Gases - AP Chemistry
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A gas is at STP
What does STP refer to?
A gas is at STP
What does STP refer to?
Standard temperature and pressure is 00C and 1 atm
Standard temperature and pressure is 00C and 1 atm
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COCl2 (g) ⇌ CO (g) + Cl2 (g)
Initially a reaction chamber contains only 2 moles of COCl2, after the reaction has reached the equilibrium shown above the pressure is measures to be 2.60 atm. What is the most likely cause for the increase in pressure?
COCl2 (g) ⇌ CO (g) + Cl2 (g)
Initially a reaction chamber contains only 2 moles of COCl2, after the reaction has reached the equilibrium shown above the pressure is measures to be 2.60 atm. What is the most likely cause for the increase in pressure?
The equilibrium expression indicates that one mole of COCl2(g) will decompose into CO (g) and Cl2 (g). One mole of gas is present on the left side of the equation, and two moles of gas are present on the right side of the equation. The increases in pressure is due to the increase in the number of gas molecules in the reaction chamber.
The equilibrium expression indicates that one mole of COCl2(g) will decompose into CO (g) and Cl2 (g). One mole of gas is present on the left side of the equation, and two moles of gas are present on the right side of the equation. The increases in pressure is due to the increase in the number of gas molecules in the reaction chamber.
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What is the osmotic pressure exerted by a solution of 2M CaCl2 at room temperature? (R=0.082 L*atm/K/mol)
What is the osmotic pressure exerted by a solution of 2M CaCl2 at room temperature? (R=0.082 L*atm/K/mol)
The van't Hoff factor for CaCl2 is 3 since the molecule dissociates in solution to 3 ions. T is in absolute temperature (25+273=298). This makes the osmotic pressure:
3*2M*0.082*298.
The van't Hoff factor for CaCl2 is 3 since the molecule dissociates in solution to 3 ions. T is in absolute temperature (25+273=298). This makes the osmotic pressure:
3*2M*0.082*298.
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What is the molar mass of a gas with a density of 
at STP?
What is the molar mass of a gas with a density of
Since we have the density of the mystery gas, we can rearrange the ideal gas law so that the remaining factors are equal to the density of the gas.
We can redefine moles of gas as mass over molar mass.
We can now rearrange the equation to solve for density (mass per volume), which is given in the question.
Using this set up and the values for standard temperature and pressure, we can solve for the molar mass.
Since we have the density of the mystery gas, we can rearrange the ideal gas law so that the remaining factors are equal to the density of the gas.
We can redefine moles of gas as mass over molar mass.
We can now rearrange the equation to solve for density (mass per volume), which is given in the question.
Using this set up and the values for standard temperature and pressure, we can solve for the molar mass.
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It can be argued that one of the greatest innovations in automobile safety is the airbag. A typical driver's side airbag expands to a volume of between 50-60L. For this problem use 
. In older airbag models sodium azide is employed to produce nitrogen gas as shown in the following decomposition reaction:
Two more reactions are employed to remove the reactive sodium metal, shift the above reaction equilibrium to the right, and produce more nitrogen gas; however for this problem assume all gas is generated from the first reaction as shown above and that the reaction proceeds to completion.
The gauge pressure is 
. This is the pressure at which the airbag is deployed. Note: gauge pressure is defined as the additional pressure in the system relative to atmospheric pressure. In this case use STP pressure as atmospheric pressure.
(It is worth noting that this pressure is the initial pressure to fill the airbag, which would provide a rock-hard cushion; therefore in the real world the air bag immediately begins to deflate so as to provide proper cushioning; that fact is not important for solving this problem).
What is the minimum mass of 
required to inflate the airbag given the above conditions?
It can be argued that one of the greatest innovations in automobile safety is the airbag. A typical driver's side airbag expands to a volume of between 50-60L. For this problem use
Two more reactions are employed to remove the reactive sodium metal, shift the above reaction equilibrium to the right, and produce more nitrogen gas; however for this problem assume all gas is generated from the first reaction as shown above and that the reaction proceeds to completion.
The gauge pressure is
(It is worth noting that this pressure is the initial pressure to fill the airbag, which would provide a rock-hard cushion; therefore in the real world the air bag immediately begins to deflate so as to provide proper cushioning; that fact is not important for solving this problem).
What is the minimum mass of
First, identify the necessary parameters.
We must use the ideal gas equation:
and since the molar gas constant given uses Kelvin in it's definition, we must convert 
to Kelvin as follows: 
Since the pressure given is gauge pressure, we must convert to the actual pressure as follow: 
So therefore the pressure we want to use is 
Now that we have listed the given quantities we can see that in order to get the mass of 
needed, we must first find the number of moles of 
needed to fill the volume of the airbag, then use the molar ratios from the above chemical equation to convert moles of 
to moles of 
, after which we can use the molar mass of 
to calculate the grams of 
needed to inflate the airbag.
In order to calculate the number of moles of 
we can employ the Ideal gas law: to find n (the number of moles of gas; which in this case the only gas involved is 
Plug in the given quantities that have been converted into the correct units and quantities:
Solve.
Now that we know the moles of 
needed we can use stoichiometry, and the information in the following equation:
To determine the number of moles of 
required and then convert the moles of 
to the grams of 
using the molar mass of 
. The molar mass of 
can be found by using a periodic table and adding the molar masses of the constituent elements as shown below:
Molar mass of
Given that we calculated the moles of 
to fill the volume of the airbag to be: 
, and the molar ratios from the chemical equation we can set up the final steps of the calculation as shown below:
Therefore in order to inflate this driver's side airbag it would need to contain at least 68.04g of
First, identify the necessary parameters.
