Gases - AP Chemistry

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Question

How much faster/slower the rate of effusion for oxygen gas compared to hydrogen gas?

Answer

Rate of effusion:

$\frac{(O_2)}{(H_2)} = \sqrt{\frac{(MWH_2}{MW O_2}} =\sqrt{\frac{2}{16}} = 1 : 4$

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.

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Question

Molecule A has twice the mass of molecule B. A sample of each molecule is released into separate, identical containers. Which compound will have a higher rate of diffusion?

Answer

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.

$\frac{Rate_{1}}{Rate_{2}}=\sqrt{\frac{mass_{2}}{mass_{1}}}$

$\frac{Rate_A}{Rate_B}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$

$1<\sqrt{2}; Rate_A<Rate_B$

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

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Question

A 20cm tube holds two cotton balls, one in each end. The left cotton ball is saturated with undiluted HCl. The right cotton ball is soaked in an undiluted mystery compound. Vapors from the two cotton balls are allowed to mix within the tube.

Let us assume that the two compounds form a precipitate in the tube 6cm to the left of the right cotton ball. What is the molar mass of the mystery compound?

Answer

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.

$\frac{rate_1}{rate_2}=\sqrt{\frac{M_2}{M_1}}$

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.

$\frac{rate_1}{rate_2}=\frac{14cm\ for\ HCl}{6cm\ for X}=2.33$

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

$2.33 = \sqrt{\frac{X}{36.45}}$

$2.33\sqrt{36.45}=14.1=\sqrt{X}$

$X=197.88\sim 198$

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.

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Question

A mixture of neon gas and argon gas is present in a container (container A). There are equal amounts of both gases in the container. A small pinhole is created in the container, allowing the gases to effuse into an empty container (container B). The effusion time is very brief, and the pinhole is eventually plugged, resulting in a mixture of both gases in both containers.

Which of the following statements is true after the pinhole is plugged?

Answer

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

$\small \frac{rate_{1}}{ rate_{2}} = \frac{\sqrt{M_{2}}}{\sqrt{M_{1}}}$

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.

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Question

A mixture of neon gas and argon gas is present in a container (container A). There are equal amounts of both gases in the container. A small pinhole is created in the container, allowing the gases to effuse into an empty container (container B). The effusion time is very brief, and the pinhole is eventually plugged, resulting in a mixture of both gases in both containers.

Suppose that after the pinhole is plugged, there are 100 argon atoms in container B. Approximately how many neon atoms would you predict to be in container B?

Answer

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

$\small \frac{ rate_{1}}{ rate_{2}} = \frac{\sqrt{M_{2}}}{\sqrt{M_{1}}}$

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.

$\frac{rate_1}{rate_2}=\frac{\sqrt{39.95\frac{g}{mol}}}{\sqrt{20.18\frac{g}{mol}}}=1.41$

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.

$100\ Ar\ atoms*\frac{1.41\ Ne\ atoms}{1\ Ar\ atom}=141\ Ne\ atoms$

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.

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Question

A glass box holds equal amounts of hydrogen, nitrogen, oxygen, and bromine. The gases are allowed to exit the container through a tiny hole. Which gas will exit the hole the fastest?

Answer

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:

$\frac{rate_1}{rate_2}=\sqrt\frac{M_2}{M_1}$

As the mass increases, the rate of effusion decreases.

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Question

Which of the following gases will have the highest rate of effusion?

Answer

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

$\frac{rate_1}{rate_2}=\sqrt{\frac{M_2}{M_1}}$

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

Oxygen: $O_2\rightarrow 2(16)=32\frac{g}{mol}$

Nitrogen: $N_2\rightarrow 2(14)=28\frac{g}{mol}$

Carbon dioxide: $CO_2\rightarrow 12+2(16)=44\frac{g}{mol}$

Sulfur dioxide: $SO_2\rightarrow 32+2(16)=64\frac{g}{mol}$

Helium: $He_2\rightarrow 2(4)=8\frac{g}{mol}$

The lightest, and therefore fastest, gas is helium.

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Question

Which of the following gases has the highest rate of effusion?

Answer

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

$\frac{\text{Rate}_1}{\text{Rate}_2}=\sqrt{\frac{M_2}{M_1}}$

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.

