Gases - AP Chemistry

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.

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.

Thus, we can see that the rate will change by a factor of two. Hence, the new rate will be . Thus, gas A will now effuse at a rate times that of gas B.

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

Because , time relates to molar mass by:

Simplifying this equation, we see that:

As a result, time relates directly to molar mass

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

From the equation, we know the following:

Thus, we can solve for the molar mass of the unknown gas. Let be the molar mass of the unknown gas.

Make sure that your answer has 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.

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.

This shows that oxygen will effuse at a rate that is about faster than carbon dioxide.

To solve this problem use Graham's Law of Effusion

By plugging in the values we can rewrite the equation as

The ideal gas law assumes the gas particles are non-interacting and small relative to the size of their container. At high temperatures the gas molecules are moving fast enough to shorten the time scale for any interactions. At high volumes, the molecular size becomes small relative to the size of the container, and the low interactions mean the molecules act more independently.

The ideal gas law assumes the gas particles are non-interacting and small relative to the size of their container. At 200K (lowest temperature in the list, and the highest pressure). This gives Ar the most time to interact due to molecular speeds and the high pressure implies the molecular size is not insignificant relative to the container.

Polar gases would have increased interactions due to their dipoles that would lead to deviations from ideal gas behavior.

is the smallest molecule in the list, and therefore the least size effects.

The ideal gas law assumes the gas particles are non-interacting and small relative to the size of their container. As the temperature is decreased the gas molecules are moving slower and allow for a greater degree of interaction.

Rate of effusion:

and must be used because they exist as bimolecular molecules. The correct answer is that Oxygen gas will effuse 4 times slower than hydrogen gas.

According to Graham's law, the rate of diffusion of a gas molecule is inversely proportional to the root square of that molecule's mass. Because molecule B has a smaller mass than molecule A, it will have a higher rate of diffusion.

Note: this also applies to finding the rate of effusion.

This question is notably difficult, as it may not be immediately apparent what concept is being tested. As the vapors of the compounds mix and react, we are able to establish the distance the each vapor has traveled from the cotton ball into the tube in the given amount of time. The tube is 20cm, and the reaction takes place 6cm from the mystery compound cotton ball. From this, we can establish that in an equal amount of time the HCl vapor traveled 14cm and the mystery compound traveled 6cm.

In order to solve this problem, we use Graham's law to compare molar masses to the rates of diffusion of the two gases.

Since HCl moved 14cm to the right before interacting with the mystery compound, we know that the mystery compound moved only 6cm to the left. As a result, the diffusion ratio is 2.33.

Now, we need to find the square root of the inversed molar masses, which equals this diffusion ratio.

So, the molar mass of the mystery compound is 198 grams per mole. This makes sense, because larger gases will move more slowly compared to lighter gases.

Use PV = nRT

n = PV/RT

= (2L)(4atm) / (0.0821 Latm/molK)(310 K) <-- must change T into K

= 8/25.451

= 0.314 mol

The rate of effusion for two gases can be compared to one another using the following equation:

Here, the effusion rates are inversely proportional to the square root of the molecular masses of the gases in question. Because the relationship is to the square roots of the molecular masses, we will not observe a 2:1 ratio of effusion for neon compared to argon.

We will, however, see that more neon effuses out of container A compared to the amount of argon because neon is the lighter gas and will thus have a faster effusion rate. As a result, there will be more argon than neon in container A after the pinhole is plugged. This results in argon having a larger partial pressure than neon in container A.

We can compare the effusion rates of these gases using the following equation.

By calling neon "gas 1" and argon "gas 2," we can compare the effusion rates of the two gases by plugging their molecular masses into the equation.

This proportion is equal to the rate of neon effusion over the rate of argon effusion, giving the ratio of neon atoms to argon atoms in container B.

As a result, 141 atoms of neon gas will effuse out of the pinhole for every 100 argon gas atoms. Keep in mind that the heavier gas will effuse at a slower rate than the lighter gas; thus, we would expect there to be more neon than argon in container B.

At a particular temperature, the average kinetic energy of all gaseous molecules is equal. Since hydrogen gas has the lowest mass out of these gases, it will have the highest average velocity. This means that it will exit out of the tiny hole at a rate faster than the other gases. Conversely, bromine, which has the most mass compared to the other gases, will exit the hole the slowest.

This relationship is mathematically represented in Graham's law:

As the mass increases, the rate of effusion decreases.

The rate of effusion for a gas is inversely proportional to the square-root of its molecular mass (Graham's Law).

The gas with the lowest molecular weight will effuse the fastest.

Oxygen:

Nitrogen:

Carbon dioxide:

Sulfur dioxide:

Helium:

The lightest, and therefore fastest, gas is helium.

The rate of effusion of a gas is inversely proportional to the square root of the molecular weight of the gas.

The lighter a gas is, the faster it will effuse; the heavier a gas is, the slower it will effuse.

Of all the choices, helium (He) has the lowest molecular weight (atomic weight in this case), so it will have the highest rate of effusion.

In order to answer this question, let's start by considering what effusion is and what things affect it. Effusion is the movement of a gas through a tiny hole that separates two different spaces. Because the gas particles move around in random directions with an average speed that is dependent on the temperature of the sample, lighter gas particle will move faster than heavier gas particles. This is because at a given temperature, all gas particles in a sample will have the same average kinetic energy. Consequently, we would expect gas particles with a higher molar mass to effuse more slowly than gases with a lower molar mass. This means that Gas B should effuse faster than Gas A.

The next step is to actually calculate how much greater Gas B effuses compared to Gas A. To do this, we'll need to use the following equation:

Since we know that Gas A is times heavier than Gas B, we can plug this into the equation to solve for the ratio of Gas B's rate of effusion to that of Gas A.

Therefore, Gas B effuses times faster than Gas A.