# Chapter 10: Gases - Houston Community College

Lecture Presentation Chapter 10 Gases 2015 Pearson Education, Inc. James F. Kirby Quinnipiac University Hamden, CT Characteristics of Gases Physical properties of gases are all similar. Composed mainly of nonmetallic elements with simple formulas and low molar masses. Unlike liquids and solids, gases expand to fill their containers. are highly compressible. have extremely low densities.

Two or more gases form a homogeneous mixture. Gases 2015 Pearson Education, Inc. 3 Elements that exist as gases at 250C and 1 atm Gases 2015 Pearson Education, Inc. Properties Which Define the State of a Gas Sample 1) 2) 3) 4) Temperature

Pressure Volume Amount of gas, usually expressed as number of moles Having already discussed three of these, we need to define pressure. Gases 2015 Pearson Education, Inc. Pressure Pressure is the amount of force applied to an area: F P= A Atmospheric pressure is the

weight of air per unit of area. 10 miles 0.2 atm 4 miles 0.5 atm Sea level 2015 Pearson Education, Inc. Composition of Earth's atmosphere by volume 1 atm

Gases Units of Pressure Pascals: 1 Pa = 1 N/m2 (SI units) Bar: 1 bar = 105 Pa = 100 kPa mm Hg or torr: These units are literally the difference in the heights measured in mm of two connected columns of mercury, as in the barometer in the figure. Normal atmospheric pressure at sea level is referred to as standard atmospheric pressure It is equal to: 1.00 atm = 760 mm Hg (760 torr) = 101.325 kP 760 mm Hg

Barometer Gases 2015 Pearson Education, Inc. Pressure Conversions: An Example The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals. 760 torr 3 = 1.9 10 torr 2.5 atm 1 atm 101,325 Pa

5 = 2.5 10 Pa 2.5 atm 1 atm Gases 2015 Pearson Education, Inc. 7 Manometer The manometer is used to measure the difference in pressure between

atmospheric pressure and that of a gas in a vessel. (The barometer seen on the last slide is used to measure the pressure in the atmosphere at any given time.) Gases 2015 Pearson Education, Inc. Boyles Law (Pressure-Volume Relationship of a Gas) The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure. P a 1/V PV = k1 k1 is a constant for a given sample of air at a

specific temperature. Gases 2015 Pearson Education, Inc. Mathematical Relationships of Boyles Law PV = a constant This means, if we compare two conditions: P1V1 = P2V2. Also, if we make a graph of V vs. P, it will not be linear. However, a graph of V vs. 1/P will result in a linear relationship! Gases 2015 Pearson Education, Inc. EXERCISE! A sample of helium gas occupies 12.4 L at 23 oC and

0.956 atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant? P1 V1 = P2 V2 P1 = 0.956 atm, V1 = 12.4L, P2 = 1.20 atm, V2 = ? V2 = P1 x V1 P2 = 0.956 atm x 12.4L/1.20 atm = 9.88L 2015 Pearson Education, Inc. Gases Charless Law (Volume-Temperature Relationship of a Gas) The volume of a fixed amount of gas at constant pressure is

directly proportional to its absolute temperature. As T increases V increases or vice versa VaT V = K2 T V/T = k2 k2 is a proportionality constant of a gas at specific pressure. K = oC + 273 (0 K is called absolute zero (-273.15C) 2015 Pearson Education, Inc. Gases Mathematical Relationships of Charless Law V = constant T This means, if we compare two conditions: V1/T1 = V2/T2.

Also, if we make a graph of V vs. T, it will be linear. Gases 2015 Pearson Education, Inc. EXERCISE! Suppose a balloon containing 1.30 L of air at 24.7 0C is placed into a beaker containing liquid nitrogen at 78.5 0C. What will the volume of the sample of air become (at constant pressure)? V1 V2 = T1 T2 or V2 = V1 T2/T1 = (1.30 L) (-78.5 + 273)K (24.7+273)K

= 0.849 L Gases 2015 Pearson Education, Inc. Gay-Lussac's law (Volume-Temperature Relationship of a Gas) Pressure and Temperature (in Kelvin) of a gas are directly proportional (at constant V and n). PaT P = k3 T P/T = k3 P1 / T1 = P2 / T2 k3 is a proportionality constant of a given gas. Gases 2015 Pearson Education, Inc.

