The Behavior of Gases

The Behavior of Gases

Compressibility

  • Define compressibility.
  • Give examples of the uses of compressed gases.


Compressing objects can help to squeeze them into small spaces

Will it all fit?

When we pack to go on vacation, there is always “one more” thing that we need to get in the suitcase. Maybe it’s another bathing suit, pair of shoes, book – whatever the item, we need to get it in. Fortunately, we can squeeze things together somewhat. There is a little space between the folds of clothing, we can rearrange the shoes, and somehow we get that last thing in and close the suitcase.

Compressibility

Scuba diving is a form of underwater diving in which a diver carries his own breathing gas, usually in the form of a tank of compressed air. The pressure in most commonly used scuba tanks ranges from 200 to 300 atmospheres. Gases are unlike other states of matter in that a gas expands to fill the shape and volume of its container. For this reason, gases can also be compressed so that a relatively large amount of gas can be forced into a small container. If the air in a typical scuba tank were transferred to a container at the standard pressure of 1 atm, the volume of that container would need to be about 2500 liters.

The pressure in a scuba tank is typically 200-300 atmospheres

Figure 14.1

Scuba diver.

Compressibility is the measure of how much a given volume of matter decreases when placed under pressure. If we put pressure on a solid or a liquid, there is essentially no change in volume. The atoms, ions, or molecules that make up the solid or liquid are very close together. There is no space between the individual particles, so they cannot pack together.

The kinetic-molecular theory explains why gases are more compressible than either liquids or solids. Gases are compressible because most of the volume of a gas is composed of the large amounts of empty space between the gas particles. At room temperature and standard pressure, the average distance between gas molecules is about ten times the diameter of the molecules themselves. When a gas is compressed, as when the scuba tank is being filled, the gas particles are forced closer together.

Compressed gases are used in many situations. In hospitals, oxygen is often used for patients who have damaged lungs to help them breathe better. If a patient is having a major operation, the anesthesia that is administered will frequently be a compressed gas. Welding requires very hot flames produced by compresses acetylene and oxygen mixtures. Many summer barbeque grills are fueled by compressed propane.

Compressed gases such as oxygen are used for a wide variety of applications

Figure 14.2

Oxygen tank.

Summary

  • Gases will compress more easily that solids or liquids because here is so much space between the gas molecules.


Practice

Questions

Use the link below to answer the following questions:

http://www.cdxetextbook.com/engines/motivePower/4gasEng/engcycle.html

  1. What brings the fuel-air mixture into the cylinder?
  2. What is the role of the compression cycle?
  3. Does the exhaust cycle compress the gases produced by ignition?


Review

Questions

  1. Why is there no change in volume when pressure is applied to liquids and solids?
  2. Why do gases compress more easily than liquids and solids?
  3. List uses for compressed gases.


  • compressibility: The measure of how much a given volume of matter decreases when placed under pressure.


Factors Affecting Gas Pressure

  • List factors that affect gas pressure.
  • Explain these effects in terms of the kinetic-molecular theory of gases.


The pressure in a basketball in a game must be kept in a certain range

How high does a basketball bounce?

The pressure of the air in a basketball has to be adjusted so that the ball bounces to the correct height. Before a game, the officials check the ball by dropping it from shoulder height and seeing how far back up it bounces. What would the official do if the ball did not bounce up as far as it is supposed to? What would he do if it bounced too high?

The pressure inside a container is dependent on the amount of gas inside the container. If the basketball does not bounce high enough, the official could remedy the situation by using a hand pump and adding more air to the ball. Conversely, if it bounces too high, he could let some air out of the ball.

Factors Affecting Gas Pressure

Recall from the kinetic-molecular theory that gas particles move randomly and in straight lines until they elastically collide with either other gas particles or with one of the walls of the container. It is these collisions with the walls of the container that defines the pressure of the gas. Four variables are used to describe the condition of a gas. They are pressure (P) , volume (V) , temperature (T) , and the amount of the gas as measured by the number moles (n) . We will examine separately how the volume, temperature, and amount of gas each affect the pressure of an enclosed gas sample.

Amount of Gas

The Figure below shows what happens when air is added to a rigid container . A rigid container is one that is incapable of expanding or contracting. A steel canister is an example of a rigid container.

An increase in the number of gas particles causes an increase in the pressure of a gas

Figure 14.3

Increase in pressure with increase in number of gas particles.

The canister on the left contains a gas at a certain pressure. The attached air pump is then used to double the amount of gas in the canister. Since the canister cannot expand, the increased number of air molecules will strike the inside walls of the canister twice as frequently as they did before. The result is that the pressure inside the canister doubles. As you might imagine, if more and more air is continually added to a rigid container, it may eventually burst. Reducing the number of molecules in a rigid container has the opposite effect and the pressure decreases.

Volume

Pressure is also affected by the volume of the container. If the volume of a container is decreased, the gas molecules have less space in which to move around. As a result, they will strike the walls of the container more often and the pressure increases.

Figure below shows a cylinder of gas whose volume is controlled by an adjustable piston. On the left, the piston is pulled mostly out and the gauge reads a certain pressure. On the right, the piston has been pushed so that the volume of the enclosed portion of the container where the gas is located has been cut in half. The pressure of the gas doubles. Increasing the volume of the container would have the opposite effect and the pressure of the gas would decrease.

A decrease in volume causes an increase in pressure for a gas

Figure 14.4

Decrease in gas volume produced increase in gas pressure.

Temperature

It would be very unadvisable to place a can of soup over a campfire without venting the can. As the can heats up, it may explode. The kinetic-molecular theory explains why. The air inside the rigid can of soup is given more kinetic energy by the heat coming from the campfire. The kinetic energy causes the air molecules to move faster and they impact the container walls more frequently and with more force. The increase in pressure inside may eventually exceed the strength of the can and it will explode. An additional factor is that the soup may begin boiling which will then aid even more gas and more pressure to the inside of the can.

Shown in the Figure below is a cylinder of gas on the left that is at room temperature (300 K). On the right, the cylinder has been heated until the Kelvin temperature has doubled to 600 K. The kinetic energy of the gas molecules increases, so collisions with the walls of the container are now more forceful than they were before. As a result, the pressure of the gas doubles. Decreasing the temperature would have the opposite effect, and the pressure of an enclosed gas would decrease.

An increase in temperature causes an increase in pressure for a gas Figure 14.5

Increase in temperature produces increase in pressure.

Summary

  • An increase in the number of gas molecules in the same volume container increases pressure.
  • A decrease in container volume increases gas pressure.
  • An increase in temperature of a gas in a rigid container increases the pressure.


Practice

Questions

Watch the video at the link below and answer the following questions:

Click on the image above for more content

  1. What causes pressure?
  2. What happens when you let gas out of the container?
  3. If you increase the temperature, what happens to the pressure?
  4. Why does the pressure drop when you increase the volume?


Review

Questions

  1. What defines the pressure of a gas?
  2. Why does an increase in the number of molecules increase the pressure?
  3. Why does an increase in temperature increase the pressure?


