PowerPoint Presentation

Gas Laws

Chemistry 9(A)Gas laws will help you understand the behavior of gases under different conditions.

1Learning ObjectivesDescribe the relationships between volume, pressure, number of moles, and temperature for an ideal gasPerform calculations that predict values for variables using Boyles lawCharles lawAvogadros law Ideal gas lawDaltons law of partial pressure

Gas LawsThis presentation will help you understand the relationships between volume, pressure, number of moles, and temperature for an ideal gas. It will also teach you to how to perform calculations that predict values for these variables using Boyles law, Charles law, Avogadro's law, the ideal gas law, and Daltons law of partial pressure.2Gases can be characterized by theirVolume ( V )Temperature ( T )Number of moles ( n )Pressure ( P )

Variables Affecting GasesThere are several variables that define the behavior of gases. Gases can be characterized by their volume, temperature, number of moles, and pressure.3Gas law law which describes the relationship between two or more variables of a gasVariables may be directly or inversely relatedDirect relationship values of variables increase or decrease together Inverse relationship as the value one variable goes up, the value of the other variable goes down Variables not included in the gas law remain constant

Gas RelationshipsThe relationship between two or more variables of a gas is given by a gas law. A gas law is a mathematical equation or written statement which describes the relationship between these variables. These variables may be directly or inversely related. When variables have a direct relationship, the values of the variables either increase or decrease together. When variables have an inverse relationship, the values of the variables have opposite trends. As the value of one variable goes up, the value of the other variable goes down. When a variable is not included in a gas law, assume that variable is being held constant, so the relationships between the other variables can be seen clearly.4Charles law describes the relationship between temperature and volume, when the pressure and amount of gas are constant

Volume and temperature are directly proportional on the Kelvin scale

Charles Law

Charles law describes the relationship between temperature and volume, when the pressure and amount of gas are constant. The formula for Charles law is the initial volume, V1, divided by the initial temperature, T1, equals the final volume, V2, divided by the final temperature, T2. Volume and temperature are directly proportional on the Kelvin scale. Therefore, as temperature increases, volume increases or as temperature decreases, volume decreases.5Temperature is directly proportional to kinetic energy Kinetic energy energy of motionAt higher temperatures, gas particlesHave more kinetic energy Move more Spread out Ex) Volume of a gas when T1 < T2

Charles Law

The reason temperature and volume have a direct relationship is because temperature is directly proportional to kinetic energy. Kinetic energy is the energy of motion. Particles have more kinetic energy at higher temperatures. The higher the temperature, the more energy particles have to move. Since gas particles have no significant attractive forces pulling them together, they spread out at higher temperatures. This expansion results in an increase in volume. For example, think about how a hot air balloon inflates when the air inside it is heated. The increase in temperature causes the kinetic energy of the gas molecules to increase. As a result, the gas molecules move further away from one another, expanding the volume of the gas and inflating the balloon.

6Celsius vs. Kelvin scaleMust use kelvin in Charles law calculationsCelsius to Kelvin scale

Ex) 25 C + 273 = 298 KTemperature Scales

There are two different temperature scales that are commonly used in science calculations, Celsius and Kelvin. When completing any calculations involving temperatures it will be important to make sure that the scale used is consistent throughout the calculation. For some calculations, it will be necessary to convert between the two scales. The unit of a kelvin and Celsius degree are the same. However, the Kelvin scale begins at absolute zero, which is theoretically the lowest temperature possible. Theoretically, absolute zero corresponds to zero volume. However, 0 C does not correspond to zero volume since it is equal to 273 K. Thus, volume is only directly proportional to temperature on the Kelvin scale. Therefore, all Charles law calculations must be preformed with Kelvin temperatures. To convert from the Celsius to Kelvin scale, simply add 273 to the Celsius temperature. For example, 25 C is equal to 298 K, because 25 plus 273 is 298. 7Charles Law CalculationEx) A 2.4 L volume of gas is heated from 298 K to 405 K. What new volume does it occupy? Write unknown and givens

