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These diagrams are meant to accompany a lecture about the Gas Laws.

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- 1. Gas Laws The gas laws are simple mathematical relationships between the volume, pressure, and amount of gas.
- 2. Boyle’s Law: Pressure-Volume Relationship Boyles Law states that the volume of a fixed mass of gas varies inversely with the pressure at constant temperature. In 1662, Robert Boyle discovered that gas volume and pressure are related mathematically. If you double the pressure on a sample of gas at constant temperature, it reduces its volume by one-half. Reducing the pressure on a gas by one-half allows the volume of the gas to double.
- 3. Boyle’s Law: Pressure-Volume Relationship This box shows the molecules of a gas at a certain pressure. The molecules have room to move around freely without bumping into the sides of the container too frequently. This box shows the same container, with the same molecules, if the volume has been reduced by one-half. The gas pressure increases because the molecules collide more frequently with the walls of the container.
- 4. Boyle’s Law: Pressure-Volume Relationship The mathematical representation of Boyle’s Law is as follows: V = k 1 P or P V = k
- 5. Boyle’s Law: Pressure-Volume Relationship k is a constant, which means it will not change. If PV = k, then the pressure times the volume will ALWAYS equal the constant “k.” Therefore, if you change the volume or pressure, it will still equal the same thing (k), so the volume times pressure will be the same before your change and after your change. P 1 = V 1 P 2 V 2 Which simplifies to: P 2 = V 2 P 1 V 1
- 6. Boyle’s Law: Pressure-Volume Relationship The first volume of a container of oxygen is 150 ml (V 1 ). The first pressure of this container is 0.947 atm (P 1 ). If you change the pressure to 0.987 atm (P 2 ), what will the volume change to (V 2 )? V 2 = V 1 P 1 P 2 First, rearrange the equation so the (V 2 ) is isolated.
- 7. Boyle’s Law: Pressure-Volume Relationship Now, plug in the information that you have, “P 1 ”, “V 1 ” and “V 2 ”. V 2 = (150 ml) (0.97 atm) 0.987 atm The “atm” on the bottom cancels out the “atm” on the top. Now, plug the numbers into the calculator, and you have your answer. 144 ml
- 8. Charles’ Law: Volume – Temperature Relationship Charles’ Law states that the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature. The quantitative relationship between volume and temperature was discovered by the French scientist Jacques Charles in 1787. Charles’ experiments showed that all gases expand to the same extent when heated through the same temperature interval. Charles found that the volume changes by 1/273 of the original volume at 0 degrees Celsius.
- 9. Charles’ Law: Volume – Temperature Relationship This box shows the molecules of a gas at a certain temperature. Let’s imagine that the walls of the container are flexible, like a balloon. This box shows the same container, with the gas inside heated to a higher temperature. The molecules of the gas move faster at the higher temperature, thus hitting the walls of the container more often and with more force. Finally, the container must increase in size, so that the molecules will hit the walls left often. The lower frequency of impact offsets the higher force of the impact.
- 10. Charles’ Law: Volume – Temperature Relationship The average kinetic energy of gas molecules is more closely related to the Kelvin temperature. The Kelvin temperature scale is a scale that starts at a temperature corresponding to -273.15 degrees Celsius. This temperature is referred to as absolute zero and is given a value of zero in the Kelvin scale. This relationship is represented by the following mathematical expression: K 273.15 = + C
- 11. Charles’ Law: Volume – Temperature Relationship Charles’ law can be expressed by the following mathematical expression: V = k T or V T = k The value of T is the Kelvin temperature, and k is a constant. The value of k depends only on the quantity of gas and the pressure.
- 12. Charles’ Law: Volume – Temperature Relationship The form of Charles’ law that can be applied directly to most volume-temperature problems involving gases is as follows: V 1 T 1 = V 2 T 2 V 1 and T 1 represent the initial conditions. V 2 and T 2 represent the new set of conditions. If three of the four variables are known, the equation can be rewritten to isolate the unknown variable and solve.
