Chapter 11 Gases

Chapter 11 Gases 11.1 Gases and Pressure Pressure and Force: We have learned that pressure is related to the number of particle collisions and temperature. The collisions cause an outward push, or force, against the walls of its container. The pressure exerted by a gas is dependent on the volume, temperature, and number of molecules present. Pressure (P) is defined as the force per unit area on a surface. The SI unit for force is the Newton, N. All objects exert a certain amount of force on the ground based on how much they weigh. Their weight is compared to the gravitational acceleration 9.8 m/s2. This force is then compared to the area of ground it is resting on. If the pressure of the same object is compared between two areas we would find that the pressure exerted by that object is different. The same force is exerted in 2 different areas by the equation: P= F/A The Earth also has pressure exerted on it, the atmospheric pressure. This is made from the individual pressures of the various gases in the atmosphere. Measuring Pressure: A barometer is used to measure atmospheric pressure. Torricelli constructed the first barometer in the early 1600’s. He was testing to see what the maximum height that water could raise to by comparing it to Hg. He noticed that each time he tested his idea with Hg it would only rise to a height of 760mm. He concluded that the Hg fell to a certain pt based on the gravitational force but then was stopped from falling by the atmospheric pressure that was around it. The maximum height of the Hg in the tube is based on the surrounding atmospheric pressure. Units of Pressure: The common unit for the atmospheric pressure is mm of Hg, (mmHg). A pressure of 1 mmHg can also be referred to as 1 torr. Pressures can also be measured in atmospheres (atm). This is defined as being exactly 760 mmHg. The SI unit for pressure is the Pascal (Pa). It is defined as the pressure exerted by a force of 1 N acting on an area of one square meter. Pressure can also be expressed as kPa.

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Sample Problem A: The average atmospheric pressure in Denver, CO is 0.830 atm. Express this pressure in mm Hg and kPa. P= 0.830 atm kPa=? mmHg=?  760 mmHg  0.830 atm    631 mmHg  1 atm   101.325 kPa  0.830atm    84.1 kPa 1 atm  

Dalton’s law of Partial Pressures: The pressure exerted by each gas in an unreacted mixture is independent of that exerted by other gases present. The pressure of each gas in a mixture is called the partial pressure of that gas. Dalton’s law of partial pressures states that the total pressure of a gas mixture is the sum of the partial pressures of the component gases. Dalton’s Law of Parital Pressures: P T= P1+P2+P3+P4… If we think about the partial pressure of each gas along with the KMT we could understand that as the number of particles increase then the number of collisions will also increase. Each gas will exert a pressure independently on the container walls. The total pressure is the result of the total number of collisions per unit of wall area in a given time. Gases collected by water displacement: Gases in lab are mostly collected over water. As the gas enters the container filled with water, the water is replaced by the gas because it is less dense than the water. This does mean that the gas that you collect will have added water vapor. This calculation will always need to be factored vapor pressure from the water. Sample Problem B: Oxygen gas from the decomposition of potassium chlorate, KClO3, was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0°C, respectively. What was the partial pressure of the oxygen collected? PT=Patm= 731.0 torr PH20= 17.5 torr (found in Table A-8)

Patm= PO2 + PH2O

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Patm PO2  PH2O  PO2  Patm  PT  731atm  17.5 torr  713.5torr

11.2 The Gas Laws In 1662, Robert Boyle discovered that gas pressure and volume are related mathematically. The gas laws are simple mathematical relationships between the volume, temperature, pressure, and amt of gas. Boyle’s law: Pressure-Volume Relationship: Boyle discovered that doubling the P on a sample of gas at constant T reduces the volume by ½. Reducing the P of a gas by ½ will double the V. As one variable increases the other one decreases. Boyle’s Law states that the volume of a fixed mass of a gas varies inversely with the pressure at constant temperature. PV=k; pressure= P, volume=V, constant=k Because this expression is equal to the constant we can express this law as P1V1=P2V2. This allows us to look at the change that a certain gas undergoes. Sample Problem C: A sample of gas has a volume of 150 mL when the pressure if 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? P1=0.947 atm V1=150 mL PV 1 1  PV 2 2  V2 

P2=0.987 atm V2=?

PV 1 1  V2 P2

 0.947atm 150mL 

0.987 atm V2  144mL O2

Charles’s Law: Volume-Temperature Relationship: We know from the KMT that if the temp of a gas is increased then the volume will also increase do to the increased KE of the particles. The volume must expand for the pressure to remain constant. This relationship was discovered by Jacques Charles in 1787. 3

He found that the volume of a gas will increase by a fraction of 1/273 for every 1°C the temp is raised in order for the pressure to remain constant. The Celsius scale can then be related to the Kelvin scale to give it a value of 0K or absolute zero is equal to -273.15°C. Gas volume and K Temps are directly proportional to each other at constant pressure. Charles’s law states that the volume of a fixed mass of gas at constant pressure varies directly with the K. V=kT or V/T=k T= temp in K, k = constant, V=volume. In the same manner as before if the ratio of these variables are related to a constant, k we can show them related to the change of itself. V1 V2  T1 T2

Sample Problem D: A sample of neon gas occupies a volume of 752 mL at 25°C. What is the volume the gas will occupy at 50°C if the pressure remains constant? V1= 752mL T1= 25°C

