Question

On average, each person in the industrialized world is responsible for the emission of $10,000kg$of carbon dioxide (data-custom-editor="chemistry" ${\mathrm{CO}}_2$) every year. This includes ${\mathrm{CO}}_2$that you generate directly, by burning fossil fuels to operate your car or your furnace, as well as ${\mathrm{CO}}_2$generated on your behalf by electric generating stations and manufacturing plants. ${\mathrm{CO}}_2$ is a greenhouse gas that contributes to global warming. If you were to store your yearly ${\mathrm{CO}}_2$ emissions in a cube at STP, how long would each edge of the cube be?

Short answer

Each edge of the required cube would be $\mathbf{17}\mathbf{.}\mathbf{2}\mathbf{}\mathit{m}$ long

Step-by-step solution

Defining the Ideal Gas law, Molar mass

The ideal gas law shows the relationship between the four state variables for an ideal gas, namely the volume V of the container, the pressure p exerted on the gas, the temperature T of the gas and the number of moles n of the gas in the container.

The ideal gas law is given by : $pV=nRT$
Here$R$ is the universal gas constant and in the unit SI its value is
$R=8.31J/mol.K$
Rewriting the ideal gas law for V in the form :

role="math" localid="1648319912293" $V=\frac{nRT}{P}$
If we know the number of molecules, we obtain the number of moles n by using the mass M and the molar mass of the gas$n=\frac{M}{M_{mol}}$

Calculating the length of the side of the said cube

Use the expression for n and insert the values for $$\begin{gathered} T=273K,p=1.013x105Pa,MandMmol=44g/mol=0.044kg/moltoobtain \\ V=\frac{MRT}{M_{mol}p} \\ =\frac{(10000kg)(8.31J/mol.K)(273K)}{(0.044kg/mol)(1.013\times {10}^{5}Pa)} \\ =5092m³ \end{gathered}$$
The cube has the volume $V=L³$, where $L$ is the side of the cube. We use the value of the volume of the cube to obtain the side by
$$\begin{gathered} L=\sqrt[3]{V}=\sqrt[3]{5092m³} \\ =17.2m \end{gathered}$$