Identify the formula and rearrange, if needed

Convert units and find intermediates, if neededPlug in and solve

Make sure the answer is reasonable

Lets work an example problem using Charles law. A 2.4 L volume of gas is heated from 298 K to 405 K. What new volume does it occupy? First write out the unknown and given values. The initial volume is 2.4 L. The initial temperature is 298 K. The final temperature is 405 K and the final volume is unknown.Now that the unknown and givens have been written, well identify the formula that well use to solve the problem and, if needed, rearrange it algebraically to solve for the unknown variable. Charles law describes the relationship between volume and temperature of a gas. The formula for Charles law is V1 divided by T1 equals V2 divided by T2. Since the final volume is unknown, well rearrange the formula to solve for V2. Multiplying both sides of the equation by T2 rearranges the formula so that it is solved for V2. Next, check to make sure that there are no conversions that need to be preformed so that your answer will be in the correct units when you finish your calculation. Since both temperatures are given in Kelvin there is no need to convert.Now, check to see if there are any intermediate calculations that need to be made before plugging in to the formula. Since we were already given a value for every variable needed to solve for the answer, there are no intermediate calculations that need to be completed.Now, well plug the values into the formula and solve algebraically. The final volume is equal to 2.4 L multiplied by 405 K, divided by 298 K. Two point four times 405 equals 972. Nine hundred and seventy two divided by 298 equals approximately 3.3.Finally, check to make sure that your answer is reasonable. We know that an increase in temperature (as seen in this problem) causes the volume of a gas to increase, because volume and temperature have a direct relationship. Our answer correctly reflects this relationship, since the value we calculated for final volume is higher than the initial volume. So, our answer is reasonable.8Boyles law describes relationship between pressure and volume when the temperature and amount of gas remain constant

Pressure and volume are inversely proportional

Boyles Law

Boyles Law describes the relationship between pressure and volume, when the temperature and amount of gas remain constant. The formula for Boyles law is the initial pressure, P1, times the initial volume, V1, equals the final pressure, P2, times the final volume, V2. Pressure and volume are inversely proportional. As volume decreases, gas molecules collide with the walls of the container more frequently increasing the pressure. The opposite is true when volume is increased, because gas molecules will collide with the walls of the container less frequently, decreasing the pressure.9Units of pressureAtmospheres (atm)Millimeters of mercury (mmHg)Kilopascals (kPa)Equivalent amounts1 atm = 760 mmHg = 101.3 kPa Ex) How many kilopascals are equal to 5.00 atm?

Pressure Units

Pressure can be measured in units of atmospheres, millimeters of mercury, or kilopascals. So, it will be important for you to be able to convert between them. The value of a pressure may be converted from one unit to another using conversion factors. Use equivalent amounts of of each unit to create conversion factors. The amount of pressure measured by 1 atm is equal to 760 mmHg and 101.3 kPa. Lets work an example. How many kilopascals are equal to 5 atm? Our given is 5 atm. Cancel out the atm unit by placing the atm unit in the denominator of the conversion factor. In the numerator, place the units you wish to convert into, kilopascals. Then fill in the equivalent amounts in the fraction. There are 101.3 kPa for every 1 atm. Now that the conversion is set up, solve by multiplying 5.00 by 101.3. We find that 5.00 atm is equal to 506.5 kPa of pressure. 10Boyles Law CalculationEx) The initial pressure of a 3.0 L sample of gas was 2.5 atm. At what pressure will the volume of the gas expand to 5.0 L? Write unknown and givens

2. Identify the formula and rearrange, if needed

Convert units and find intermediates, if neededPlug in and solve

Make sure the answer is reasonable

Lets work out an example. The initial pressure of a 3.0 L sample of gas was 2.5 atm. At what pressure will the volume of the gas expand to 5.0 L? First, write the unknown and given values. The initial pressure is 2.5 atm. The initial volume is 3.0 L and the final volume is 5.0 L. We are asked to solve for the final pressure so it is the unknown value. Next, we will identify the formula and rearrange it if necessary. The formula for Boyles law is P1 times V1 equals P2 times V2. Rearrange the formula algebraically to solve for P2 by dividing each side of the equation by V2. Now, check so see if there are any conversions that need to be completed or intermediates that need to be solved for. Since all of values for volume are in units of liters and and all values for pressure are in units of atm, no conversions are necessary. There are also no intermediates to solve for, so we can move on to the next step.Next, plug in the values and solve. The initial pressure, 2.5 atm, is multiplied by the initial volume 3.0 L, then divided by the final volume, 5.0 L. The value of the final pressure, P2, is approximately 1.5 atm. Finally, lets make sure the answer is reasonable. Boyles law describes the relationship between pressure and volume as being inversely proportional. As the volume increased from 3.0 L to 5.0 L, the pressure should have decreased proportionally from 2.5 atm to 1.5 atm. The answer, therefore, is reasonable according to Boyles law. 11Avogadros law describes the relationship between number of moles and volume, when temperature and pressure remain constant

Number of moles of gas are directly proportional to volume

Avogadro's Law

Avogadros Law describes the relationship between number of moles and volume, when temperature and pressure remain constant. The formula for Avogadros law is the initial volume, V1, divided by the initial number of moles of gas, n1, equals the final volume, V2, divided by the final number of moles of gas, n2. The relationship is directly proportional. Therefore, the greater the amount of a gas present, the larger the volume occupied and vice versa.