  • rigid container: One that is incapable of expanding or contracting.


Boyle's Law

  • State Boyle’s Law.
  • Use Boyle’s Law to calculate volume-pressure relationships.


Weather balloons expand and eventually burst as they travel to higher altitudes

How important is it to check the weather?

Each day, hundreds of weather balloons are launched. Made of a synthetic rubber and carrying a box of instruments, the helium-filled balloon rises up into the sky. As it gains altitude, the atmospheric pressure becomes less and the balloon expands. At some point the balloon bursts due to the expansion, the instruments drop (aided by a parachute) to be retrieved and studied for information about the weather.

Boyle’s Law

Robert Boyle (1627-1691), an English chemist, is widely considered to be one of the founders of the modern experimental science of chemistry. He discovered that doubling the pressure of an enclosed sample of gas while keeping its temperature constant caused the volume of the gas to be reduced by half. Boyle’s law states that the volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant. An inverse relationship is described in this way. As one variable increases in value, the other variable decreases.

Physically, what is happening? The gas molecules are moving and are a certain distance apart from one another. An increase in pressure pushes the molecules closer together, reducing the volume. If the pressure is decreased, the gases are free to move about in a larger volume.

Robert Boyle states that PV is constant at a given temperature

Figure 14.6

Robert Boyle.

Mathematically, Boyle’s law can be expressed by the equation:

P times V = k

The  k is a constant for a given sample of gas and depends only on the mass of the gas and the temperature. The Table below shows pressure and volume data for a set amount of gas at a constant temperature. The third column represents the value of the constant  (k) for this data and is always equal to the pressure multiplied by the volume. As one of the variables changes, the other changes in such a way that the product of  P times V always remains the same. In this particular case, that constant is 500 atm · ml.

Pressure-Volume Data
Pressure (atm) Volume (mL) P times V = k (text{atm} cdot text{mL})
0.5 1000 500
0.625 800 500
1.0 500

500

2.0 250

500

5.0 100 500
8.0 62.5 500
10.0 50 500

A graph of the data in the table further illustrates the inverse relationship nature of Boyle’s Law (see Figure below ). Volume is plotted on the x -axis, with the corresponding pressure on the y -axis.

The pressure of a gas decreases as its volume increases

Figure 14.7

The pressure of a gas decreases as the volume increases, making Boyle’s law an inverse relationship.

Boyle’s Law can be used to compare changing conditions for a gas. We use  P_1 and  V_1 to stand for the initial pressure and initial volume of a gas. After a change has been made,  P_2 and  V_2 stand for the final pressure and volume. The mathematical relationship of Boyle’s Law becomes:

P_1 times V_1=P_2 times V_2

This equation can be used to calculate any one of the four quantities if the other three are known.

Sample Problem: Boyle’s Law

A sample of oxygen gas has a volume of 425 mL when the pressure is equal to 387 kPa. The gas is allowed to expand into a 1.75 L container. Calculate the new pressure of the gas.

Step 1: List the known quantities and plan the problem.

Known

  • P_1=387 text{kPa}
  • V_1=425 text{mL}
  • V_2=1.75 text{L}=1750 text{mL}


Unknown

  • P_2=? text{kPa}


Use Boyle’s Law to solve for the unknown pressure (P_2) . It is important that the two volumes ( V_1 and V_2 ) are expressed in the same units, so  V_2 has been converted to mL.

Step 2: Solve.

First, rearrange the equation algebraically to solve for P_2 .

P_2=frac{P_1 times V_1}{V_2}

Now substitute the known quantities into the equation and solve.

P_2=frac{387 text{ kPa} times 425 text{ mL}}{1750 text{ mL}}=94.0 text{ kPa}

Step 3: Think about your result.

The volume has increased to slightly over 4 times its original value and so the pressure is decreased by about frac{1}{4}{th} . The pressure is in kPa and the value has three significant figures. Note that any pressure or volume units can be used as long as they are consistent throughout the problem.

Summary

  • The volume of a gas is inversely proportional to temperature.


Practice

Do the problems at the link below:

http://www.concord.org/~ddamelin/chemsite/g_gasses/handouts/Boyle_Problems.pdf

Review

Questions

  1. What does “inversely” mean in this law?
  2. Explain Boyle’s law in terms of the kinetic-molecular theory of gases.
  3. Does it matter what units are used?


  • Boyle’s law: The volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant.


Charles's Law

  • State Charles’ Law.
  • Use this law to perform calculations involving volume-temperature relationships.


When bread is baked, the carbon dioxide inside expands

How do you bake bread?

Everybody enjoys the smell and taste of freshly-baked bread. It is light and fluffy as a result of the action of yeast on sugar. The yeast converts the sugar to carbon dioxide, which at high temperatures causes the dough to expand. The end-result is an enjoyable treat, especially when covered with melted butter.

Charles’s Law

As a gas is heated at constant pressure, its volume increases

Figure 14.8

As a container of confined gas is heated, its molecules increase in kinetic energy and push the movable piston outward, resulting in an increase in volume.

French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles’s law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stoppage of molecular motion.

Mathematically, the direct relationship of Charles’s law can be represented by the following equation:

frac{V}{T}=k

As with Boyle’s law, k is constant only for a given gas sample. The Table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature.

Temperature-Volume Data
Temperature (K) Volume (mL) frac{V}{T}=kleft (frac{mL}{K}right)
50 20 0.40
100 40 0.40
150 60 0.40
200 80 0.40
300 120 0.40
500 200 0.40
1000 400 0.40

When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in Figure below .

The volume of a gas increases as the Kelvin temperature increases

Figure 14.9

The volume of a gas increases as the Kelvin temperature increases.

Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases.

Charles’s Law can also be used to compare changing conditions for a gas. Now we use  V_1 and  T_1 to stand for the initial volume and temperature of a gas, while V_2 and  T_2 stand for the final volume and temperature. The mathematical relationship of Charles’s Law becomes:

frac{V_1}{T_1}=frac{V_2}{T_2}

This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work. Recall the relationship that K = °C + 273.

Sample Problem: Charles’s Law

A balloon is filled to a volume of 2.20 L at a temperature of 22°C. The balloon is then heated to a temperature of 71°C. Find the new volume of the balloon.

Step 1: List the known quantities and plan the problem.

Known

  • V_1=2.20 text{ L}
  • T_1=22^circ text{C}=295 text{ K}
  • T_2=71^circ text{C}=344 text{ K}


Unknown

  • V_2= ? text{ L}


Use Charles’s law to solve for the unknown volume (V_2) . The temperatures have first been converted to Kelvin.

Step 2: Solve.

First, rearrange the equation algebraically to solve for V_2 .

V_2=frac{V_1 times T_2}{T_1}

Now substitute the known quantities into the equation and solve.

V_2=frac{2.20 text{ L} times 344 text{ K}}{295 text{ K}}=2.57 text{ L}

Step 3: Think about your result.

