\(\displaystyle \frac{Rate \: 1}{Rate \: 2} = \sqrt{\frac{Molar\: Mass\, \, 2}{Molar\: Mass\: 1} }\)

This equation explicitly shows how the rate of effusion is inversely proportional to the molar mass of a gas in a gaseous solution. 

Because \(\displaystyle Rate = \frac{Amount \, \, \, of \, \, Substance}{Time}\), time relates to molar mass by: 

\(\displaystyle \frac{\frac{Amount \, \, of \, \, Substance \, \, 1}{Time\: 1}}{\frac{Amount\, \, of \, \, Subtance\, \, 2}{Time \: 2}}=\sqrt{\frac{Molar\, \, Mass\, \, 2}{Molar\, \, Mass\, \, 1}}\)

Simplifying this equation, we see that: 

\(\displaystyle \frac{Time \: 2}{Time\: 1}=\sqrt{\frac{Molar\, \, Mass\, \, 2}{Molar\, \, Mass\, \, 1}}\) 

As a result, time relates directly to molar mass

Recall Graham's Law of Effusion for two gases, A and B:

\(\displaystyle \frac{\text{Rate}_A}{\text{Rate}_B}=\sqrt{\frac{\text{Molar Mass of B}}{\text{Molar mass of A}}}\)

From the equation, we know the following:

\(\displaystyle \frac{\text{Rate}_\text{unknown}}{\text{Rate}_\text{oxygen}}=\sqrt{\frac{\text{Molar Mass of oxygen}}{\text{Molar mass of uknown}}}=0.291\)

Thus, we can solve for the molar mass of the unknown gas. Let \(\displaystyle x\) be the molar mass of the unknown gas.

\(\displaystyle \sqrt{\frac{32.00}{x}}=0.291\)

\(\displaystyle \frac{32.00}{x}=0.084681\)

\(\displaystyle x=377.889\)

Make sure that your answer has \(\displaystyle 3\) significant figures.

We're being asked to compare the effusion rates of oxygen and carbon dioxide.

Remember that effusion is the spontaneous movement of a gas through a small hole from one area to another. It's worth noting that at a given temperature, the average speed of all gas molecules in a system is used to calculate the average kinetic energy of the gas particles. This dependence of kinetic energy on temperature means that at a given temperature, any gas particle will have the same kinetic energy.

In this case, we can say that the kinetic energy of oxygen molecules in one system is equal to the kinetic energy of carbon dioxide molecules in another system. Furthermore, since mass is inversely proportional to velocity, identical kinetic energies would mean that as the mass of the gas particles in a system decreases, their velocity (and thus, effusion rates) would increase.

We can use this information to solve for the relative effusion rates between oxygen and carbon dioxide. By setting their kinetic energies equal to each other, we can derive an expression that relates their relative speeds to their relative masses.

\(\displaystyle KE_{O_{2}}=KE_{CO_{2}}\)

\(\displaystyle \frac{1}{2}m_{O_{2}}v^{2}_{O_{2}}=\frac{1}{2}m_{CO_{2}}v^{2}_{CO_{2}}\)

\(\displaystyle m_{O_{2}}v^{2}_{O_{2}}=m_{CO_{2}}v^{2}_{CO_{2}}\)

\(\displaystyle \frac{v^{2}_{O_{2}}}{v^{2}_{CO_{2}}}=\frac{m_{CO_{2}}}{m_{O_{2}}}\)

\(\displaystyle \frac{v_{O_{2}}}{v_{CO_{2}}}=\sqrt{\frac{m_{CO_{2}}}{m_{O_{2}}}}\)

Generally speaking, this expression shows how the velocity of any two gasses depends on their mass. In this case, the gasses are oxygen and carbon dioxide.

We can use the periodic table of the elements to find out the mass of each gas, and use that information to calculate the relative effusion rates.

\(\displaystyle v_{O_{2}}=v_{CO_{2}} \sqrt{\frac{m_{CO_{2}}}{m_{O_{2}}}}\)

\(\displaystyle v_{O_{2}}=v_{CO_{2}} \sqrt{\frac{44\:\frac{grams}{mol}}{32\:\frac{grams}{mol}}}=1.17v_{CO_{2}}\)

This shows that oxygen will effuse at a rate that is about \(\displaystyle 17\, \%\) faster than carbon dioxide.

For this question, we're given the relative effusion rates for two gasses. We're then told how the mass of each of these gasses is changed, and then we're asked to determine the new relative effusion rates of the two gasses.

First, we can recall the expression that describes the dependence of the effusion rates of two gasses on their mass. Since we're told that the rate of gas A is twice that of gas B, we can write the following expression.

\(\displaystyle \frac{Rate_{A}}{Rate_{B}}=\sqrt\frac{Mass_{B}}{Mass_{A}}=2\)

Furthermore, since we're told that the mass of gas B is doubled and the mass of gas A is halved, we can determine how the rate will change.

\(\displaystyle \sqrt\frac{2}{0.5}=\sqrt{4}=2\)

Thus, we can see that the rate will change by a factor of two. Hence, the new rate will be \(\displaystyle 2\cdot2=4\). Thus, gas A will now effuse at a rate \(\displaystyle 4\) times that of gas B.

To solve this problem use Graham's Law of Effusion

\(\displaystyle \frac{\textrm{rate of effusion of A}}{\textrm{rate of effusion of B}}=\sqrt{\frac{M_{\textrm{B}}}{M_{\textrm{A}}}}\)

\(\displaystyle \textrm{A}=\textrm{Ne}\)

\(\displaystyle \textrm{B}=\textrm{unknown gas}\)

By plugging in the values we can rewrite the equation as

\(\displaystyle \frac{60.7\textrm{ s}}{45.6\textrm{ s}}=\sqrt{\frac{M_{\textrm{unknown}}}{20.2\frac{\textrm{g}}{\textrm{mol}}}}\)

\(\displaystyle (\frac{60.7\textrm{ s}}{45.6\textrm{ s}})^{2}=\frac{M_{\textrm{unknown}}}{20.2\frac{\textrm{g}}{\textrm{mol}}}\)

\(\displaystyle M_{\textrm{unknown}}=35.8\frac{\textrm{g}}{\textrm{mol}}\)