Chemical Thermodynamics: Raoult's Law 1

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Frances Docherty

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Created this as a copy of Raoult's Law.

Name	Status	Author	Last Modified
Chemical Thermodynamics: Raoult's Law 1	Ready to use	Frances Docherty	30/03/2025 18:52
Chemical Thermodynamics: Raoult's Law 2	Ready to use	Frances Docherty	30/03/2025 18:53

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			Name	Type	Generated Value

px	number	0.7
py	number	1.8
nx	number	1.8
ny	number	1.5
xx	number	0.5454545455
xy	number	0.4545454545
partialx	number	0.3818181818
partialy	number	0.8181818182
pressure	number	1.2

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Substances $\mathrm{X}$ and $\mathrm{Y}$ have similar molecular masses and structures. At $\mathrm{100^\circ C}$ , pure $\mathrm{X}$ boils at $\mathrm{\var{px}\space atm}$ , pure $\mathrm{Y}$ at $\mathrm{\var{py}\space atm}$ . If $\mathrm{\var{nx}}$ moles of $\mathrm{X}$ are mixed with $\mathrm{\var{ny}}$ moles of $\mathrm{Y}$ at $\mathrm{100^\circ C}$ , what is the vapour pressure of this mixture?

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Raoult's law is used to relate the equilibrium vapour pressure of the pure form of a substance to the partial vapour pressure of that same substance in a mixture (i.e. the substance's contribution to the total vapour pressure of the mixture):

$\mathrm{p_i=p_i^*x_i}$ ,

where $\mathrm{p_i}$ is the partial vapour pressure of the $\mathrm{i^{th}}$ substance in the mixture; $\mathrm{p_i^*}$ is the equilibrium vapour pressure of the pure form of this substance; and $\mathrm{x_i}$ is the mole fraction of the $\mathrm{i^{th}}$ substance in the mixture.

We are given the atmospheric pressures at which substances X and Y boil. At a substance's boiling point, its saturated vapour pressure is equal to the surrounding atmospheric pressure, so the atmospheric pressures at which substances X and Y boil are simply equal to their equilibrium vapour pressures. Thus, we know from what we are told that $\mathrm{p_{_{X}}^*=\var{px} \space atm}$ and $\mathrm{p_{_{Y}}^*=\var{py} \space atm}$ .

We are also given the number of moles of each of the two substances in the mixture, $\mathrm{n_{_{X}}=\var{nx} \space mol}$ and $\mathrm{n_{_{Y}}=\var{ny} \space mol}$ ; we will use these to calculate their corresponding mole fractions, $\mathrm{x_{_{X}}}$ and $\mathrm{x_{_{Y}}}$ , which we will in turn use with Raoult's law to calculate the partial vapour pressure of each substance, $\mathrm{p_{_{X}}}$ and $\mathrm{p_{_{Y}}}$ .

Recall that mole fraction is given by the number of moles of the $\mathrm{i^{th}}$ substance, divided by the total number of moles of all substances in the mixture:

$\mathrm{x_i=\frac{n_i}{n_{tot}}}$ .

Thus:

$\mathrm{x_{_{X}}=\frac{n_{_{X}}}{n_{_{X}}+n_{_{Y}}}=\frac{\var{nx}}{\var{nx}+\var{ny}}=\var{siground(xx,3)}}$ ,

$\mathrm{x_{_{Y}}=\frac{n_{_{Y}}}{n_{_{Y}}+n_{_{X}}}=\frac{\var{ny}}{\var{ny}+\var{nx}}=\var{siground(xy,3)}}$ .

We can now apply Raoult's law to calculate our partial vapour pressures:

$\mathrm{p_{_X}=p_{_X}^*x_{_X}=\var{px}\times\var{xx}=\var{sigformat(partialx,3)}\space atm}$ ,

$\mathrm{p_{_Y}=p_{_Y}^*x_{_Y}=\var{py}\times\var{xy}=\var{sigformat(partialy,3)}\space atm}$ .

Lastly, to find the total vapour pressure of the mixture, we simply sum up the partial vapour pressures of the mixture's components:

$\mathrm{p_{tot}=p_{_X}+p_{_Y}=\var{sigformat(partialx,3)}+\var{sigformat(partialy,3)}=\var{sigformat(pressure,3)}\space atm}$ .

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Chemical Thermodynamics: Raoult's Law 1

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

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Frances Docherty 2 weeks, 4 days ago

Frances Docherty 2 weeks, 4 days ago

Frances Docherty 2 weeks, 4 days ago

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