V2=? T2= 50°C

V1 V2 VT   V2  1 2 T1 T2 T1 V2 

 752mL  323K   298K 

V2  815mL Ne

Gay-Lussac’s law: Pressure-Temp Relationship: the energy and frequency of gas at constant KE of molecules. At a constant volume the pressure should be directly proportional to the K temp, which is directly related to the KE. Joseph Gay- Lussac discovered this relationship in 1802. Gay-Lussac Law: the pressure of a fixed mass of gas at constant volume varies directly with the K temp. P=kT or P/T=k P=pressure, T=Temperature, k=constant

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Sample Problem E: The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container warn the user to keep it in a place where the temp exceeds 52°C. What would the gas pressure in the container be at 52°C. P1=3.00 atm T1= 25°C

P2 = ? T2= 52°C

P1 P2 PT   P2  1 2 T1 T2 T1 P2 

 3.00atm  325K   298K 

P2  3.27 atm

The Combined Gas Law: gases can undergo a change of temp, pressure, and volume all at the same time. We would have to combine the 3 laws that we just learned about. PV/T=k

PV PV 1 1  2 2 T1 T2

Sample Problem F: A helium-filled balloon has a volume of 50.0L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C. P1=1.08atm V1=50.0L T1=25°C=298K

P2=0.855 atm V2=? T2=10°C=283K

PV PV PV T 1 1  2 2  V2  1 1 2 T1 T2 T1 P2 V2 

1.08atm  50.0 L  283K   298K  0.855atm 

V2  60.0 L

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11.3 Gas Volumes and the Ideal Gas Law Measuring and Comparing the Volumes of Reacting Gases: While Guy-Lussac was studying the relationship between pressure and temperature he also noticed a simple and definite proportion by volume. Guy-Lussac’s law of combined volumes of gases states that at constant temp and pressure with volumes of gaseous rcts and pdts can be expressed as ratios of small whole numbers. The ratio of the combo of H and O to form H20 is a ratio of 2:1:2. By the law it would state that 2mL of H combined with 1mL of O will form 2 mL of H20, or 400 L, 800 L, 400 L… Avogadro’s Law: Some of the volume relationships observed by GayLussac could not be accounted for by Dalton’s theory. In 1811 Avogadro found a way to satisfy both observations. He thought that molecules that combine could contain more than one atom. His law states that equal volumes of gases at the same temp and pressure contain equal number of molecules. If the temp and pressure remain constant then the volume will vary directly to the number of molecules. Avogadro’s Law: V=kn;

V1 V2  n is the amount of moles and V is the n1 n2

volume. From this law we will look at the coefficients of chemical reactions involving gases as the relative number of moles, molecules, and volume. Molar Volume of a Gas: Remember that one mole of a substance contains the number of molecules equal to Avogadro’s number. According to Avogadro’s law states that one mole of any gas will occupy the same volume as one mole of any other gas at the same temp and pressure no matter the difference in masses.

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The volume of gas occupied ay one mole of gas at STP is 22.4 L. The conversion that can be used for calculations is 1 mol/22.4 L. This can be used to find the mass or moles of any gas at a particular volume. Sample Problem G: What volume does a 0.685 mol of gas occupy at STP? What quantity of gas, in moles, is contained in 2.21 l at STP?  22.4 L  0.0685 mol    1.53L  1mol   1mol  2.21L    0.0987 mol  22.4 L  Gas Stoichiometry:

Sample Problem H: Propane, C3H8, is a gas that is sometimes used as a fuel for cooking and heating. The complete combustion of propane occurs according to the following balanced equation. C3H8 + 5O2  3CO2 + 4H2O What will the volume, in L, of oxygen required for the complete combustion of 0.350L of propane? What will the volume of carbon dioxide produced in the reaction? Assume that all volume measurements are made at the same temp and pressure.  5O2  0.350 L C3 H 8    1.75L O2 1 L C H 3 8    3L CO2  0.350 L C3 H 8    1.05L CO2 1 L C H 3 8  

Ideal Gas Law: The mathematical relationship among T, P, V, and n. This is the equation of state for an ideal gas because the state of a gas can be defined by its P, V, T, and n. This equation also involves the constant R. Ideal Gas Law: PV=nRT This law shows that the increase in volume keeps that collision rate constant.

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The Ideal Las Constant: R is the ideal gas constant. The units are derived from the formula. R= 0.0821 (L atm/mol K)

Sample Problem I: What is the pressure in atm exerted by a 0.500 mol sample of N2 in a 10.0L container at 298K? PV  nRT  P 

P

nRT V

 0.500mol   0.0821

Latm    298K  molK    1.22atm 10.0 L

11.4 Diffusion and Effusion The constant motion of particles causes the random mixing of gas particles also known as diffusion. Effusion is the process where gas molecules are confined to a container and randomly pass through a tiny opening. Rates of diffusion and effusion depend on the velocities of particles. Lighter molecules will move faster than smaller ones. This gives us the relationship that the velocities of gas particles vary inversely with the square root of their mass. Remember that the KE = 1/2mv2 which is also dependent on temp. We can bring these two equations together to give a mathematical relationship for the rate of effusion.

MB rate of effusion gas A  rate of effusion gas B MA This formula is known as Graham’s Law. It states that the rates of effusion of gases at the same temp and pressure are inversely proportional to the square root of the molar masses.

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Sample Problem J: Compare the rates of effusion of H gas and O gas at the same temp and pressure.

32.00 g / mol Rate of Effusion of H 2 32.00 g / mol    3.98 Rate of Effusion of O2 2.016 g / mol 2.016 g / mol This tells us that H will effuse almost 4 times faster than O.

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