The volume increases as the temperature increases. The result has three significant figures.

Summary

  • Increasing the temperature of a gas at constant pressure will produce and increase in the volume.


Practice

Perform the calculations at the web site below:

http://mmsphyschem.com/chuckL.pdf

Review

Questions

  1. Explain Charles’ Law in terms of the kinetic molecular theory.
  2. Why does the temperature need to be in Kelvin?
  3. Does Charles’ law hold when the gas becomes a liquid?


  • Charles’s law: The volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant.


Gay-Lussac's Law

  • State Gay-Lussac’s law.
  • Use this law to perform calculations involving pressure-temperature relationships.


The temperature of a tank of gas needs to be taken into account when measuring its pressure

How much propane is in the tank?

Propane tanks are widely used with barbeque grills. But it’s not fun to find out half-way through your grilling that you’ve run out of gas. You can buy gauges that measure the pressure inside the tank to see how much is left. The gauge measures pressure and will register a higher pressure on a hot day than it will on a cold day. So you need to take the air temperature into account when you decide whether or not to refill the tank before your next cook-out.

Gay-Lussac’s Law

When the temperature of a sample of gas in a rigid container is increased, the pressure of the gas increases as well. The increase in kinetic energy results in the molecules of gas striking the walls of the container with more force, resulting in a greater pressure. The French chemist Joseph Gay-Lussac (1778-1850) discovered the relationship between the pressure of a gas and its absolute temperature. Gay-Lussac’s law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. Gay-Lussac’s law is very similar to Charles’s law, with the only difference being the type of container. Whereas the container in a Charles’s law experiment is flexible, it is rigid in a Gay-Lussac’s law experiment.

In Gay-Lussac's Law, the pressure of a gas varies directly with the temperature of a gas

Figure 14.10

Joseph Louis Gay-Lussac.

The mathematical expressions for Gay-Lussac’s law are likewise similar to those of Charles’s law:

frac{P}{T}=k quad text{and} quad frac{P_1}{T_1}=frac{P_2}{T_2}

A graph of pressure vs. temperature also illustrates a direct relationship. As a gas is cooled at constant volume its pressure continually decreases until the gas condenses to a liquid.

Sample Problem: Gay-Lussac’s Law

The gas in an aerosol can is under a pressure of 3.00 atm at a temperature of 25°C. It is dangerous to dispose of an aerosol can by incineration. What would the pressure in the aerosol can be at a temperature of 845°C?

Step 1: List the known quantities and plan the problem.

Known

  • P_1=3.00 text{atm}
  • T_1=25^circ text{C}=298 text{ K}
  • T_2=845^circ text{C}=1118 text{ K}


Unknown

  • P_2=? text{atm}


Use Gay-Lussac’s law to solve for the unknown pressure (P_2) . The temperatures have first been converted to Kelvin.

Step 2: Solve.

First, rearrange the equation algebraically to solve for V_2 .

P_2=frac{P_1 times T_2}{T_1}

Now substitute the known quantities into the equation and solve.

P_2=frac{3.00 text{ atm} times 1118 text{ K}}{298 text{ K}}=11.3 text{ atm}

Step 3: Think about your result.

The pressure increases dramatically due to large increase in temperature.

Summary

  • Pressure and temperature at constant volume are directly proportional.


Practice

Work on the problems found at the web site below:

http://www.chemteam.info/GasLaw/WS-Gay-Lussac.html

Review

Questions

  1. Explain Gay-Lussac’s Law in terms of the kinetic-molecular theory.
  2. What would a graph of pressure vs. temperature show us?
  3. What is the difference in containers in Charles’ Law and Gay-Lussac’s Law?


  • Gay-Lussac’s law: The pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant.


Combined Gas Law

  • State the combined gas law.
  • Use the law to calculate parameters in general gas problems.


A refrigerator uses the combined gas law to dissipate heat

What keeps things cold?

The modern refrigerator takes advantage of the gas laws to remove heat from a system. Compressed gas in the coils (see above) is allowed to expand. This expansion lowers the temperature of the gas and transfers heat energy from the material in the refrigerator to the gas. As the gas is pumped through the coils, the pressure on the gas compresses it and raises the gas temperature. This heat is then dissipated through the coils into the outside air. As the compressed gas is pumped through the system again, the process repeats itself.

Combined Gas Law

To this point, we have examined the relationships between any two of the variables of P , V , and T , while the third variable is held constant. However, situations arise where all three variables change. The combined gas law expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas. For a combined gas law problem, only the amount of gas is held constant.

frac{P times V}{T}=k quad text{and} quad frac{P_1 times V_1}{T_1}=frac{P_2 times V_2}{T_2}

Sample Problem: Combined Gas Law

2.00 L of a gas at 35°C and 0.833 atm is brought to standard temperature and pressure (STP). What will be the new gas volume?

Step 1: List the known quantities and plan the problem.

Known

  • P_1=0.833 text{ atm}
  • V_1=2.00 text{ L}
  • T_1=35^circ text{C}=308 text{ K}
  • P_2=1.00 text{ atm}
  • T_2=0^circ text{C}=273 text{ K}


Unknown

  • V_2=? text{ L}


Use the combined gas law to solve for the unknown volume (V_2) . STP is 273 K and 1 atm. The temperatures have been converted to Kelvin.

Step 2: Solve

First, rearrange the equation algebraically to solve for V_2 .

V_2=frac{P_1 times V_1 times T_2}{P_2 times T_1}

Now substitute the known quantities into the equation and solve.

V_2=frac{0.833 text{ atm} times 2.00 text{ L} times 273 text{ K}}{1.00 text{ atm} times 308 text{ K}}=1.48 text{ L}

Step 3: Think about your result.

Both the increase in pressure and the decrease in temperature cause the volume of the gas sample to decrease.  Since both changes are relatively small, the volume does not decrease dramatically.

It may seem challenging to remember all the different gas laws introduced so far. Fortunately, Boyle’s, Charles’s, and Gay-Lussac’s laws can all be easily derived from the combined gas law. For example, consider a situation where a change occurs in the volume and pressure of a gas while the temperature is being held constant. In that case, it can be said that  T_1 = T_2 . Look at the combined gas law and cancel the  T variable out from both sides of the equation. What is left over is Boyle’s law:

P_1 times V_1 = P_2 times V_2 . Likewise, if the pressure is constant, then  P_1 = P_2 and canceling  P out of the equation leaves Charles’s law. If the volume is constant, then  V_1 = V_2 and canceling  V out of the equation leaves Gay-Lussac’s law.

Summary

  • The combined gas law shows the relationships among temperature, volume, and pressure.
  • frac{P_1V_1}{T_1} = frac{P_2V_2}{T_2}


Practice

Work on the problems at the link below:

http://misterguch.brinkster.net/WKS001_007_146637.pdf

Review

Questions

  1. What is the only thing held constant in a combined gas law problem?
  2. If you want to solve for the volume of a gas  (V_2) and  P_1 is greater than P_2 , would you expect  V_2 to be larger or smaller than V_1 ?
  3. What would be the equation for finding  P_2 given all the other parameters?


