Gaseous State of Matter

Gaseous State of Matter
Gases Gaseous State of Matter

States of matter Solids Liquids Gases Definite Shape, Rigid structure
Definite volume Least energetic molecules Liquids Indefinite shape, No fixed structure More energetic than solids, mobile Gases Indefinite shape Indefinite volume Most energetic of the phases Low density

Kinetic-Molecular Theory (KMT)
Gases consist of tiny particles Distance between the particles is large No attraction between the particles Particles move in straight lines in all directions, frequently colliding No energy is lost in collisions, elastic Average KE for particles is the same for all gases

Kinetic energy Expressed as KE = ½ mv2 m is mass v is velocity
All gases have the same kinetic energy at the same temperature. Velocities differ – higher mass, slower velocity lower mass, faster velocity

Diffusion vs. Effusion Diffusion – the ability of two or more gases to mix spontaneously until they form a uniform mixture Effusion – the process by which gas molecules pass through a very small opening from a container at higher pressure to one at lower pressure

Pressure Force per square unit area, F/A
Results from the collisions of gas particles

ATMOSPHERIC PRESSURE Gravitational forces hold the gas molecules in air relatively close to Earth which prevents air from going into space Atmospheric pressure is at any point is due to the mass of the atmosphere pressing downward at that point Concentration of gas molecules decrease with altitude thus lower pressure at higher altitudes

Barometer

Standard Units of Pressure

Example 11.1 Converting between Pressure Units
A high-performance road bicycle tire is inflated to a total pressure of 125 psi. What is this pressure in millimeters of mercury? SORT You are given a pressure in psi and .asked to convert it to mm Hg. STRATEGIZE Begin the solution map with the given units of psi. Use the conversion factors to convert first to atm and then to mm Hg. GIVEN: 125 psi FIND: mm Hg SOLUTION MAP RELATIONSHIPS USED 1 atm = 14.7 psi (Table 11.1) 760 mm Hg = 1 atm (Table 11.1)

Example 11.1 Converting between Pressure Units
Continued SOLVE Follow the solution map to solve the problem. SOLUTION The answer has the correct units, mm Hg. The answer is reasonable because the mm Hg is a smaller unit than psi; therefore the value of the pressure in mm Hg should be greater than the value of the same pressure in psi. CHECK Check your answer. Are the units correct? Does the answer make physical sense? SKILLBUILDER 11.1 Converting between Pressure Units Convert a pressure of 173 in. Hg into pounds per square inch. Answer: 85.0 psi SKILLBUILDER PLUS Convert a pressure of 23.8 in. Hg into kilopascals. Answer: 80.6 kPa For More Practice Example 11.13; Problems 23, 24, 25, 26, 29, 30, 31, 32.

A barometer reads 1.12 atm. Calculate the corresponding pressure in (a) torr and (b) mm Hg.

Boyles’ Law At constant temperature (T), the volume (V) of a fixed mass ( number of molecules/ moles is constant too) of a gas is inversely proportional to the pressure (P). V 1/P or P1V1 = P2V2 PV = constant; or PV = k P1V1 is the pressure-volume for one set of conditions P2V2 is the pressure-volume for another set of conditions

Example 11.2 Boyle’s Law SORT STRATEGIZE
A cylinder equipped with a moveable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume of the cylinder if the applied pressure is decreased to 1.0 atm? SORT You are given an initial pressure, an initial volume, and a final pressure. You are asked to find the final volume. STRATEGIZE Draw a solution map beginning with the given quantities. Boyle’s law shows the relationship necessary to get to the find quantity. GIVEN: P1 = 4.0 atm V1 = 6.0 L P2 = 1.0 atm FIND: V2 SOLUTION MAP RELATIONSHIPS USED P1V1 = P2V2 (Boyle’s law, presented in this section)

Example 11.2 Boyle’s Law SOLVE CHECK Continued SOLUTION
Solve the equation for the quantity you are trying to find (V2), and then substitute the numerical quantities into the equation to compute the answer. SOLUTION The answer has units of volume (L) as expected. The answer is reasonable because we expect the volume to increase as the pressure decreases. CHECK Check your answer. Are the units correct? Does the answer make physical sense? SKILLBUILDER 11.2 Boyle’s Law A snorkeler takes a syringe filled with 16 mL of air from the surface, where the pressure is 1.0 atm, to an unknown depth. The volume of the air in the syringe at this depth is 7.5 mL. What is the pressure at this depth? If the pressure increases by an additional 1 atm for every 10 m of depth, how deep is the snorkeler? Answer: P2 = 2.1 atm; depth is approximately 11 m For More Practice Example 11.14; Problems 33, 34, 35, 36.