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Question

Gas A has a molar mass that is times greater than that of Gas B. Which of these gases would be expected to effuse through a small hole faster? By how much?

Answer

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:

$\frac{Rate_{B}}{Rate_{A}}=\sqrt{\frac{molar: mass_{A}}{molar: mass_{B}}}$

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.

$\frac{Rate_{B}}{Rate_{A}}=\sqrt{36}=6$

Therefore, Gas B effuses times faster than Gas A.

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Question

Per Graham's law of effusion, how does the molar mass relate to both the rate and time of effusion?

Answer

$\frac{Rate : 1}{Rate : 2} = \sqrt{\frac{Molar: Mass, , 2}{Molar: Mass: 1} }$

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

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

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

Simplifying this equation, we see that:

$\frac{Time : 2}{Time: 1}=\sqrt{\frac{Molar, , Mass, , 2}{Molar, , Mass, , 1}}$

As a result, time relates directly to molar mass

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Question

At the same temperature, an unknown gas effuses at a rate that is times that of oxygen. Find the molar mass, in grams per mole, of the unknown gas.

Answer

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

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

From the equation, we know the following:

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

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

$\sqrt{\frac{32.00}{x}}=0.291$

$\frac{32.00}{x}=0.084681$

Make sure that your answer has significant figures.

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Question

Which of the following is a true statement with regards to the relative effusion rates of oxygen and carbon dioxide?

Answer

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.

$KE_{O_{2}}=KE_{CO_{2}}$

$\frac{1}{2}m_{O_{2}}v^{2}_{O_{2}}=\frac{1}{2}m_{CO_{2}}v^{2}_{CO_{2}}$

$m_{O_{2}}v^{2}_{O_{2}}=m_{CO_{2}}v^{2}_{CO_{2}}$

$\frac{v^{2}_{O_{2}}}{v^{2}_{CO_{2}}}=\frac{m_{CO_{2}}}{m_{O_{2}}}$

$\frac{v_{O_{2}}}{v_{CO_{2}}}=\sqrt{\frac{m_{CO_{2}}}{m_{O_{2}}}}$

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.

$v_{O_{2}}=v_{CO_{2}} \sqrt{\frac{m_{CO_{2}}}{m_{O_{2}}}}$

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

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

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Question

A sample of Ne(g) effusses through a tiny hole in 60.7 s. An unknown gas, under identical conditions, effusses in 45.6 s.

What is the molar mass of the unknown gas?

Answer

To solve this problem use Graham's Law of Effusion

$\frac{\textrm{rate of effusion of A}}{\textrm{rate of effusion of B}}=\sqrt{\frac{M_{\textrm{B}}}{M_{\textrm{A}}}}$

$\textrm{A}=\textrm{Ne}$

$\textrm{B}=\textrm{unknown gas}$

By plugging in the values we can rewrite the equation as

$\frac{60.7\textrm{ s}}{45.6\textrm{ s}}=\sqrt{\frac{M_{\textrm{unknown}}}{20.2\frac{\textrm{g}}{\textrm{mol}}}}$

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

$M_{\textrm{unknown}}=35.8\frac{\textrm{g}}{\textrm{mol}}$

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Question

Suppose that gas A effuses at a rate that is twice that of gas B. If the mass of gas A is halved and the mass of gas B is doubled, which of the following correctly describes the new relative effusion rates of these two gasses?

Answer

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.

$\frac{Rate_{A}}{Rate_{B}}=\sqrt\frac{Mass_{B}}{Mass_{A}}=2$

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.

$\sqrt\frac{2}{0.5}=\sqrt{4}=2$

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

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Question

A sample of gas at a constant temperature has an initial pressure of $14.5\text{L}$ at a pressure of $720\text{mmHg}$ . If the volume of gas is decreased to $10.5\text{L}$ , what is its pressure?

Answer

Since we are given the volume and the pressure of this sample of gas, we will need to use Boyle's Law, which states that the pressure and volume of a gas, at a constant temperature, are inversely related. As thus, we can then write the following equation:

$\text{P}_1\text{V}_1=\text{P}_2\text{V}_2$

Since all the answer choices are in units of atmospheres, we will need to convert the given units into atmospheres.

