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

Question

Natalie Anton · Accepted Answer

I get about #"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)#...

#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")#

Kai Bowman · Answer

undefined The volume of 20 g of a gas with a vapor density of 20 at NTP is 22.4 liters.