Avogadros Law (Volume-Moles Relationship of a Gas) The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Also, at STP, one mole of gas occupies 22.4 L. Mathematically: V = constant n, or V1/n1 = V2/n2 Gases 2015 Pearson Education, Inc. EXERCISE! If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure? V1 / n1 = V2 / n2 or V2 = V1 n2 / n1

= (89.0L) (2.10 mol)/(2.45 mol) = 76.3 L Gases 2015 Pearson Education, Inc. 17 Ideal-Gas Equation So far weve seen that V 1/P (Boyles law). V T (Charless law). V n (Avogadros law). Combining these, we get nT V P

Finally, to make it an equality, we use a constant of proportionality (R) and reorganize; this gives the Ideal-Gas Equation: PV = nRT. where R = 0.08206 Latm/molK, the universal gas constant 2015 Pearson Education, Inc. Gases Ideal Gas A hypothetical gas whose molecules occupy negligible space and have no interactions between them but with the walls of the container, and that consequently obeys the gas laws exactly. No known gas is an ideal gas, but actual gases approximate this behavior at relatively low pressure and high temperature. Gases 2015 Pearson Education, Inc.

EXERCISE! An automobile tire at 23 0C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire. PV = nRT or n = PV/RT n = (3.18 atm)(25.0 L) / (0.08206)(23+273)K = 3.27 mol (R = 0.08206 Latm/molK) Gases 2015 Pearson Education, Inc. EXERCISE! What is the pressure in a 304.0 L tank that contains 5.670 kg of helium at 25 0C? convert kg of helium into moles of helium before using the

ideal gas law (n = moles = mass/molar mass). n = (5.6701000)/ 4.003 = 1416.44 PV =nRT or P = nRT/V = 1416.44(0.0821)(25+273)/304.0 = 114 atm Gases 2015 Pearson Education, Inc. EXERCISE! At what temperature (in 0C) does 121 mL of CO2 at 27 0C and 1.05 atm occupy a volume of 293 mL at a pressure of 1.40 atm? compare two conditions: PV = nRT or PV / T = nR (constant) or P1V1 / T 1 = P2V2 / T2 or T2 = P2V2 T1 / P1V1

= (1.40 atm) (293mL)(27+273)K (1.05 atm) (121 mL) = 969K or 0C = K-273 = 696 0C Gases 2015 Pearson Education, Inc. Molar Volume of an Ideal Gas For 1 mole of an ideal gas at 0 0C and 1 atm, the volume of the gas is 22.42 L. V= nRT P = 1.000 mol 0.08206 L atm/K mol 273.2 K

1.000 atm = 22.42 L The conditions 0 0C and 1 atm are called standard temperature and pressure = STP Therefore, the molar volume is 22.42 L at STP. Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. 2015 Pearson Education, Inc. Gases 24 What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K PV = nRT

nRT V= P P = 1 atm 1 mol HCl n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.0821 V= Latm molK x 273.15 K 1 atm

V = 30.7 L Gases 2015 Pearson Education, Inc. 25 Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR = P = constant T

V P1 P2 = T1 T2 P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K T2 = 1.20 atm x 358 K = 1.48 atm P2 = P1 x 291 K T1 Gases

2015 Pearson Education, Inc. EXERCISE! A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O2 are present? PV = nRT or n = PV/RT n = (1.00 atm)(2.50 L) / (0.08206)(273) n = 0.112 mol g of O2 = 0.112 mol 32.00 g/mol = 3.57 g Gases 2015 Pearson Education, Inc. Density (d) and Molar Mass (M) of a Gas n = m/M and d = m/V

PV = nRT nRT P= V mRT P= MV dRT P= M PM d= RT M=

dRT P g L atm K dRT L mol K g Molar mass = = = P mol atm

d = density of gas (g/L) T = temperature in Kelvin P = pressure of gas in atm R = universal gas constant (0.08206 Latm/molK) Gases 2015 Pearson Education, Inc. Density & Molar Mass of a Gas To recap: One needs to know only the molecular mass, the pressure, and the temperature to calculate the density of a gas. d = MP/RT

Also, if we know the mass, volume, pressure and temperature of a gas, we can find its molar mass. M = mRT/PV Gases 2015 Pearson Education, Inc. What is the density of HCl gas in grams per liter at 700 mmHg and 25 oC? PM d= RT P=700mmHg=700/760atm=0.92atm T= 25 oC=25+273.15K=298.15K d= 0.92 atm x 36.45g/mol

0.0821 Latm molK x 298.15 K =1.37g/L Gases 2015 Pearson Education, Inc. What is the molar mass (g/mol) of 7.10 grams of gas whose volume is 5.40 L at 741 torr and 40 oC? dRT M= P T=313.15K

1.31 M= 7.10 g m = = 1.31 d= V 5.40 L g L P= 741torr=741/760atm=0.975atm g L

x 0.0821 Latm molK x 313.15 K 0.975 atm M = 34.6 g/mol Gases 2015 Pearson Education, Inc. Volume and Chemical Reactions The balanced equation tells us relative amounts of moles in a reaction, whether the compared materials are products or reactants. PV = nRT So, we can relate volume for gases, as well.