- 13. Charles’ Law: Volume – Temperature Relationship The first volume (V 1 ) of the element Ne is 752 mL. The first temperature (T 1 ) of the element is 25 degrees C + 273 = 298K If you change the temperature (T 2 ) of the element to 50 degrees C + 273 = 323 K, what will the new volume be? V 2 = T 2 V 1 T 1 First, rearrange the equation so the (V 2 ) is isolated.
- 14. Charles’ Law: Volume – Temperature Relationship Now, plug in the information that you have, “V 1 ”, “T 1 ” and “T 2 ”. V 2 = (323 K) (752 mL) 298 K The “K” on the bottom cancels out the “K” on the top. Now, plug the numbers into the calculator, and you have your answer. 815 mL
- 15. Gay-Lussac’s Law: Pressure-Temperature Relationship Gay-Lussac’s Law states that the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature In 1802, Joseph Gay-Lussac discovered that for every Kelvin of temperature change, the pressure of a confined gas changes by 1/273 of the pressure at 0 degrees C.
- 16. Gay-Lussac’s Law: Pressure-Temperature Relationship This box shows the molecules of a gas at a certain temperature. The molecules have room to move around freely without bumping into the sides of the container too frequently. This box shows the same container, with the same molecules, if the temperature has been increased. The gas pressure increases because the movement of the molecules become faster and have more kinetic energy, so they hit the walls of the container with more force and more often.
- 17. Gay-Lussac’s Law: Pressure-Temperature Relationship Gay-Lussac’s law can be expressed by the following mathematical expression: P = k T or P T = k The value of T is the Kelvin temperature, and k is a constant that depends on the quantity of gas and the volume.
- 18. Gay-Lussac’s Law: Pressure-Temperature Relationship If k is always the same, then the P/T ratio will remain the same before and after you make a change. Therefore, you can set the initial P/T ratio to the changed P/T ratio. P 1 T 1 = P 2 T 2
- 19. Gay-Lussac’s Law: Pressure-Temperature Relationship The first pressure (P 1 ) of a gas is 3.00 atm. The first temperature (T 1 ) of the gas is 25 degrees C + 273 = 298K If you change the temperature (T 2 ) of the gas to 52 degrees C + 273 = 325 K, what will the new pressure be? Find the second pressure (P 2 ). P 2 = T 2 P 1 T 1 First, rearrange the equation so the (P 2 ) is isolated.
- 20. Gay-Lussac’s Law: Pressure-Temperature Relationship Now, plug in the information that you have, “P 1 ”, “T 1 ” and “T 2 ”. P 2 = (325 K) (3.00 atm) 298 K The “K” on the bottom cancels out the “K” on the top. Now, plug the numbers into the calculator, and you have your answer. 3.27 atm
- 21. The Combined Gas Law The Combined Gas Law expresses the relationship between pressure, volume, and temperature of a fixed amount of gas. The Combined Gas Law is used when TWO or THREE variables are being changed for a fixed amount of gas. For example, if we change both the pressure AND the temperature, what will happen to the volume? Because of the complication of two or more changes, we have no way to visually predict what will happen to the gas – we must go straight to the mathematical expression.
- 22. The Combined Gas Law The Combined Gas Law can be shown using the following mathematical expression: PV T = k In the equation, k is constant and depends on the amount of gas.
- 23. The Combined Gas Law If k is always the same, then the PV/T ratio will remain the same before and after you make a change. Therefore, you can set the initial PV/T ratio to the changed PV/T ratio. P 1 V 1 T 1 = P 2 V 2 T 2 Students often find this law the most difficult because of all the math involved. To simplify it, we will show you the equation with all six variables isolated.