12Avogadros Law CalculationEx) How many moles of carbon monoxide gas are present in a 9.6 L sample if 4.2 moles were contained in 5.0 L sample? Write unknown and givens

Identify the formula and rearrange, if needed

Convert units and find intermediates, if neededPlug in and solve

Make sure the answer is reasonable

Lets work an example problem. How many moles of carbon monoxide gas are present in a 9.6 L sample if 4.2 moles were contained in 5.0 L sample? First, write out the unknown and given values. Volume one is 5.0 L. The number of moles in sample one is 4.2 moles. Volume two is 9.6 L and the number of moles in sample two is unknown. Now, identify the formula that well use to solve the problem and, if needed, rearrange it algebraically to solve for the unknown variable. The formula for Avogadros law is V1 divided by n1 equals V2 divided by n2. To rearrange the formula to solve for n2, first multiply both sides of the equation by n2 to get it out of the denominator. Then, multiply both sides by n1 and divide both sides of the equation by V1. When solved for n2, the equation for Avogadro's law is n2 equals n1 times V2 divided by V1. Next, check to make sure that there are no conversions that need to be preformed so that your answer will be in the correct units when you finish your calculation. Since all amounts of gas are given in moles and all volumes are given in liters, no intermediate calculations are needed. Now, plug the values into the formula and solve algebraically. To find n2, multiply 4.2 moles times 9.6 L and then divide by 5.0 L . The number of moles contained in the second sample is approximately 8.1 moles.Finally, check to make sure that your answer is reasonable. Avogadros law describes the relationship between volume and the number of moles as being directly proportional. Since the volume increased, the number of moles should also increase. Therefore, our value of 8.1 moles for n2 is reasonable because it is larger than the number of moles in sample one. 13Combining Charles, Boyles, and Avogadros laws gives the following equation:

A gas always has the same relationships between its variablesCharles volume and temperature are directly proportionalBoyle pressure and volume are indirectly proportionalAvogadro number of moles and volume are directly proportional

R is the ideal gas constantIdeal Gas Law

Combining Charles, Boyles, and Avogadros laws gives the following equation: pressure 1 times volume 1 divided by the initial number of moles and temperature 1 is equal to pressure 2 times volume 2 divided by the final number of moles and temperature 2. As you can see, a gas always has the same relationships between its variables: volume and Kelvin temperature are always directly proportional; pressure and volume are always indirectly proportional; and the number of moles and volume are always directly proportional, etc. Experiments performed on gases where the values for each variable were known, have shown that the pressure times the volume divided by the number of moles and temperature of an ideal gas is equal to a constant. This constant is symbolized by the capital letter R and is known as the ideal gas constant.14Ideal gas law describes the relationship between all four variables: temperature, volume, number of moles, and pressure

Potential values for R depending on the pressure unit used0.0821 = 8.31 = 62.4Ideal Gas Law

The equation derived from these observations and experiments (PV = nRT) is known as the ideal gas law. So, the ideal gas law describes the relationship between all four variables temperature, volume, number of moles, and pressure. As mention before, pressure can be given in several different units. Since the ideal gas constant is partially dependent on pressure, the unit of pressure used in the calculation will affect the value used for R. If pressure is given in atmospheres use the value 0.0821 as the gas constant. If pressure is given in kilopascals use the value 8.31 as the gas constant. If pressure is given in millimeters of mercury use the value 62.4 as the gas constant.