  • combined gas law: Expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas.


Avogadro's Law

  • State Avogadro’s Law.
  • Use this law to perform calculations involving quantities of gases.


Avogadro's Law tells us that putting gas into a tire increases its pressure

How much air do you put into a tire?

A flat tire is not very useful. It does not cushion the rim of the wheel and creates a very uncomfortable ride. When air is added to the tire, the pressure increases as more molecules of gas are forced into the rigid tire. How much air should be put into a tire depends on the pressure rating for that tire. Too little pressure and the tire will not hold its shape. Too much pressure and the tire could burst.

Avogadro’s Law

You have learned about Avogadro’s hypothesis: equal volumes of any gas at the same temperature and pressure contain the same number of molecules. It follows that the volume of a gas is directly proportional to the number of moles of gas present in the sample. Avogadro’s law states that the volume of a gas is directly proportional to the number of moles of gas when the temperature and pressure are held constant. The mathematical expression of Avogadro’s law is

V=k times n quad text{and} quad frac{V_1}{n_1}=frac{V_2}{n_2}

where  n is the number of moles of gas and  k is a constant. Avogadro’s law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up.

If the container holding the gas is rigid rather than flexible, pressure can be substituted for volume in Avogadro’s law. Adding gas to a rigid container makes the pressure increase.

Sample Problem: Avogadro’s Law

A balloon has been filled to a volume of 1.90 L with 0.0920 mol of helium gas. If 0.0210 mol of additional helium is added to the balloon while the temperature and pressure are held constant, what is the new volume of the balloon?

Step 1: List the known quantities and plan the problem.

Known

  • V_1=1.90 text{ L}
  • n_1=0.0920 text{ mol}
  • n_2=0.0920+0.0210=0.1130 text{ mol}


Unknown

  • V_2=? text{ L}


Note that the final number of moles has to be calculated by adding the original number of moles to the moles of added helium. Use Avogadro’s law to solve for the final volume.

Step 2: Solve.

First, rearrange the equation algebraically to solve for V_2 .

V_2=frac{V_1 times n_2}{n_1}

Now substitute the known quantities into the equation and solve.

V_2=frac{1.90 text{ L} times 0.1130 text{ mol}}{0.0920 text{ mol}}=2.33 text{ L}

Step 3: Think about your result.

Since a relatively small amount of additional helium was added to the balloon, its volume increases slightly.

Summary

  • Calculations are shown for relationships between volume and number of moles of a gas.


Practice

Work on the problems at the site below:

http://www.gst-d2l.com/homework/hwavogadroslaw.html

Review

Questions

  1. What is held constant in the Avogadro’s Law relationship?
  2. What happens if you add gas to a rigid container?
  3. Why does a balloon expand when you add air to it?


  • Avogadro’s law: The volume of a gas is directly proportional to the number of moles of gas when the temperature and pressure are held constant.


Ideal Gas Law

  • Derive the ideal gas law from the combined gas law and Avogadro’s law.
  • Calculate the value of the ideal gas constant.
  • Use the ideal gas law to calculate parameters for ideal gases.


The ideal gas law can be used to determine the amount of gas present in a system

What chemical reactions require ammonia?

There are a number of chemical reactions that require ammonia. In order to carry out the reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system.

Ideal Gas Law

The combined gas law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro’s law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these together leaves us with the following equation:

frac{P_1 times V_1}{T_1 times n_1}=frac{P_2 times V_2}{T_2 times n_2}

As with the other gas laws, we can also say that  frac{left(P times V right)}{left(T times n right)} is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.

The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable  R for the constant, the equation becomes:

frac{P times V}{T times n}=R

The ideal gas law is conventionally rearranged to look this way, with the multiplication signs omitted:

PV=nRT

The variable  R in the equation is called the ideal gas constant .

Evaluating the Ideal Gas Constant

The value of R , the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: kPa, atm, or mmHg. Therefore,  R can have three different values.

We will demonstrate how  R is calculated when the pressure is measured in kPa. Recall that the volume of 1.00 mol of any gas at STP is measured to be 22.414 L. We can substitute 101.325 kPa for pressure, 22.414 L for volume, and 273.15 K for temperature into the ideal gas equation and solve for R .

R=frac{PV}{nT}=frac{101.325 text{kPa} times 22.414 text{ L}}{1.000 text{mol} times 273.15 text{ K}}=8.314 text{kPa} cdot text{L/K} cdot text{mol}

This is the value of  R that is to be used in the ideal gas equation when the pressure is given in kPa. The Table below shows a summary of this and the other possible values of R . It is important to choose the correct value of  R to use for a given problem.

Values of the Ideal Gas Constant
Unit of P Unit of V Unit of n Unit of T Value and unit of R
kPa L mol K 8.314 text{J/K} cdot text{mol}
atm L mol K 0.08206 text{L} cdot text{atm/K} cdot text{mol}
mmHg L mol K 62.36 text{ L} cdot text{mmHg/K} cdot text{mol}

Notice that the unit for  R when the pressure is in kPa has been changed to J/K • mol. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule (J).

Sample Problem: Ideal Gas Law

What volume is occupied by 3.760 g of oxygen gas at a pressure of 88.4 kPa and a temperature of 19°C? Assume the oxygen is ideal.

Step 1: List the known quantities and plan the problem.

Known

  • P = 88.4 text{kPa}
  • T = 19^circ text{C} = 292 text{ K}
  • text{mass} O_2 = 3.760 text{ g}
  • O_2 = 32.00 text{ g/mol}
  • R = 8.314 text{ J/K} cdot text{mol}


Unknown

  • V = ? text{ L}


In order to use the ideal gas law, the number of moles of O 2   (n) must be found from the given mass and the molar mass. Then, use  PV = nRT to solve for the volume of oxygen.

Step 2: Solve .

3.760 text{ g} times frac{1 text{mol} O_2}{32.00 text{ g} O_2}=0.1175 text{mol} O_2

Rearrange the ideal gas law and solve for V .

V=frac{nRT}{P}=frac{0.1175 text{mol} times 8.314 text{ J/K} cdot text{mol} times 292 text{ K}}{88.4 text{kPa}}=3.23 text{ L } O_2

Step 3: Think about your result

The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume (22.4 L/mol) since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for  T and P . Since a joule (J) = kPa • L, the units cancel correctly, leaving a volume in liters.

Summary

  • The ideal gas constant is calculated.
  • An example of calculations using the ideal gas law is shown.


Practice

Work on the problems at the link below:

http://chemsite.lsrhs.net/gasses/handouts/Ideal_Problems.pdf

Review

Questions

  1. Which value of  R will you use if the pressure is given in atm?
  2. You are doing a calculation where the pressure is given in mm Hg. You select 8.314 J/K • mol as your value for R . Will you get a correct answer?
  3. How would you check that you have chosen the correct value of  R for your problem?


