Question

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

Answer 1

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

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

`rho_V = rho_"gas"/rho_(H_2)`

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

`PM = rhoRT`,

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

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

• `T` is temperature in `"K"`.

Thus, `rho = (PM)/(RT)`, 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"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")`