If the vapour density for a gas is 2020, then what is the volume of "20 g"20 g of this gas at NTP?

1 Answer
Truong-Son N.
Jun 27, 2017

I get about "11.93 L"11.93 L, assuming ideality all the way through.

This is an old term for the ratio of the density to the density of "H"_2(g)H2(g)...

rho_V = rho_"gas"/rho_(H_2)ρV=ρgasρH2

Densities here are in "g/L"g/L. From the ideal gas law,

PM = rhoRTPM=ρRT,

where MM is molar mass in "g/mol"g/mol and the remaining variables should be well-known...

  • PP is pressure in "atm"atm as long as R = "0.082057 L"cdot"atm/mol"cdot"K"R=0.082057 L⋅atm/mol⋅K.

  • TT is temperature in "K"K.

Thus, rho = (PM)/(RT)ρ=PMRT, 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))ρgasρH2≈PMgas/RTPMH2/RT≈MgasMH2

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"g gas" xx "1 mol"/(40.316 cancel"g") = "0.4961 mols"

So, at 20^@ "C" and "1 atm", i.e. NTP, assuming ideality again:

color(blue)(V_(gas)) ~~ (nRT)/P

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

~~ color(blue)("11.93 L")

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