Charles’ Law At constant pressure the volume of a fixed mass of any gas is directly proportional to the absolute temperature, which may be expressed as V  T or V1/T1 = V2/T2 (T is temperature in Kelvin) V/T = constant Kelvin = 0C K Absolute Zero is -2730C; zero point on the Kelvin scale. At Absolute Zero, the volume of an ideal, or perfect, gas would be zero.

Example 11.3 Charles’s Law SORT STRATEGIZE
A sample of gas has a volume of 2.80 L at an unknown temperature. When the sample is submerged in ice water at t = 0 °C, its volume decreases to 2.57 L. What was its initial temperature (in kelvins and in Celsius)? Assume a constant pressure. (To distinguish between the two temperature scales, use t for temperature in C, and T for temperature in K.) SORT You are given an initial volume, a final volume, and a final temperature. You are asked to find the intitial temperature in both kelvins (T1) and degrees Celsius (t1). STRATEGIZE Draw a solution map beginning with the given quantities. Charles’s law shows the relationship necessary to get to the find quantity. GIVEN: V1 = 2.80 L V2 = 2.57 t2 = 0 °C FIND: T1 and t1 SOLUTION MAP RELATIONSHIPS USED

Example 11.3 Charles’s Law SOLVE CHECK Continued SOLUTION
Solve the equation for the quantity you are trying to find (T1). Before you substitute in the numerical values, you must convert the temperature to kelvins. Remember, gas law problems must always be worked using Kelvin temperatures. Once you have converted the temperature to kelvins, substitute into the equation to find T1. Convert the temperature to degrees Celsius to find t1. SOLUTION The answers have the correct units, K and C. The answer is reasonable because the initial volume was larger than the final volume; therefore the initial temperature must be higher than the final temperature. CHECK Check your answer. Are the units correct? Does the answer make physical sense?

Example 11.3 Charles’s Law Continued SKILLBUILDER 11.3 Charles’s Law
A gas in a cylinder with a moveable piston has an initial volume of 88.2 mL and is heated from 35 C to 155 C. What is the final volume of the gas in milliliters? Answer: 123 mL For More Practice Problems 39, 40, 41, 42.

Gay-Lussac’s Law Pressure and Temperature at constant volume and constant amount The pressure of a fixed mass of a gas, at constant volume, is directly proportional to the Kelvin temperature. PT; P/T = constant P1/T1 = P2/T2

Pressure Dependent upon the number of molecules and the temperature.
Pressure is directly proportional to temperature (T) Pressure is directly proportional to number of molecules (in moles, n)

Pressure and Temperature

Combined Gas Laws To compare volumes of gas, standard conditions were selected. STP – standard temperature and pressure K or 00C and 1 atm or 760 torr or 760 mmHg or kPa. The Combined Gas Law is a combination of Boyles Law, Charles’ Law and Gay-Lussac’s Law.

Related to the Boyle’s Law Charles’ Law Gay Lussac’s Law
T constant  P1V1 = P2V2 Charles’ Law P constant  V1/T1 = V2/T2 Gay Lussac’s Law V constant  P1/T1 = P2/T2

Using P, V, T and hold n constant
PV/T  constant Given initial and final conditions can solve for an unknown variable P1, V1, and T1 are initial conditions P2,V2, and T2 are final conditions P1V1/T1 = P2V2/T2

Example 11.4 The Combined Gas Law
A sample of gas has an initial volume of 158 mL at a pressure of 735 mm Hg and a temperature of 34 C. If the gas is compressed to a volume of 108 mL and heated to a temperature of 85 C, what is its final pressure in millimeters of mercury? SORT You are given an initial pressure, temperature, and volume as well as a final temperature and volume. You are asked to find the final pressure. STRATEGIZE Draw a solution map beginning with the given quantities. The combined gas law shows the relationship necessary to get to the find quantity. GIVEN: P1 = 735 mm Hg t1 = 34 C t2 = 85 C V1 = 158 mL V2 = 108 mL FIND: P2 SOLUTION MAP RELATIONSHIPS USED