  • ideal gas constant: The variable  R in the ideal gas law equation.
  • ideal gas law: A single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.


Calculating the Molar Mass of a Gas

  • Calculate the molar mass of a gas.
  • Calculate the density of a gas.


Gases with low molar masses such as hydrogen and helium can be used to float balloons

What makes it float?

Helium has long been used in balloons and blimps. Since it is much less dense than air, it will float above the ground. We can buy small balloons filled with helium at stores, but large ones (such as the balloon seen above) are much more expensive and take up a lot more helium.

Calculating Molar Mass and Density of a Gas

A chemical reaction, which produces a gas, is performed. The produced gas is then collected and its mass and volume are determined. The molar mass of the unknown gas can be found using the ideal gas law, provided the temperature and pressure of the gas are also known.

Sample Problem: Molar Mass and the Ideal Gas Law

A certain reaction occurs, producing an oxide of nitrogen as a gas. The gas has a mass of 1.211 g and occupies a volume of 677 mL. The temperature in the laboratory is 23°C and the air pressure is 0.987 atm. Calculate the molar mass of the gas and deduce its formula. Assume the gas is ideal.

Step 1: List the known quantities and plan the problem .

Known

  • text{mass}=1.211text{ g}
  • V = 677 text{ ml} = 0.677 text{ L}
  • T= 23^circ text{C}=296 text{ K}
  • P =0.987 text{atm}
  • R =0.08206 text{ L} cdot text{atm/K} cdot text{mol}


Unknown

  • n= ? text{mol}
  • text{molar mass} = ? text{g/mol}


First the ideal gas law will be used to solve for the moles of unknown gas (n) . Then the mass of the gas divided by the moles will give the molar mass.

Step 2: Solve .

n=frac{PV}{RT}=frac{0.987 text{atm} times 0.677 text{ L}}{0.08206 text{ L} cdot text{atm/K} cdot text{mol} times 296 text{ K}}=0.0275 text{mol}

Now divide g by mol to get the molar mass.

text{molar mass}=frac{1.211 text{ g}}{0.0275 text{mol}}=44.0 text{g/mol}

Since N has a molar mass of 14 g/mol and O has a molar mass of 16 g/mol, the formula N 2 O would produce the correct molar mass.

Step 3: Think about your result

The  R value that corresponds to a pressure in atm was chosen for this problem. The calculated molar mass gives a reasonable formula for dinitrogen monoxide.

Calculating Density of a Gas

The ideal gas law can be used to find the density of a gas at conditions that are not standard. For example, we will determine the density of ammonia gas (NH 3 ) at 0.913 atm and 20°C, assuming the ammonia is ideal. First, the molar mass of ammonia is calculated to be 17.04 g/mol. Next, assume exactly 1 mol of ammonia  (n = 1) and calculate the volume that such an amount would occupy at the given temperature and pressure.

V = frac{ nRT}{P}=frac{1.00 text{mol} times 0.08206 text{ L} cdot text{atm/K}cdot text{mol} times 293 text{ K}}{0.913 text{atm}}=26.3 text{ L}

Now the density can be calculated by dividing the mass of one mole of ammonia by the volume above.

text{density}=frac{17.04 text{ g}}{26.3 text{ L}}=0.647 text{g/L}

As a point of comparison, this density is slightly less than the density of ammonia at STP, which is equal to frac{(17.04 text{g/mol})}{(22.4 text{L/mol})} = 0.761 text{g/L} . It makes sense that the density should be lower compared to that at STP since both the increase in temperature (from 0°C to 20°C) and the decrease in pressure (from 1 atm to 0.913 atm) would cause the NH 3 molecules to spread out a bit further from one another.

Summary

  • Calculations of molar mass and density of an ideal gas are described.


Practice

Answer questions and perform calculations of problems at the following link:

http://www.mybookezz.com/ebook.php?u=aHR0cDovL2dvLmhydy5jb20vcmVzb3VyY2VzL2dvX3NjL21jL0hDMlNSMTEzLlBERgpTZWN0aW9uIDM=

Review

Questions

  1. Why do you need the volume, temperature, and pressure of the gas to calculate molar mass?
  2. What assumption about the gas is made in all these calculations?
  3. Why do you need the mass of the gas to calculate the molar mass?


Gas Stoichiometry

  • Use the ideal gas law to calculate stoichiometry problems for gases.


Knowing the amount of a gas needed is important for many applications

How is fertilizer produced?

The Haber cycle reaction of gaseous nitrogen and hydrogen to form ammonia is a critical step in the production of fertilizer from ammonia. It is important to have an excess of the starting materials so that a maximum yield of ammonia can be achieved. By knowing how much ammonia is needed for manufacture of a batch of fertilizer, the proper amounts of nitrogen and hydrogen gases can be incorporated into the process.

Gas Stoichiometry

You have learned how to use molar volume to solve stoichiometry problems for chemical reactions involving one or more gases at STP. Now, we can use the ideal gas law to expand our treatment of chemical reactions to solve stoichiometry problems for reactions that occur at any temperature and pressure.

Sample Problem: Gas Stoichiometry and the Ideal Gas Law

What volume of carbon dioxide is produced by the combustion of 25.21 g of ethanol (C 2 H 5 OH) at 54°C and 728 mmHg? Assume the gas is ideal.

Before using the ideal gas law, it is necessary to write and balance the chemical equation. Recall that most combustion reactions, the given substance reacts with O 2 to form CO 2 and H 2 O. Here is the balanced equation for the combustion of ethanol.

text{C}_2text{H}_5text{OH}(l)+3text{O}_2(g) rightarrow 2text{CO}_2(g)+3text{H}_2text{O}(l)

Step 1: List the known quantities and solve the problem.

Known

  • text{mass} text{C}_2text{H}_5text{OH}=25.21 text{ g}
  • text{molar mass} text{C}_2text{H}_5text{OH}=46.08 text{ g/mol}
  • P=728 text{ mmHg}
  • T=54^circ text{C}=327 text{ K}


Unknown

  • text{Volume} text{CO}_2=? text{ L}


The number of moles of carbon dioxide gas is first calculated by stoichiometry. Then the ideal gas law is used to calculate the volume of CO 2 produced.

Step 2: Solve.

25.21 text{ g } text{C}_2text{H}_5text{OH} times frac{1 text{ mol } text{C}_2text{H}_5text{OH}}{46.08 text{ g } text{C}_2text{H}_5text{OH}} times frac{2 text{ mol } text{CO}_2}{1 text{ mol } text{C}_2text{H}_5text{OH}}=1.094 text{ mol } text{C}_2text{H}_5text{OH}

The moles of ethanol  (n) is now substituted into  PV=nRT to solve for the volume.

V=frac{nRT}{P}=frac{1.094 text{ mol} times 62.36 text{ L} cdot text{mmHg/K} cdot text{mol} times 327 text{ K}}{728 text{ mmHg}}=30.6 text{ L}

Step 3: Think about your result.