We must use the ideal gas equation:
Since the pressure given is gauge pressure, we must convert to the actual pressure as follow:
So therefore the pressure we want to use is
Now that we have listed the given quantities we can see that in order to get the mass of
In order to calculate the number of moles of
Plug in the given quantities that have been converted into the correct units and quantities:
Solve.
Now that we know the moles of
To determine the number of moles of
Molar mass of
Given that we calculated the moles of
Therefore in order to inflate this driver's side airbag it would need to contain at least 68.04g of
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Which of the following will not be found in the atmosphere for an extended period of time?
Which of the following will not be found in the atmosphere for an extended period of time?
Most gases exist as diatomic molecules in nature. 
, 
, and 
are all common gases found in the atmosphere and the air we breathe.
Ozone, 
, is a rare exception to the diatomic bonding principle, and is also found in significant quantities in the atmosphere.
Monatomic oxygen, 
, has only six valence electrons and is relatively unstable; this compound would not be found naturally in the atmosphere.
Most gases exist as diatomic molecules in nature.
Ozone,
Monatomic oxygen,
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A flexible box contains four moles of octane gas at STP with an excess of oxygen. If the octane is combusted according to the following reaction and the box is allowed to expand, what is the change in volume?
A flexible box contains four moles of octane gas at STP with an excess of oxygen. If the octane is combusted according to the following reaction and the box is allowed to expand, what is the change in volume?
This question does not acually deal with the ideal gas law, but only with the standard volume of a gas at STP. At standard temperature and pressure, one mole of gas has a volume of 22.5L.
We know that initially we have four moles of octane and excess oxygen. This means that octane will be the limiting reagent for the combustion reaction, since oxygen is in excess.
The total moles of reactant gas when utilizing all four moles of octane is 54, while the total moles of product gas is 68.
There is a total change in moles of:
Since we end the reaction with 14 more moles of gas than we started, the volume must increase:
The change in volume is 
.
This question does not acually deal with the ideal gas law, but only with the standard volume of a gas at STP. At standard temperature and pressure, one mole of gas has a volume of 22.5L.
We know that initially we have four moles of octane and excess oxygen. This means that octane will be the limiting reagent for the combustion reaction, since oxygen is in excess.
The total moles of reactant gas when utilizing all four moles of octane is 54, while the total moles of product gas is 68.
There is a total change in moles of:
Since we end the reaction with 14 more moles of gas than we started, the volume must increase:
The change in volume is
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At 
the vapor pressure of pure water is 
. What is the minimum amount of liquid water we need to put in a 
flask to achieve this vapor pressure. 
At
We use the ideal gases equation to know the mol of water we need in the gas phase to generate a pressure of 
in a 
flask at 
.
And the mass is:
If we have less than that amount of water the liquid will evaporate completely but there will not be enough molecules to reach the vapor pressure. More than 
of water will allow to reach the vapor pressure and there will be also a liquid leftover. Remember that the vapor pressure of a liquid at a fixed temperature is a constant.
We use the ideal gases equation to know the mol of water we need in the gas phase to generate a pressure of
And the mass is:
If we have less than that amount of water the liquid will evaporate completely but there will not be enough molecules to reach the vapor pressure. More than
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of 
are introduced in a 
flask. A vacuum is applied to the flask and its temperature raised to 
. At this temperature ammonium nitrate decomposes according to the reaction:
If the percent yield of this reaction is 
, which will be the total pressure of the flask after the reaction concludes. 
If the percent yield of this reaction is
Using the molecular mass we calculate the number of mol of 
:
At this moment we could apply the percentage yield, however let's live it for the end. Each mole of 
decomposed will generate 
of gas. Then the total pressure assuming 
yield will be:
Since only 
of the 
decomposes, the real total pressure will be:
Using the molecular mass we calculate the number of mol of
At this moment we could apply the percentage yield, however let's live it for the end. Each mole of
Since only
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Suppose that a container contains 
of a gaseous sample of unknown hydrocarbon at STP. If this sample occupies a volume of 
, what is the identity of the hydrocarbon?
Suppose that a container contains
In this question, we're given the mass and volume of an unknown hydrocarbon filling a container. We're asked to determine the identity of this compound.
To solve this problem, it's important to realize that for any ideal gas at STP (standard temperature and pressure), 
of the gas will equate to 
of that gas. For this question, we're told that there is present.
Now that we know how many moles of gas are in the container, we can use this information, together with the mass provided to us in the question stem, to determine the molecular mass of the unknown compound.
Looking at the answer choices, the only hydrocarbon that matches this molecular mass is ethane, 
.
In this question, we're given the mass and volume of an unknown hydrocarbon filling a container. We're asked to determine the identity of this compound.
To solve this problem, it's important to realize that for any ideal gas at STP (standard temperature and pressure),
Now that we know how many moles of gas are in the container, we can use this information, together with the mass provided to us in the question stem, to determine the molecular mass of the unknown compound.
Looking at the answer choices, the only hydrocarbon that matches this molecular mass is ethane,
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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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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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A gas is at STP
What does STP refer to?
A gas is at STP
What does STP refer to?
Standard temperature and pressure is 00C and 1 atm
Standard temperature and pressure is 00C and 1 atm
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