Continued SOLUTION The answer has the correct units, mm Hg. The answer is reasonable because the decrease in volume and the increase in temperature should result in a pressure that is higher than the initial pressure. SOLVE Solve the equation for the quantity you are trying to find (P2). Before you substitute in the numerical values, you must convert the temperatures to kelvins. Once you have converted the temperature to kelvins, substitute into the equation to find P2. CHECK Check your answer. Are the units correct? Does the answer make physical sense?

Continued SKILLBUILDER 11.4 The Combined Gas Law A balloon has a volume of 3.7 L at a pressure of 1.1 atm and a temperature of 30 C. If the balloon is submerged in water to a depth where the pressure is 4.7 atm and the temperature is 15 C, what will its volume be (assume that any changes in pressure caused by the skin of the balloon are negligible)? Answer: 0.82 L For More Practice Example 11.15; Problems 51, 52, 53, 54, 55, 56.

Given 20. 0L of ammonia gas at 5. 00oC and 730
Given 20.0L of ammonia gas at 5.00oC and 730. torr, calculate the volume at 50.00C and 800. torr Organize your data Initial Final Solve the equation for the unknown. Plug in your values and calculate

Avogadro’s Law Equal volumes of different gases at the same temperature and pressure contain the same number of molecules. Since a mole contains 6.22 x 1023 molecules, a mole of any gas will have the same volume as a mole of any other gas at the same temperature and pressure.

Avogadro’ Law

Example 11.5 Avogadro’s Law
A 4.8-L sample of helium gas contains 0.22 mol of helium. How many additional moles of helium gas must be added to the sample to obtain a volume of 6.4 L? Assume constant temperature and pressure. SORT You are given an initial volume, an initial number of moles, and a final volume. You are (essentially) asked to find the final number of moles. STRATEGIZE Draw a solution map beginning with the given quantities. Avogadro’s law shows the relationship necessary to get to the find quantity. GIVEN: V1 = 4.8 L n1 = 0.22 mol V2 = 6.4 L FIND: n2 SOLUTION MAP RELATIONSHIPS USED

Continued SOLVE Solve the equation for the quantity you are trying to find (n2) and substitute the appropriate quantities to calculate n2. Since the balloon already contains 0.22 mol, subtract this quantity from the final number of moles to determine how much must be added. SOLUTION The answer has the correct units, moles. The answer is reasonable because the increase in the number of moles is proportional to the given increase in the volume. CHECK Check your answer. Are the units correct? Does the answer make physical sense?

Continued SKILLBUILDER 11.5 Avogadro’s Law A chemical reaction occurring in a cylinder equipped with a moveable piston produces 0.58 mol of a gaseous product. If the cylinder contained 0.11 mol of gas before the reaction and had an initial volume of 2.1 L, what was its volume after the reaction? Answer: 13 L For More Practice Problems 45, 46, 47, 48.

Ideal Gas Law PV = nRT R = .0821 L-atm/mol-K
Used to calculate any one of the four variables when three are known

PV = nRT Derived from Boyle’s Law, Charles’s Law (and Gay-Lussac’s Law) and Avogadro’s Law

Example 11.6 The Ideal Gas Law
Calculate the volume occupied by mol of nitrogen gas at a pressure of 1.37 atm and a temperature of 315 K. SORT You are given the number of moles, the pressure, and the temperature of a gas sample. You are asked to find the volume. STRATEGIZE Draw a solution map beginning with the given quantities. The ideal gas law shows the relationship necessary to get to the find quantity. GIVEN: n = mol P = 1.37 atm T = 315 K FIND: V SOLUTION MAP RELATIONSHIPS USED PV = nRT (Ideal gas law, presented in this section)

