33 Practice Exercises
Exercise: Phase Change – Carbon Dioxide
Identify the phase changes that carbon dioxide will undergo as its temperature is increased from −100 °C while holding its pressure constant at 1,000 kPa. At what approximate temperatures do these phase changes occur? The phase diagram of carbon dioxide is shown below.
Image obtained from OpenStax Chemistry/ CC BY 4.0
Note: This diagram is missing the “vapour” phase. Usually, we call the gaseous phase under the liquid region “vapor” because it is able to condense to a liquid, while the “gas” phase is in a temperature that is too high to condense. Sometimes the terms are not strictly distinguished.
Solution
The phase change from solid to liquid at ~-55°C; from liquid to gas at ~-40°C.
To represent holding pressure constant while increasing temperature, we can draw a line moving horizontally to the right on the phase diagram.
A phase change occurs every time when the horizontal line hits a curve on the diagram.
Exercise: Gibb’s Phase Rule
How many degrees of freedom exist for a vapour-liquid (V-L) mixture of acetone ([latex](CH_3)_2CO[/latex]) and methyl ethyl ketone ([latex]CH_3C(O)CH_2CH_3[/latex])? Assume no chemical reaction happens.
Solution
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]
$$D\!F=2+c-\pi-r=2+2-2-0=2$$
Exercise: Ideal Gas Law
Calculate the density (mass per cubic metre) of dry air (a) under IUPAC standard conditions and (b) in a hot air balloon at a temperature of 120°C under the same pressure. Assuming ideal gas behaviour.[latex]^{[1]}[/latex]
Take average molar mass =29.0 mol/L for dry air. [latex]R=8.314\frac{J}{molK}[/latex]
Solution
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.
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