The mass of ethanol is slightly more than one half mole, meaning that the mole ratio results in slightly more than one mole of carbon dioxide being produced. Because of the elevated temperature and reduced pressure compared to STP, the resulting volume is larger than 22.4 L.

Summary

  • The ideal gas law is used to calculate stoichiometry problems for gases.


Practice

Solve the problems on the worksheet at this site:

http://misterguch.brinkster.net/PRA036.pdf

Review

Questions

  1. Do we need gas conditions to be at STP to calculate stoichiometry problems?
  2. Why do we want to determine the stoichiometry of these reactions?
  3. What assumption are we making about the gases involved?


Real and Ideal Gases

  • Define a real gas.
  • Describe differences between real gases and ideal gases.


Intermolecular forces can cause deviations from ideality in gases

Location, Location, Location

The behavior of a molecule depends a lot on its structure. We can have two compounds with the same number of atoms and yet they act very differently. Ethanol (C 2 H 5 OH) is a clear liquid that has a boiling point of about 79°C. Dimethylether (CH 3 OCH 3 ) has the same number of carbons, hydrogens, and oxygens, but boils at a much lower temperature (-25°C). The difference lies in the amount of intermolecular interaction (strong H-bonds for ethanol, weak van der Waals force for the ether).

Real and Ideal Gases

An ideal gas is one that follows the gas laws at all conditions of temperature and pressure. To do so, the gas would need to completely abide by the kinetic-molecular theory. The gas particles would need to occupy zero volume and they would need to exhibit no attractive forces what so ever toward each other. Since neither of those conditions can be true, there is no such thing as an ideal gas. A real gas is a gas that does not behave according to the assumptions of the kinetic-molecular theory. Fortunately, at the conditions of temperature and pressure that are normally encountered in a laboratory, real gases tend to behave very much like ideal gases.

Under what conditions then, do gases behave least ideally? When a gas is put under high pressure, its molecules are forced closer together as the empty space between the particles is diminished. A decrease in the empty space means that the assumption that the volume of the particles themselves is negligible is less valid. When a gas is cooled, the decrease in kinetic energy of the particles causes them to slow down. If the particles are moving at slower speeds, the attractive forces between them are more prominent. Another way to view it is that continued cooling the gas will eventually turn it into a liquid and a liquid is certainly not an ideal gas anymore (see liquid nitrogen in the Figure below ). In summary, a real gas deviates most from an ideal gas at low temperatures and high pressures. Gases are most ideal at high temperature and low pressure.

Liquid nitrogen is at such low temperatures that it is no longer a gas

Figure 14.11

Nitrogen gas that has been cooled to 77 K has turned to a liquid and must be stored in a vacuum insulated container to prevent it from rapidly vaporizing.

The Figure below shows a graph of  frac{PV}{RT} plotted against pressure for 1 mol of a gas at three different temperatures - 200 K, 500 K, and 1000 K. An ideal gas would have a value of 1 for that ratio at all temperatures and pressures and the graph would simply be a horizontal line. As can be seen, deviations from an ideal gas occur. As the pressure begins to rise, the attractive forces cause the volume of the gas to be less than expected and the value of  frac{PV}{RT} drops under 1. Continued pressure increase results in the volume of the particles to become significant and the value of  frac{PV}{RT} rises to greater than 1. Notice, that the magnitude of the deviations from ideality is greatest for the gas at 200 K and least for the gas at 1000 K.

Real gases deviate from ideal gases at high pressures and at low temperatures

Figure 14.12

Real gases deviate from ideal gases at high pressures and at low temperatures.

The ideality of a gas also depends on the strength and type of intermolecular attractive forces that exist between the particles. Gases whose attractive forces are weak are more ideal than those with strong attractive forces. At the same temperature and pressure, neon is more ideal than water vapor because neon’s atoms are only attracted by weak dispersion forces, while water vapor’s molecules are attracted by relatively stronger hydrogen bonds. Helium is a more ideal gas than neon because its smaller number of electrons means that helium’s dispersion forces are even weaker than those of neon.

Summary

  • The properties of real gases and their deviations from ideality are described.


Practice

Questions

Use the link below to answer the following questions:

http://www.adichemistry.com/physical/gaseous/deviation/van-der-waals-equation.html

  1. What is the compressibility factor for a perfect (ideal) gas?
  2. What does it mean if Z>1 ?
  3. What does it mean if Z<1 ?


Review

Questions

  1. What becomes more significant as the pressure increases?
  2. Do the attractive forces between gas particles become more prominent at higher or lower temperatures?
  3. Would HCl gas be more or less ideal than helium?


  • real gas: A gas that does not behave according to the assumptions of the kinetic-molecular theory.


Dalton's Law of Partial Pressures

  • Define partial pressure.
  • State Dalton’s law of partial pressures.
  • Use this law to calculate pressures of gas mixtures.


Venus has a high partial pressure of carbon dioxide and nitrogen

Is there oxygen available on Venus?

The atmosphere of Venus is markedly different from that of Earth. The gases in the Venusian atmosphere are 96.5% carbon dioxide and 3% nitrogen. The atmospheric pressure on Venus is roughly 92 times that of Earth, so the amount of nitrogen on Venus would contribute a pressure well over 2700 mm Hg. And there is no oxygen present, so we couldn’t breathe there. Not that we would want to go to Venus – the surface temperature is usually over 460°C.

Dalton’s Law of Partial Pressures

Gas pressure results from collisions between gas particles and the inside walls of their container. If more gas is added to a rigid container, the gas pressure increases. The identities of the two gases do not matter. John Dalton, the English chemist who proposed the atomic theory, also studied mixtures of gases. He found that each gas in a mixture exerts a pressure independently of every other gas in the mixture. For example, our atmosphere is composed of about 78% nitrogen and 21% oxygen, with smaller amounts of several other gases making up the rest. Since nitrogen makes up 78% of the gas particles in a given sample of air, it exerts 78% of the pressure. If the overall atmospheric pressure is 1.00 atm, then the pressure of just the nitrogen in the air is 0.78 atm. The pressure of the oxygen in the air is 0.21 atm.

The partial pressure of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture. The partial pressure of a gas is indicated by a  P with a subscript that is the symbol or formula of that gas. The partial pressure of nitrogen is represented by  P_{N_2} . Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases. Dalton’s law can be expressed with the following equation:

P_{text{total}}=P_1+P_2+P_3+ldots

The Figure below shows two gases that are in separate, equal-sized containers at the same temperature and pressure. Each exerts a different pressure,  P_1 and P_2 , reflective of the number of particles in the container. On the right, the two gases are combined into the same container, with no volume change. The total pressure of the gas mixture is equal to the sum of the individual pressures. If  P_1 = 300 text{ mmHg} and P_2 = 500 text{ mmHg} , then P_{text{Total}}=800 text{ mmHg} .

Dalton's law states that the pressure of a gas mixture is equal to the partial pressure of the components

Figure 14.13

Dalton’s law says that the pressure of a gas mixture is equal to the partial pressures of the combining gases.