- 24. The Combined Gas Law Find the first pressure (P 1 ). P 1 = P 2 V 2 T 1 T 2 V 1 Find the first Volume (V 1 ). V 1 = P 2 V 2 T 1 T 2 P 1
- 25. The Combined Gas Law Find the first Temperature (T 1 ). T 1 = P 2 V 2__ T 2 P 1 V 1 Find the second pressure (P 2 ). P 2 = P 1 V 1 T 2 T 1 V 2
- 26. The Combined Gas Law Find the second volume (V 2 ). V 2 = P 1 V 1 T 2 T 1 P 2 Find the second temperature (T 2 ). T 2 = P 1 V 1__ T 1 P 1 V 2
- 27. The Combined Gas Law The first volume of a Helium is 50.0 L. The first temperature of Helium is 25 degrees C + 273 = 298K. The second temperature of Helium is 10 degrees C + 273 = 283 K. The first pressure of Helium is 1.08 atm. The second pressure of Helium is 0.855 atm. Find the second volume (V 2 ) of the gas. V 2 = First, rearrange the equation so the (V 2 ) is isolated. P 1 V 1 T 2 T 1 P 2
- 28. The Combined Gas Law Now, plug in the information that you have, “P 1 ”, “V 1 ” and “T 1 ”, P 2 , & T 2 . V 2 = (283 K) (1.08 atm) The “K” on the bottom cancels out the “K” on the top The “atm” on the cancels out the “atm” on the top.. Now, plug the numbers into the calculator, and you have your answer. 60.0 L (50.0 L) (0.855 atm) (298 K)
- 29. Dalton’s Law of Partial Pressures The partial pressure of a gas is the pressure of that gas in a mixture. Dalton’s Law of Partial Pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. John Dalton, the English chemist who proposed the atomic theory, also studied gas mixtures. He found that in the absence of a chemical reaction, the pressure of a gas mixture is the sum of the individual pressures of each gas alone.
- 30. Dalton’s Law of Partial Pressures This box shows the molecules of a gas at a pressure of 0.12 atm, and at 0 degrees Celsius. This box shows another type of gas at the pressure of 0.12 atm and at 0 degrees Celsius. Finally, this box shows the combined gases at a pressure of 0.24 atm, and at 0 degrees Celsius. The rapidly moving particles of each gas have an equal chance to collide with the container walls. Therefore, each gas exerts a pressure independent of that exerted by the other gases present. The total pressure is the result of the total number of collisions per unit of wall area in a given time.
- 31. Dalton’s Law of Partial Pressures Gay-Lussac’s law can be expressed by the following mathematical expression: P T = P 1 + P T is the total pressure of the mixture, P 1 , P 2 , P 3 , etc. are the partial pressures of component gases 1, 2, 3, and so on. + P 3 P 2 + …
- 32. Dalton’s Law of Partial Pressures Gases produced in the laboratory are often collected over water. The gas produced by the reaction displaces the water, which is more dense. A gas collected by water displacement is not pure, but is always mixed with water vapor. That is because water molecules at the liquid surface evaporate and mix with the gas. Water vapor exerts water-vapor pressure . To determine the total pressure of the gas and water vapor, you would raise the bottle and water levels inside and outside the bottle until the water levels inside and outside the bottle were the same. At this point, the total pressure inside the bottle would be the same as the atmospheric pressure.
- 33. Dalton’s Law of Partial Pressures The mathematical expression used to calculate the pressure of the gas with the water vapor: P atm = P gas + If you want to calculate the pressure of the gas, alone, you would find out the atmospheric pressure, using a barometer, and then subtract the vapor pressure of the water at the given temperature from the atmospheric (total) pressure. P H 2 O
- 34. Dalton’s Law of Partial Pressures The total pressure (P T , or, P atm ) is 731.0 torr. The water vapor pressure (P H 2 O ) at 20.0 degrees Celsius is 17.5 torr. Find the pressure of the Oxygen (P O 2 ) in the container. P O 2 = First, rearrange the equation so the (P O 2 ) is isolated. P atm - P H 2 O
- 35. Dalton’s Law of Partial Pressures Now, plug in the information that you have, P atm & P H 2 O . P O 2 = 17.5 torr 731.0 torr Now, do the math, and you have your answer: 713.5 torr -

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