Example 11.6 The Ideal Gas Law
Continued SOLVE Solve the equation for the quantity you are trying to find (V) and substitute the appropriate quantities to compute V. SOLUTION The answer has the correct units for volume, liters. The value of the answer is a bit more difficult to judge. However, at standard temperature and pressure (T = 0 C or K and P = 1 atm), 1 mol gas occupies 22.4 L (see Section 11.10). Therefore our answer of 16.0 L seems reasonable for the volume of 0.85 mol of gas under conditions that are not too far from standard temperature and pressure. CHECK Check your answer. Are the units correct? Does the answer make physical sense? SKILLBUILDER 11.6 The Ideal Gas Law An 8.5-L tire is filled with 0.55 mol of gas at a temperature of 305 K. What is the pressure of the gas in the tire? Answer: 1.6 atm For More Practice Example 11.16; Problems 59, 60, 61, 62.

Example 11.7 The Ideal Gas Law Requiring Unit Conversion
Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25 C. SORT You are given the pressure, the volume, and the temperature of a gas sample. You are asked to find the number of moles. STRATEGIZE Draw a solution map beginning with the given quantities. The ideal gas law shows the relationship necessary to get to the find quantity. GIVEN: P = 24.2 psi V = 3.2 L t = 25 C FIND: n SOLUTION MAP RELATIONSHIPS USED PV = nRT (Ideal gas law, presented in this section)

Continued SOLVE Solve the equation for the quantity you are trying to find (n). Before substituting into the equation, you must convert P and t into the correct units. (Since atm is an intermediate answer, mark the least significant digit, but don’t round until the end.) Finally, substitute into the equation to calculate n. SOLUTION The answer has the correct units, moles. The value of the answer is a bit more difficult to judge. Again, knowing that at standard temperature and pressure (T = 0 C or K and P = 1 atm), 1 mol of gas occupies 22.4 L can help (see Check step in Example 11.6). A 3.2 L sample of gas at STP would contain about 0.15 mol; therefore at a greater pressure, the sample should contain a bit more than 0.15 mol, which is consistent with the answer. CHECK Check your answer. Are the units correct? Does the answer make physical sense?

Continued SKILLBUILDER 11.7 The Ideal Gas Law Requiring Unit Conversion How much volume does mol of gas occupy when its pressure is 715 mm Hg and its temperature is 58 C? Answer: 16.1 L SKILLBUILDER PLUS Find the pressure in millimeters of mercury of a g sample of helium gas at 32 C and contained in a 648-mL container. Answer: 977 mm Hg For More Practice Problems 63, 64, 67, 68.

Example 11.8 Molar Mass Using the Ideal Gas Law and Mass Measurement
A sample of gas has a mass of g. Its volume is L at a temperature of 55 C and a pressure of 886 mm Hg. Find its molar mass. SORT You are given the mass, the volume, the temperature, and the pressure of a gas sample. You are asked to find the molar mass of the gas. STRATEGIZE In the first part of the solution map, use the ideal gas law to find the number of moles of gas from the other given quantities. In the second part, use the number of moles from the first part, as well as the given mass, to find the molar mass. GIVEN: m = g V = L t = 55 C P = 886 mm Hg FIND: Molar mass (g/mol) SOLUTION MAP RELATIONSHIPS USED

Continued SOLVE First, solve the ideal gas law for n. Before substituting into the equation, you must convert the pressure to atm and temperature to K. Now, substitute into the equation to calculate n, the number of moles. Finally, use the number of moles just found and the given mass (m) to find the molar mass. SOLUTION The answer has the correct units, g/mol. The answer is reasonable because its value is within the range of molar masses for common compounds. CHECK Check your answer. Are the units correct? Does the answer make physical sense?

Continued SKILLBUILDER 11.8 Molar Mass Using the Ideal Gas Law and Mass Measurement A sample of gas has a mass of 827 mg. Its volume is L at a temperature of 88 C and a pressure of 975 mm Hg. Find its molar mass. Answer: 70.8 g/mol For More Practice Problems 69, 70, 71, 72.

Real gases – DO NOT ALWAYS BEHAVE IDEALLY!!!!

Dalton’s Law of Partial Pressures
The total of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture. PTotal = P1 + P2 + P If 3 gases are present, the total pressure can be calculated by adding the pressures exerted by the three gases. To determine the pressure of one of the gases, solve for that gas.

