_(H_2) ~~ (PM_"gas""/"RT)/(PM_(H_2)"/"RT) ~~ M_("gas")/(M_(H_2))

Thus, the molar mass of the gas is apparently...

M_("gas") ~~ overbrace(20)^(rho_V) xx overbrace("2.0158 g/mol")^(M_(H_2)) ~~ "40.316 g/mol"

And the mols of this would be...

$20 \cancel{\text{g gas" xx "1 mol"/(40.316 cancel"g") = "0.4961 mols}}$

So, at ${20}^{\circ} \text{C}$ and $\text{1 atm}$, i.e. NTP, assuming ideality again:

$\textcolor{b l u e}{{V}_{g a s}} \approx \frac{n R T}{P}$

~~ (("0.4961 mols")("0.082057 L"cdot"atm/mol"cdot"K")("293.15 K"))/("1 atm")

$\approx$ $\textcolor{b l u e}{\text{11.93 L}}$