Gases - AP Chemistry
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When does a gas behave most like an ideal gas?
When does a gas behave most like an ideal gas?
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 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.
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Under which conditions would you expect Ar to deviate the most from ideal behavior?
Under which conditions would you expect Ar to deviate the most from ideal behavior?
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.
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.
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Would you expect a polar or non polar gas to deviate most from ideal gas behavior?
Would you expect a polar or non polar gas to deviate most from ideal gas behavior?
Polar gases would have increased interactions due to their dipoles that would lead to deviations from ideal gas behavior.
Polar gases would have increased interactions due to their dipoles that would lead to deviations from ideal gas behavior.
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Which of the following would behave most like an ideal gas?
Which of the following would behave most like an ideal gas?
is the smallest molecule in the list, and therefore the least size effects.
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Why do gases deviate from ideal behavior as the temperature is decreased?
Why do gases deviate from ideal behavior as the temperature is decreased?
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.
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.
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How much faster/slower the rate of effusion for oxygen gas compared to hydrogen gas?
How much faster/slower the rate of effusion for oxygen gas compared to hydrogen gas?
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.
Rate of effusion:
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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?
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?
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.
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.
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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?
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?
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.
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.
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How many moles will a gas that is behaving ideally be if it takes up 2L at 4atm at 37o C?
How many moles will a gas that is behaving ideally be if it takes up 2L at 4atm at 37o C?
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
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
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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?
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?
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.
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.
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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?
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?
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.
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.
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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?
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?
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.
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.
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Which of the following gases will have the highest rate of effusion?
Which of the following gases will have the highest rate of effusion?
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 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.
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Which of the following gases has the highest rate of effusion?
Which of the following gases has the highest rate of effusion?
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.
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.
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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?
Gas A has a molar mass that is
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.
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
Therefore, Gas B effuses
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What is the final pressure of a gas initially has a pressure of 10 atm at 50 L if the volume s now 25 L?
What is the final pressure of a gas initially has a pressure of 10 atm at 50 L if the volume s now 25 L?
Use P1V1 = P2V2
P1 = 10atm; V1 = 50L
P2 = X; V2 = 25L
(10atm)(50L) = (x)(25L)
500 = 25x
x = 20
Use P1V1 = P2V2
P1 = 10atm; V1 = 50L
P2 = X; V2 = 25L
(10atm)(50L) = (x)(25L)
500 = 25x
x = 20
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How many moles of Oxygen gas are present at a volume of 10 L at 1 atm and 25o C? (MW Oxygen gas = 32 g/mol)
How many moles of Oxygen gas are present at a volume of 10 L at 1 atm and 25o C? (MW Oxygen gas = 32 g/mol)
use PV = nRT
n = PV/ RT
P = 1 atm; V = 10 L; R = 0.0821 Latm/molK; T = 298 K (MUST switch temperature to K)
n = moles of gas
n = (1atm)(10L)/(0.0821Latm/molK)(298K)
n= 0.41 mol
use PV = nRT
n = PV/ RT
P = 1 atm; V = 10 L; R = 0.0821 Latm/molK; T = 298 K (MUST switch temperature to K)
n = moles of gas
n = (1atm)(10L)/(0.0821Latm/molK)(298K)
n= 0.41 mol
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An ideal gas takes up 60L at 2 atm. If the gas is compressed to 30L, what will the new pressure be?
An ideal gas takes up 60L at 2 atm. If the gas is compressed to 30L, what will the new pressure be?
Ideal gas law (modified)
P1V1 = P2V2
P1 = 2 atm; V1 = 60L; P2 = ?; V2 = 30L
(2)(60) = (X)(30)
P2 = 4 atm
Ideal gas law (modified)
P1V1 = P2V2
P1 = 2 atm; V1 = 60L; P2 = ?; V2 = 30L
(2)(60) = (X)(30)
P2 = 4 atm
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Each of the following compounds are contained in a closed container with a volume of 100mL. Which of the following will exert the most pressure?
Each of the following compounds are contained in a closed container with a volume of 100mL. Which of the following will exert the most pressure?
The pressure exerted is dependent on molar concentration only (not mass). As the containers are all the same volume, the container with the most moles will exert the most pressure. 60g of 44g/mole CO2 gives 1.4mol, 50g of O2 gives 1.4mol, 46g N2 gives 1.6mol, and 54g F2 gives 1.4mol, 40g H2 gives 40mol.
The pressure exerted is dependent on molar concentration only (not mass). As the containers are all the same volume, the container with the most moles will exert the most pressure. 60g of 44g/mole CO2 gives 1.4mol, 50g of O2 gives 1.4mol, 46g N2 gives 1.6mol, and 54g F2 gives 1.4mol, 40g H2 gives 40mol.
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A mixture of three gases in a balloon is able to generate 2 atm of pressure. The identies and amounts of each gas are as follows. 2 moles of O2, 3 moles of N2, 5 moles of H2.
What is the partial pressure of Oxygen gas?
A mixture of three gases in a balloon is able to generate 2 atm of pressure. The identies and amounts of each gas are as follows. 2 moles of O2, 3 moles of N2, 5 moles of H2.
What is the partial pressure of Oxygen gas?
Partial pressure is the amount of pressure that is the result of one gas. This is calculated by mutiplying the mole fraction of the gas by the overall pressure. The mole fraction is calculated as the moles of the compound of interest divided by the total amount of moles present. 2 moles O2/ 2 moles O2 + 3 moles N2 + 5 moles H2 = 2/10 = mole fraction of O2. 2/10 multiplied by the total pressure gives us the answer as 0.4 atm.
Partial pressure is the amount of pressure that is the result of one gas. This is calculated by mutiplying the mole fraction of the gas by the overall pressure. The mole fraction is calculated as the moles of the compound of interest divided by the total amount of moles present. 2 moles O2/ 2 moles O2 + 3 moles N2 + 5 moles H2 = 2/10 = mole fraction of O2. 2/10 multiplied by the total pressure gives us the answer as 0.4 atm.
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