If the pressure of P2 is unknown, but Ptotal is 760 torr, P1 is 480 torr and P3 is 150 torr, solve for P2. P2 = Ptotal – P1 – P3 P2 = 760 torr – 480 torr – 150 torr P2 = 130 torr

Example 11.9 Total Pressure and Partial Pressure
A mixture of helium, neon, and argon has a total pressure of 558 mm Hg. If the partial pressure of helium is 341mm Hg and the partial pressure of neon is 112 mm Hg, what is the partial pressure of argon? You are given the total pressure of a gas mixture and the partial pressures of two (of its three) components. You are asked to find the partial pressure of the third component. To solve this problem, solve Dalton’s law for the partial pressure of argon and substitute the correct values to calculate it. GIVEN: Ptot = 558 mm Hg PHe = 341 mm Hg PNe = 112 mm Hg FIND: PAr SOLUTION MAP Ptot = PHe + PNe + PAr PAr = Ptot – PHe – PNe = 558 mm Hg – 341 mm Hg – 112 mm Hg = 105 mm Hg SKILLBUILDER 11.9 Total Pressure and Partial Pressure A sample of hydrogen gas is mixed with water vapor. The mixture has a total pressure of 745 torr, and the water vapor has a partial pressure of 24 torr. What is the partial pressure of the hydrogen gas? Answer: 721 torr For More Practice Example 11.17; Problems 73, 74, 75, 76.

Example 11.10 Partial Pressure, Total Pressure, and Percent Composition
Calculate the partial pressure of oxygen that a diver breathes with a heliox mixture containing 2.0% oxygen at a depth of 100 m where the total pressure is 10.0 atm. GIVEN: O2 percent = 2.0% Ptot = 10.0 atm FIND: PO2 SOLUTION You are given the percent oxygen in the mixture and the total pressure. You are asked to find the partial pressure of oxygen. The partial pressure of a component in a gas mixture is equal to the fractional composition of the component multiplied by the total pressure. Calculate the fractional composition of O2 by dividing the percent composition by Calculate the partial pressure of O2 by multiplying the fractional composition by the total pressure. SKILLBUILDER Partial Pressure, Total Pressure, and Percent Composition A diver breathing heliox with an oxygen composition of 5.0% wants to adjust the total pressure so that PO2 = 0.21 atm. What must the total pressure be? Answer: Ptot = 4.2 atm For More Practice Problems 79, 80, 81, 82.

VAPOR PRESSURE Always a certain amount of molecules of vapor over a liquid; called vapor pressure. As temperature increases so does vapor pressure Low boiling point = high vapor pressure

Gases in chemical equations
Using stoichiometry, the Ideal Gas law and molar volume

Example 11.11 Gases in Chemical Reactions
How many liters of oxygen gas form when 294 g of KClO3 completely react in this reaction (which is used in the ignition of fireworks)? 2 KClO3(s)  2 KCl(s) + 3 O2(g) Assume that the oxygen gas is collected at P = 755 mm Hg and T = 305 K. GIVEN: 294 g KClO3 P = 755 mm Hg (of oxygen gas) T = 305 K FIND: Volume of O2 in liters SOLUTION MAP RELATIONSHIPS USED 1 mol KClO3 = g (molar mass of KClO3) 2 mol KClO3 : 3 mol O2 (from balanced equation given in problem) PV = nRT (ideal gas law, Section 11.8) SORT You are given the mass of a reactant in a chemical reaction. You are asked to find the volume of a gaseous product at a given pressure and temperature. STRATEGIZE The solution map has two parts. In the first part, convert from g KClO3 to mol KClO3 and then to mol O2. In the second part, use mol O2 as n in the ideal gas law to find the volume of O2. You will need the molar mass of KClO3 and the stoichiometric relationship between KClO3 and O2 (from the balanced chemical equation). You will also need the ideal gas law.