Summary

  • The total pressure in a system is equal to the sums of the partial pressures of the gases present.


Practice

Review the concepts at the link below and work the sample problems:

http://www.kentchemistry.com/links/GasLaws/dalton.htm

Review

Questions

  1. What is the foundation for Dalton’s law?
  2. Argon makes up about 0.93% of our atmosphere. If the atmospheric pressure is 760 mm Hg, what is the pressure contributed by argon?
  3. On a given day, the water vapor in the air is 2.5%. If the partial pressure of the vapor is 19.4 mm Hg, what is the atmospheric pressure?


  • Partial pressure: The contribution that gas makes to the total pressure when the gas is part of a mixture.
  • Dalton’s law of partial pressures: The total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases.


Mole Fraction

  • Define mole fraction.
  • Perform calculations involving mole fractions.


The changing mole fraction of sulfur dioxide is a mixed blessing

The mixed blessing of sulfur dioxide

Sulfur dioxide is a by-product of many processes, both natural and human-made. Massive amounts of this gas are released during volcanic eruptions such as the one seen above on the Big Island (Hawaii). Humans produce sulfur dioxide by burning coal. The gas has a cooling effect when in the atmosphere by reflecting sunlight back away from the earth. However, sulfur dioxide is also a component of smog and acid rain, both of which are harmful to the environment. Many efforts have been made to reduce SO 2 levels to lower acid rain production. An unforeseen complication: as we lower the concentration of this gas in the atmosphere, we lower its ability to cool and then we have global warming concerns.

Mole Fraction

One way to express relative amounts of substances in a mixture is with the mole fraction. Mole fraction   (X) is the ratio of moles of one substance in a mixture to the total number of moles of all substances. For a mixture of two substances,  A and B , the mole fractions of each would be written as follows:

X_A= frac{text{mol} A}{text{mol} A+text{mol} B} quad text{and} quad X_B=frac{text{mol} B}{text{mol} A+text{mol} B}

If a mixture consists of 0.50 mol  A and 1.00 mol B , then the mole fraction of  A would be X_A=frac{0.5}{1.5} = 0.33 .  Similarly, the mole fraction of  B would be X_B =frac{1.0}{1.5} = 0.67 .

Mole fraction is a useful quantity for analyzing gas mixtures in conjunction with Dalton’s law of partial pressures. Consider the following situation: A 20.0 liter vessel contains 1.0 mol of hydrogen gas at a pressure of 600 mmHg. Another 20.0 liter vessel contains 3.0 mol of helium at a pressure of 1800 mmHg. These two gases are mixed together in an identical 20.0 liter vessel. Because each will exert its own pressure according to Dalton’s law, we can express the partial pressures as follows:

P_{H_2}=X_{H_2} times P_{text{Total}} quad text{and} quad P_{He}=X_{He} times P_{text{Total}}

The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. For our mixture of hydrogen and helium:

X_{H_2}=frac{1.0 text{mol}}{1.0 text{mol}+3.0 text{mol}}=0.25 quad text{and} quad X_{He}=frac{3.0 text{mol}}{1.0 text{mol} + 3.0 text{mol}}=0.75

The total pressure according to Dalton’s law is 600 text{ mmHg} + 1800 text{ mmHg} = 2400 text{ mmHg} . So, each partial pressure will be:

& P_{H_2}=0.25 times 2400 text{ mmHg}=600 text{ mmHg} \& P_{He}=0.75 times 2400 text{ mmHg}=1800 text{ mmHg}

The partial pressures of each gas in the mixture don’t change since they were mixed into the same size vessel and the temperature was not changed.

Sample Problem: Dalton’s Law

A flask contains a mixture of 1.24 moles of hydrogen gas and 2.91 moles of oxygen gas. If the total pressure is 104 kPa, what is the partial pressure of each gas?

Step 1: List the known quantities and plan the problem .

Known

  • 1.24 mol H 2
  • 2.91 mol O 2
  • P_{text{Total}}=104 text{kPa}


Unknown

  • P_{H_2}=? text{kPa}
  • P_{O_2}=? text{kPa}


First, the mole fraction of each gas can be determined. Then, the partial pressure can be calculated by multiplying the mole fraction by the total pressure.

Step 2: Solve .

& X_{H_2}=frac{1.24 text{mol}}{1.24 text{mol} + 2.91 text{mol}}=0.299 && X_{O_2}=frac{2.91 text{mol}}{1.24 text{mol} + 2.91 text{mol}}=0.701 \& P_{H_2}=0.299 times 104 text{ kPa}=31.1 text{ kPa} && P_{O_2}=0.701 times 104 text{ kPa}=72.9 text{ kPa}

Step 3: Think about your result .

The hydrogen is slightly less than one third of the mixture, so it exerts slightly less than one third of the total pressure.

Summary

  • Use of the mole fraction allows calculation to be made for mixtures of gases.


Practice

Questions

Watch the video at the link below and answer the following questions:

Click on the image above for more content

  1. What is mole percent?
  2. Do the mole fractions add up to 1.00?
  3. What other way could you calculate the mole fraction of oxygen once you have the mole fraction of nitrogen?


Review

Questions

  1. What is mole fraction?
  2. How do you determine partial pressure of a gas when given the mole fraction and the total pressure?
  3. In a gas mixture containing equal numbers of moles of two gases, what can you say about the partial pressures of each gas?


  • Mole fraction (X) : The ratio of moles of one substance in a mixture to the total number of moles of all substances.


Gas Collection by Water Displacement

  • Calculate volumes of dry gases obtained after collecting over water.


The pressure of gases collected over water can be determined by using the atmospheric pressure in the room

What is the pressure?

You need to do a lab experiment where hydrogen gas is generated. In order to calculate the yield of gas, you have to know the pressure inside the tube where the gas is collected. But how can you get a barometer in there? Very simple: you don’t. All you need is the atmospheric pressure in the room. As the gas pushed out the water, it is pushing against the atmosphere, so the pressure inside is equal to the pressure outside.

Gas Collection by Water Displacement

Gases that are produced in laboratory experiments are often collected by a technique called water displacement (see Figure below ). A bottle is filled with water and placed upside-down in a pan of water. The reaction flask is fitted with rubber tubing which is then fed under the bottle of water. As the gas is produced in the reaction flask, it exits through the rubber tubing and displaces the water in the bottle. When the bottle is full of the gas, it can be sealed with a lid.

A gas produced in a chemical reaction can be collected by water displacement.

Figure 14.14

A gas produced in a chemical reaction can be collected by water displacement.

Because the gas is collected over water, it is not pure but is mixed with vapor from the evaporation of the water. Dalton’s law can be used to calculate the amount of the desired gas by subtracting the contribution of the water vapor.

P_{text{Total}} &=P_g+P_{H_2 O} qquad P_g text{ is the pressure of the desired gas}\P_g &=P_{text{Total}}- P_{H_2 O}

In order to solve a problem, it is necessary to know the vapor pressure of water at the temperature of the reaction (see Table below). The sample problem illustrates the use of Dalton’s law when a gas is collected over water.

