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Does vapour density change with temperature or is it solely dependent on molecular mass

as per the formula VD=molar mass/2

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Vapour density is the density of gas with respect to the density of hydrogen at same temperature and pressure. You measure density of the gas at, say $p$ and at temperature $T$ and divide it by the density of Hydrogen at the same temperature and pressure.

The formula you used was only valid for an ideal gas.

$$\rho_{gas} = \frac{pM_{gas}}{RT}$$ $$\rho_{\ce H_2} = \frac{pM_{\ce H_2}}{RT}$$ Thus $$\frac{\rho_{gas}}{\rho_{\ce H_2}} = \frac{M_{gas}}{M_{\ce H_2}}$$ And we know that $M_{\ce H_2}$ is $2\pu{g/mol}$ $$\frac{\rho_{gas}}{\rho_{\ce H_2}} = \frac{M_{gas}}{2}$$ So vapour density of an ideal gas will be independent of temperature.

But for real gases after if you set the pressure of the gas and Hydrogen gas in such a way that after van Der Waal correction, the ideal pressure of both comes out to be the same, then your formula will ve valid for real gases as well.

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  • $\begingroup$ I see a clash of definitions, or rather their terminology for the relative definition of vapour density versus the absolute density definition with the unit kg/m3 in sense of "density of vapour". As it is easy not to be aware of "vapour density" is not "density of vapour" but an independent term. Personally, I have never met the term and meaning "vapour density" yet. $\endgroup$ – Poutnik Oct 8 '20 at 12:24
  • $\begingroup$ "Mass of a certain volume of a substance divided by mass of same volume of hydrogen". This is just another way to define it. Wiki has defined it by stating that volume of both gases is same. But it still is the ratio of densities. It is defined in such a way because it is easier to calculate masses of the two gases in similar container than to calculate the absolute densities of the two gases in container of different volume. $\endgroup$ – Eyy boss Oct 8 '20 at 12:33
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    $\begingroup$ Sure, I have linked the Wiki article as well, and modified my answer. I do not have any objection about the term nature, but the term itself. What I wanted to say is that the choice of term "vapour density" is unlucky, leading to confusion with "substance density" as mass per volume. As non-native, learning English after chemistry, it is easier for me to fall into a terminology trap. $\endgroup$ – Poutnik Oct 8 '20 at 12:41
  • $\begingroup$ Yes, they did mess up the terminology. $\endgroup$ – Eyy boss Oct 8 '20 at 12:46
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    $\begingroup$ In fact, it has both meanings, depending if you consider "vapour" as adjective or as a part on the term. $\endgroup$ – Poutnik Oct 8 '20 at 12:47
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For ideal gas vapour behaviour, the vapour density is the ratio of its density and of the density of hydrogen at the same conditions

$$\frac { \frac {pM}{RT}}{ \frac {pM_{\ce{H2}}}{RT}} \simeq \frac M2 $$.

There for assumption of vapour ideal gas begaviour, "vapour density" does not depend either on temperature either on pressure.

I see a clash of definitions, or rather their terminology for the relative definition of vapour density versus the absolute density definition with the unit kg/m3 in sense of "density of vapour" or substance density.

It is easy not to be aware of "vapour density" is not "density of vapour". In fact, lack of context gives the reader the confusing choice if vapour is an adjective to density or a part of the term.

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  • $\begingroup$ so it is dependent? $\endgroup$ – JustJohan Oct 6 '20 at 13:46
  • $\begingroup$ Is y in y=2x dependent on x ? $\endgroup$ – Poutnik Oct 6 '20 at 13:50
  • $\begingroup$ BTW, when you ask a question, always write down also the context or background. It prevents the need of clarifications and helps to provide relevant answers. If constex/bckground is omitted, it often leads to clarification ping-pong and/or helper may blindly shoot useless answers as they do not know the needed context. Last but not least, provide you resolution effort first, otherwise the questions may get closed. $\endgroup$ – Poutnik Oct 6 '20 at 14:23
  • $\begingroup$ understood my bad , what kind of context should i have added? $\endgroup$ – JustJohan Oct 6 '20 at 15:08
  • $\begingroup$ At least what scenario you consider. And it is expected you search through available resources before asking. $\endgroup$ – Poutnik Oct 6 '20 at 15:32

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