For example: use (PV = nRT) for substance A to get moles A; use the mole ratio from the balanced equation to get moles B; and (PV = nRT) for substance B to get volume of B. Gases 2015 Pearson Education, Inc. 32 Gas Stoichiometry What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) g C6H12O6

mol C6H12O6 5.60 g C6H12O6 x 6 mol CO2 1 mol C6H12O6 x = 0.187 mol CO2 180 g C6H12O6 1 mol C6H12O6 V= nRT = P 2015 Pearson Education, Inc.

mol CO2 V CO2 Latm x 310.15 K molK 1.00 atm 0.187 mol x 0.0821 = 4.76 L Gases Gas Stoichiometry The combustion process for methane is CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) If 15.0 moles of methane are reacted, what is the

volume of carbon dioxide (in L) produced at 23.0 oC and 0.985 atm? x 1CO2/1CH4 15 mole CH4 ---------------- 15 mole CO2 nRT V= = P Latm x 296.15 K molK = 369.8 L 0.985 atm 15mol x 0.0821 Gases 2015 Pearson Education, Inc.

Daltons Law of Partial Pressures If two gases that dont react are combined in a container, they act as if they are alone in the container. The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. In other words, Ptotal = p1 + p2 + p3 + Gases 2015 Pearson Education, Inc. Pressure and Mole Fraction Because each gas in a mixture acts as if it is alone, we can relate amount in a mixture to partial pressures: That ratio of moles of a substance to total moles is called the mole fraction, .

The end result is Gases 2015 Pearson Education, Inc. Pi = Xi PT ni mole fraction (Xi) = nT A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm

Xpropane = 0.116 8.24 + 0.421 + 0.116 = 0.0132 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm 2015 Pearson Education, Inc. Gases Effusion & Diffusion Effusion is the escape of gas molecules through a tiny hole into an evacuated space.

Diffusion is the spread of one substance throughout a space or a second substance. Gases 2015 Pearson Education, Inc. Grahams Law Describes Diffusion & Effusion Grahams Law relates the molar mass of two gases to their rate of speed. The lighter gas always has a faster rate of speed. Gases 2015 Pearson Education, Inc. Postulates of the Kinetic Molecular

Theory 1) The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero). 2) The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. 3) The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other. 4) The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin Gases 2 temperature of the gas. KE = (1/2)mv 2015 Pearson Education, Inc.

An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law. We must correct for non-ideal gas behavior when: Pressure of the gas is high. Temperature is low. Under these conditions: Concentration of gas particles is high. Attractive forces become important. Gases 2015 Pearson Education, Inc. Real Gases In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure. Even the same gas will show wildly different behavior under high pressure at different temperatures.

Gases 2015 Pearson Education, Inc. Nitrogen Gas Deviations from Ideal Behavior The assumptions made in the kinetic-molecular model (negligible volume of gas molecules themselves, no attractive forces between gas molecules, etc.) break down at high pressure Gases and/or low temperature. 2015 Pearson Education, Inc. Corrections for Nonideal Behavior The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account.

The corrected ideal-gas equation is known as the van der Waals equation. The additive pressure adjustment is due to the fact that molecules attract and repel each other. For a real gas, the actual observed pressure is lower. The volume adjustment is due to the fact that molecules occupy some space on their own. Therefore the volume occupied by the molecules needs to be subtracted. Gases 2015 Pearson Education, Inc. The van der Waals Equation corrected pressure corrected volume (Pideal ) (Videal ) Gases 2015 Pearson Education, Inc.

Air Pollution Two main sources: Transportation Production of electricity Combustion of petroleum produces CO, CO2, NO, and NO2, along with unburned molecules from petroleum. Gases 2015 Pearson Education, Inc. Nitrogen Oxides (Due to Cars and Trucks) At high temperatures, N2 and O2 react to form

NO, which oxidizes to NO2. The NO2 breaks up into nitric oxide and free oxygen atoms. Oxygen atoms combine with O2 to form ozone (O3). Radiant energy NO2 ( g ) NO(g ) O( g ) O(g ) O2 (g ) O3 (g ) 2015 Pearson Education, Inc.

Gases Concentration for Some Smog Components vs. Time of Day Gases 2015 Pearson Education, Inc. Sulfur Oxides (Due to Burning Coal for Electricity) Sulfur produces SO2 when burned. SO2 oxidizes into SO3, which combines with water droplets in the air to form sulfuric acid. S(in coal) O2 ( g ) SO2 ( g ) 2SO2 ( g ) O2 ( g ) 2SO3 ( g )

SO3 ( g ) H2O( l ) H2SO 4 (aq ) Sulfuric acid is very corrosive and produces acid rain. Use of a scrubber removes SO2 from the exhaust gas when burning coal. 2015 Pearson Education, Inc. Gases A Schematic Diagram of a Scrubber Gases 2015 Pearson Education, Inc.

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