Continued SOLVE Begin by converting mass KClO3 to mol KClO3 and then to mol O2. Then solve the ideal gas equation for V. Before substituting the values into this equation, you must convert the pressure to atm. Finally, substitute the given quantities along with the number of moles just calculated to calculate the volume. SOLUTION The answer has the correct units, liters. The value of the answer is a bit more difficult to judge. Again, knowing that at standard temperature and pressure (T = 0 C or K and P = 1 atm), 1 mol of gas occupies 22.4 L can help (see Check step in Example 11.6). A 90.7 L sample of gas at STP would contain about 4 mol; since we started with a little more than 2 mol KClO3, and since 2 mol KClO3 forms 3 mol O2, an answer that corresponds to about 4 mol O2 is reasonable. CHECK Check your answer. Are the units correct? Does the answer make physical sense?

Continued SKILLBUILDER Gases in Chemical Reactions In this reaction, 4.58 L of O2 were formed at 745 mm Hg and 308 K. How many grams of Ag2O decomposed? 2 Ag2O(s)  4 Ag(s) + O2 (g) Answer: 82.3 g For More Practice Problems 89, 90, 91, 92, 93, 94.

One mole of gas occupies 22.4L at STP
MOLAR VOLUME Experimentally determined that One mole of gas occupies 22.4L at STP V = nRT/P = 22.4 L If the mass and volume of a gas at STP are known, we can calculate the molar mass. Standard temperature = K or 0.000C Standard pressure = 1 atm

Conversion factor 22.4 L = 1 mole If conditions are not at STP change them to STP.

Example 11.12 Using Molar Volume in Calculations
How many grams of water form when 1.24 L of H2 gas at STP completely reacts with O2? 2 H2(g) + O2(g)  2 H2O(g) SORT You are given the volume of a reactant at STP and asked to find the mass of the product formed. STRATEGIZE In the solution map, use the molar volume to convert from volume H2 to mol H2. Then use the stoichiometric relationship to convert to mol H2O and finally the molar mass of H2O to get to mass H2O. GIVEN: 1.24 L H2 FIND: g H2O SOLUTION MAP RELATIONSHIPS USED 1 mol = 22.4 L (molar volume at STP, presented in this section) 2 mol H2 : 2 mol H2O (from balanced equation given in problem) 18.02 g H2O = 1 mol H2O (molar mass of H2O)

Continued SOLVE Begin with the volume of H2 and follow the solution map to arrive at mass H2O in grams. SOLUTION The answer has the correct units, g H2O. The answer is reasonable because 1.24 L of a gas is about mol of reactant (at STP) and 1 g H2O is about 0.05 mol product (1 g/18 g/mol ≈ 0.05 mol). Since the reaction produces 1 mol H2O to 1 mol H2, we expect the number of moles of H2O produced to be equal to the number of moles of H2 that react. CHECK Check your answer. Are the units correct? Does the answer make physical sense?

Continued SKILLBUILDER Using Molar Volume in Calculations How many liters of oxygen (at STP) are required to form 10.5 g of H2O? 2 H2(g) + O2(g)  2 H2O(g) Answer: 6.53 L O2 For More Practice Problems 95, 96, 97, 98.

How many moles O2 will occupy a volume of 1.75L at STP?
Liters  moles 1.75 L___1 mol = mol O2 22.4 L

What volume will each of the following occupy at STP?
6.022 X 1023 molecules of CO2 Molecules  moles  volume 2.5 mol CH4 Mol  vol 12.5 g Oxygen g  mol  vol

Density of Gases Density is measured in g/L.

What volume at STP of oxygen can be formed from 0
What volume at STP of oxygen can be formed from mol of potassium chlorate? Balanced chemical equation 2KClO3(s)  2KClO(s) O2(g) Starting with mol potassium chlorate Going to volume of oxygen mol KClO3  mol O2  L O2 Use mole ratio to go to mol O2 Use molar volume to go to liters of O2

Consider the combustion reaction of propane with oxygen,
A) How many liters of oxygen are required to react with 7.2 L of propane? Both gases are at STP. B) How many grams of CO2 will be produced from 35 L of C3H8 if both gases are at STP?

Volume to Volume Calculations
Can make the following assumptions when in the gaseous state at STP: H2(g) Cl2(g)  2 HCl(g) 1 mol 1 mol mol 22.4 L L x 22.4 L 1 volume 1 volume volumes For reacting ases at constant temperature and pressure, volume-volume relationships are the same as mole-mole relationships.

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