33 Practice Exercises

vapour-liquid mixture: [latex]\pi=2[/latex]

mixture of acetone and methyl ethyl ketone: [latex]c=2[/latex]

no chemical reactions: [latex]r=0[/latex]

$$D\!F=2+c-\pi-r=2+2-2-0=2$$

a) To find the mass per cubic metre, we need the number of moles contained in a cubic metre under the IUPAC standard temperature and pressure. Calculate n by rearranging the ideal gas law:
\begin{align*}
PV & = nRT \\
n & =\frac{PV}{RT}\\
n & =\frac{1×10^5Pa×1m^3}{8.314\frac{J}{molK}×273.15K}\\
n & = 44.1mol
\end{align*}

The mass of air is equal to the number of moles contained in a cubic metre multiplied by molar mass:
$$m=n×MW=44.1mol×29.0g/mol=1278.9g=1.28kg$$

The density of air is the mass divide by volume:
$$\rho=\frac{m}{V}=\frac{1.28kg}{1m^3}=1.28kg/m^3$$

b) Here, we also have the same volume (1 [latex]m^3[/latex]) but different temperature. We can once again find the number of moles this will hold.
\begin{align*}
PV & = nRT \\
n & =\frac{PV}{RT}\\
n & =\frac{1×10^5Pa×1m^3}{8.314\frac{J}{molK}×(273.15+120)K}\\
n & = 30.6mol
\end{align*}

We do a similar calculation as before to find the mass of the gas in the cubic metre
$$m=n×MW=30.6mol×29.0g/mol=887.4g=0.89kg$$

We can then use this to find density under this new condition:
$$\rho=\frac{m}{V}=\frac{0.89kg}{1m^3}=0.89kg/m^3$$

This is less dense, which makes sense as the temperature is higher, so we would expect less gas molecules in the same volume at a similar pressure.

One other way to solve this is to observe that the density is proportional to the number of moles, which is inversely proportional to the temperature:
\begin{align*}
\frac{\rho_{2}}{\rho_{1}} & = \frac{T_{1}}{T_{2}} \\
\rho_{2} & =\frac{T_{1}}{T_{2}}\rho_{1}\\
\rho_{2} & =\frac{273.15K}{(120+273.15)K}×1.28kg/m^3\\
\rho_{2} & = 0.89 kg/m^3
\end{align*}