Vapor Pressure of Water (mmHg) at Selected Temperatures (°C)
Temperature (°C) Vapor Pressure (mmHg) Temperature (°C) Vapor Pressure (mmHg)
0 4.58 40 55.32
5 6.54 45 71.88
10 9.21 50 92.51
15 12.79 55 118.04
20 17.54 60 149.38
25 23.76 65 187.54
30 31.82 70 233.7
35 42.18

Sample Problem: Gas Collected by Water Displacement

A certain experiment generates 2.58 L of hydrogen gas, which is collected over water. The temperature is 20°C and the atmospheric pressure is 98.60 kPa. Find the volume that the dry hydrogen would occupy at STP.

Step 1: List the known quantities and plan the problem.

Known

  • V_{text{Total}} =2.58 text{ L}
  • T=20^ circ text{C}=293 text{ K}
  • P_{text{Total}} =98.60 text{ kPa}=739.7 text{ mmHg}


Unknown

  • V_{H_2} text{at} STP= ? text{ L}


The atmospheric pressure is converted from kPa to mmHg in order to match units with the table. The sum of the pressures of the hydrogen and the water vapor is equal to the atmospheric pressure. The pressure of the hydrogen is found by subtraction. Then, the volume of the gas at STP can be calculated by using the combined gas law.

Step 2: Solve.

P_{H_2}=P_{text{Total}}-P_{H_2 O}=739.7 text{ mmHg} -17.54 text{ mmHg}=722.2 text{ mmHg}

Now the combined gas law is used, solving for V_2 , the volume of hydrogen at STP.

V_2=frac{P_1 times V_1 times T_2}{P_2 times T_1}=frac{722.2 text{ mmHg} times 2.58 text{ L} times 273 text{ K}}{760 text{ mmHg} times 293 text{ K}}=2.28 text{ L } H_2

Step 3: Think about your result.

If the hydrogen gas were to be collected at STP and without the presence of the water vapor, its volume would be 2.28 L. This is less than the actual collected volume because some of that is water vapor. The conversion using STP is useful for stoichiometry purposes.

Summary

  • The vapor pressure due to water in a sample can be corrected for in order to get the true value for the pressure of the gas.


Practice

Questions

Watch the video at the link below and answer the following questions:

Click on the image above for more content

  1. What was the thistle tube used for?
  2. How did the instructor tests for oxygen?
  3. Did you observe any unsafe lab practices in the video?
  4. What would have happened to the splint if carbon dioxide had been collected?


Review

Questions

  1. Why is gas collected over water not pure?
  2. Why would we want to correct for water vapor?
  3. A student wants to collect his gas over diethyl ether (vapor pressure of 530 mm Hg at 25°C). Is this a good idea? Explain your answer.


  • water displacement: Collection of a gas over water.


Diffusion and Effusion and Graham's Law

  • Define diffusion and effusion.
  • State Graham’s law.
  • Use Graham’s law to perform calculations involving movement of gases.


A classic experiment to find the rate of diffusion for gases uses hydrochloric acid and ammonia

How do we know how fast a gas moves?

We usually cannot see gases, so we need ways to detect their movements indirectly. The relative rates of diffusion of ammonia to hydrogen chloride can be observed in a simple experiment. Cotton balls are soaked with solutions of ammonia and hydrogen chloride (hydrochloric acid) and attached to two different rubber stoppers. These are simultaneously plugged into either end of a long glass tube. The vapors of each travel down the tube at different rates. Where the vapors meet, they react to form ammonium chloride (NH 4 Cl), a white solid that appears in the glass tube as a ring.

Graham’s Law

When a person opens a bottle of perfume in one corner of a large room, it doesn’t take very long for the scent to spread throughout the entire room. Molecules of the perfume evaporate and the vapor spreads out to fill the entire space. Diffusion is the tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform. While gases diffuse rather quickly, liquids diffuse much more slowly. Solids essentially do not diffuse.

Video of bromine diffusion: 

Click on the image above for more content

A related process to diffusion is the effusion. Effusion is the process of a confined gas escaping through a tiny hole in its container. Effusion can be observed by the fact that a helium-filled balloon will stop floating and sink to the floor after a day or so. This is because the helium gas effuses through tiny pores in the balloon. Both diffusion and effusion are related to the speed at which various gas molecules move. Gases that have a lower molar mass effuse and diffuse at a faster rate than gases that have a higher molar mass.

Scottish chemist Thomas Graham (1805-1869) studied the rates of effusion and diffusion of gases. Graham’s law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Graham’s law can be understood by comparing two gases ( A and B ) at the same temperature, meaning the gases have the same kinetic energy. The kinetic energy of a moving object is given by the equation KE =frac{1}{2}mv^2 ,

where  m is mass and  v is velocity. Setting the kinetic energies of the two gases equal to one another gives:

frac{1}{2}m_Av^2_A=frac{1}{2}m_Bv^2_B

The equation can be rearranged to solve for the ratio of the velocity of gas  A to the velocity of gas  Bleft(frac{v_A}{v_B}right) .

frac{v^2_A}{v^2_B}=frac{m_B}{m_A} text{ which becomes } frac{v_A}{v_B}=sqrt{frac{m_B}{m_A}}

For the purposes of comparing the rates of effusion or diffusion of two gases at the same temperature, the molar masses of each gas can be used in the equation for m .

Sample Problem: Graham’s Law

Calculate the ratio of diffusion rates of ammonia gas (NH 3 ) to hydrogen chloride (HCl) at the same temperature and pressure.

Step 1: List the known quantities and plan the problem.

Known

  • molar mass NH 3 = 17.04 g/mol
  • molar mass HCl = 36.46 g/mol


Unknown

  • velcoity ratio  dfrac{v_{text{NH}_3}}{v_{text{HCl}}}


Substitute the molar masses of the gases into Graham’s law and solve for the ratio.

Step 2: Solve.

frac{v_{NH_3}}{v_{HCl}}=sqrt{frac{36.46 text{ g/mol}}{17.04 text{ g/mol}}}=1.46

The rate of diffusion of ammonia is 1.46 times faster than the rate of diffusion of hydrogen chloride.

Step 3: Think about your result

Since ammonia has a smaller molar mass than hydrogen chloride, the velocity of its molecules is greater and the velocity ratio is larger than 1.

Summary

  • The processes of gas diffusion and effusion are described.
  • Graham’s law relates the molecular mass of a gas to its rate of diffusion or effusion.


Practice

Read the material on the link below and do the practice problems:

http://www.kentchemistry.com/links/GasLaws/GrahamsLaw.htm

Review

Questions

  1. Why can you smell food cooking when you are in the next room?
  2. Why does a helium-filled balloon gradually sink?
  3. What does temperature have to do with gas kinetic energies?


  • diffusion: The tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform.
  • effusion: The process of a confined gas escaping through a tiny hole in its container.
  • Graham’s law: The rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas.






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