15Ideal Gas Law CalculationEx) What is the volume of a 1.42 mol sample of O2 gas at 25 C and 1.25 atm of pressure? Write unknown and givens

Identify the formula and rearrange, if needed

Convert units and find intermediates, if needed

4. Plug in and solve

Make sure the answer is reasonable

Lets work an example problem. What is the volume of a 1.42 mol sample of O2 gas at 25 C and 1.25 atm of pressure?First write out the unknown and given values. The number of moles is 1.42 mol. The temperature is 25 C. The pressure is 1.25 atm. Well use 0.0821 as the value for the gas constant since the pressure is in units of atm. The volume is the unknown.Now that the unknown and givens have been written, well identify the formula that well use to solve the problem and, if needed, rearrange it algebraically to solve for the unknown variable. Since this problem involves all four gas variables, the ideal gas law, PV = nRT, equation is best suited to solve it. Rearrange the formula to solve for the unknown, V, by dividing both sides by P. Next, check to make sure that there are no conversions that need to be preformed so that your answer will be in the correct units when you finish your calculation. The temperature given is in Celsius and must be converted to kelvin in order to match the units of the ideal gas constant. To convert to kelvin, well add 273 to the Celsius temperature. Twenty five plus 273 is 298. Now, plug the values into the formula and solve algebraically. The volume equals 1.42 mol times 0.0821 Latm/molK times 298 K divided by 1.25 atm. Therefore, the volume of the ideal gas is approximately 27.8 L. Finally, check to make sure that your answer is reasonable. All the original units have cancelled out except for the desired units for the answer, L. Therefore, we can be confident that the mathematical steps we used to find the answer were correct.16Daltons law of partial pressure describes the relationships in a mixture of gases

Total pressure of a gas mixture is equal to the sum of the partial pressures of the individual gases

Daltons Law of Partial Pressure

Daltons law of partial pressure describes the relationships in a mixture of gases. It states that the total pressure of a gas mixture is equal to the sum of the partial pressures of the individual gases. Therefore, the partial pressures or the total pressure may be calculated using the equation Ptotal equals P1 plus P2 plus P3 and so on.For example, if the pressure inside a container that holds gas A increases from 46 kPa to 138 kPa when gas B is added, then the partial pressure of gas B can be determined using Daltons law of partial pressure. The partial pressure of gas B can be found by subtracting the total pressure, 138 kPa, by the partial pressure of gas A, 46 kPa. Therefore, the partial pressure of gas B is 92 kPa.

17Daltons Law CalculationEx) A mixture of gases containing methane, CH4, ethane, C2H6, and propane, C3H8, gases has a total pressure of 975 mmHg. If the partial pressures of CH4 and C2H6 are 235 mmHg and 450mmHg, respectively, what is the partial pressure of C3H8 ?Write unknown and givens

Identify the formula and rearrange, if needed

Convert units and find intermediates, if neededPlug in and solve

Make sure the answer is reasonable

Lets work an example problem. A mixture of gases containing methane, CH4, ethane, C2H6, and propane, C3H8, gases has a total pressure of 975 mmHg. If the partial pressures of CH4 and C2H6 are 235 mmHg and 450 mmHg, respectively, what is the partial pressure of C3H8? First write out the unknown and given values. The total pressure is 975 mmHg. The pressure of CH4 is 235 mmHg. The pressure of C2H6 gas is 450 mmHg and the pressure of C3H8 gas is the unknown. Now that the unknown and givens have been written, well identify the formula that well use to solve the problem and, if needed, rearrange it algebraically to solve for the unknown variable. Since this problem involves partial pressures of gases in a mixture, Daltons law is used. Well rearrange the formula to solve for the pressure of C3H8 gas. The pressure of C3H8 gas is equal to the total pressure minus the pressure of the CH4 gas and the C2H6 gas.Next, check to make sure that there are no conversions or intermediate calculations that need to be preformed. We have all of the information we need to solve for the unknown so no intermediate calculations are needed. All pressure units are given in mmHg so no conversions are necessary.Now, plug the values into the formula and solve algebraically. The pressure of the C3H8 gas equals 975 mmHg minus 235 mmHg minus 450 mmHg. Therefore, the pressure of propane, C3H8, is 290 mmHg. Finally, check to make sure that your answer is reasonable. The sum of the all the partial pressures should equal the total pressure, according to Daltons law, so 235 mmHg plus 450 mmHg plus 290 mmHg should equal 975 mmHg, which is does. The answer is reasonable. 18Learning ObjectivesDescribe the relationships between volume, pressure, number of moles, and temperature for an ideal gasPerform calculations that predict values for variables using Boyles lawCharles lawAvogadros law Ideal gas lawDaltons law of partial pressure

Gas LawsThis concludes our presentation on the relationships between volume, pressure, number of moles, and temperature for an ideal gas. You now know how to perform calculations that predict values for these variables using Boyless law, Charles law, Avogadro's law, the ideal gas law, and Daltons law of partial pressure.19