$720\text{mmHg}\cdot\frac{1\text{ atm}}{760\text{mmHg}}=0.947\text{atm}$

Plug in the given pressures and volume into the equation, and solve for $\text{P}_2$ .

$(14.5)(0.947)=\text{P}_2(10.5)$

$\text{P}_2=1.31\text{atm}$

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Question

What is the pressure of a $5.20\text{L}$ cylinder filled with moles of helium gas at $317\text{K}$ ?

Answer

Recall the ideal gas law:

$\text{PV=nRT}$ ,

where $\text{P=Pressure}$ ,

$\text{V=Volume}$ ,

$\text{n=moles of gas}$ ,

$\text{R=Ideal gas constant}$ ,

and $\text{T=temperature}$ .

Since the question asks for pressure, rearrange the equation to solve for $\text{P}$ .

$\text{P}=\frac{\text{nRT}}{V}$

Plug in the given values. Remember that $\text{R=0.0821}\text{L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mole}^{-1}$

$\text{P}=\frac{(0.681)(0.0821)(317)}{5.20}=3.41\text{atm}$

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Question

Consider the following chemical reaction:

$6\text{Li}(s)+\text{N}_2(g)\rightarrow2\text{Li}_3\text{N}(s)$

How many grams of lithium is needed to react with $12.6\text{L}$ of nitrogen gas measured at $745\text{mmHg}$ and $28.5^\circ \text{C}$ ?

Answer

Start by finding the number of moles of nitrogen gas by using the ideal gas law:

$\text{PV}=\text{nRT}$

Rearrange the equation to solve for the variable $\text{n}$ :

$\text{n}=\frac{\text{PV}}{\text{RT}}$

In order to use the ideal gas law, the pressure must have units of atmospheres and the temperature must have units of Kelvin.

Convert the given pressure into atmospheres:

$745\text{mmHg}\cdot\frac{1\text{atm}}{760\text{mmHg}}=0.980\text{atm}$

$28.5\text{C}+273.15=301.65\text{K}$

Now, substitute in all the given values in order to find the number of moles of nitrogen gas.

$\text{n}=\frac{(0.980)(12.6)}{(0.0821)(301.65)}=0.499\text{ moles of nitrogen gas}$

Next, use the stoichiometric ratio given by the chemical equation to find the number of moles of lithium needed to react completely with the nitrogen gas.

$0.499\ \text{moles of N}_2\cdot\frac{6\text{ moles of Li}}{1\text{ mole of N}_2}=2.99\text{ moles of Li}$

Finally, convert the number of moles of lithium to number of grams of lithium.

$2.99\text{ moles of Li}\cdot\frac{6.94\text{g of Li}}{1\text{ mole of Li}}=20.8\text{g of Li}$

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Question

What is the pressure exerted by $0.50 mol N_{2}$ in a container at ?

Answer

$P=\frac{nRT}{V}$

$P=\frac{(0.50 mol N_{2}) (0.0821 L\cdot atm) (303K)}{13.0 L}=0.947 atm$

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Question

Avogadro's Laws relate which of the following variables?

Answer

Avogadro's law states that volume is proportional to moles of gas. Temperature and pressure are fixed. Boyle law states that volume of gas is inversely proportional to pressure. Charles' law states that volume of gas is proportional to temperature.

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Question

Ideal gas law calculation

$CO(g)+2H_{2}(g)\rightarrow CH_{3}OH(g)$

The given equation shows the synthesis of methanol. Find the volume in liters of $H_{2}(g)$ required to synthesize of methanol at $395^{\circ}K$ and a pressure of .

Answer

Use the ideal gas equation

Rearrange to solve for the volume

$V_{H2}=\frac{nRT}{P}$

$103.6gCH_{3}OH(g)\frac{1molCH_{3}OH}{32.04gCH_{3}OH}=3.23mol CH_{3}OH\frac{2molH_{2}}{1molCH_{3}OH}=6.47molH_{2}$

$V_{H2}=\frac{6.47molH_{2}(.08206\frac{Latm}{molK})395K}{\frac{725mmHg}{760}}$

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