General Physics: Ideal Gas Law and Kinetic Theory 1

Temperature and Heat Temperature: determined by the average KE of molecules. Heat: a way to transfer energy other than work. Heat flows from high temperature to low temperature. Internal Energy: total energy of the molecules of a body.

Temperature and Heat Total Internal Energy= Kinetic Energy of Molecules + Potential Energy of Molecules

Temperature Scales Measuring temperature. Thermometer: using expansion of the volume of a column of liquid. Scale: set of numbers associated to the value of the temperature.

Scales of Temperature Celsius Kelvin 100 373 Boiling point of water 273 Freezing point of water -273 Absolute zero

Temperature Scales Celsius-Kelvin K = C + 273.15

Temperature Scales Fahrenheit C = (5/9)(F-32)

Aside: The Periodic Table 05 1

The Periodic Table Explained ? Look carefully proton neutron electron 05 1

Energy vs Mass E = mc2 He (m=4.0026 u) O (M=15.9995 u) 4 x He = 16.01 u Mass difference = 0.01 u = binding energy So energy is the same as mass somehow ?? E = mc2 06 1

The Mole, Avogadro's Number, and Molecular Mass

Atomic Mass Unit, U By international agreement, the reference element is chosen to be the most abundant type of carbon, called carbon-12, and its atomic mass is defined to be exactly twelve atomic mass units, or 12 u.

Molecular Mass The molecular mass of a molecule is the sum of the atomic masses of its atoms. For instance, hydrogen and oxygen have atomic masses of 1.007 94 u and 15.9994 u, respectively. The molecular mass of a water molecule (H2O) is: 2(1.007 94 u) + 15.9994 u = 18.0153 u.

Avogadro's Number NA The number of atoms per mole is known as Avogadro's number NA, after the Italian scientist Amedeo Avogadro (1776–1856): Molar Volume Vμ = 22.42 l: Volume one mole of ideal gas will occupy at STP (To=273.15K, po = 1atm)

Number of Moles, n The number of moles n contained in any sample is the number of particles N in the sample divided by the number of particles per mole NA (Avogadro's number): The number of moles contained in a sample can also be found from its mass.

Molecular Picture of Gas Gas is made up of many individual molecules Number density is number of molecules/volume: N/V = r/m r is the mass density m is the mass for one molecule Number of moles: n = N / NA NA = Avogadro’s Number = 6.022x1023 mole-1 1 mole of a substance = molecular mass in  in grams e.g., 1 mole of N2 has mass of 2x14=28 grams

Atomic Act I Which contains the most molecules ? 1. A mole of water (H2O) 2. A mole of oxygen gas (O2) 3. Same H2O O2 correct 12

Atomic Act II Which contains the most atoms ? 1. A mole of water (H2O) 2. A mole of oxygen gas (O2) 3. Same H2O (3 atoms) O2 (2 atoms) correct 15

Atomic Act III Which weighs the most ? 1. A mole of water (H2O) 2. A mole of oxygen gas (O2) 3. Same correct H2O (M = 16 + 1 + 1) O2 (M = 16 + 16) 17

Thermodynamic variables. Pressure. Pressure is an important thermodynamic variable. Pressure is defined as the force per unit area: p = F/A The SI unit is pressure is the Pascal: 1 Pa = 1 N/m2. Another common unit is the atm (atmospheric pressure) which is the pressure exerted by the atmosphere on us (1 atm = 1.013 x 105 N/m2). A pressure of 1 atm will push a mercury column up by 76 cm.

Physical Characteristics of Gases Physical Characteristics Typical Units Volume, V liters (L) Pressure, P atmosphere (1 atm = 1.015x105 N/m2) Temperature, T Kelvin (K) Number of atoms or molecules, n mole (1 mol = 6.022x1023 atoms or molecules)

“Father of Modern Chemistry” Chemist & Natural Philosopher Boyle’s Law Pressure and volume are inversely related at constant temperature. PV = constant As one goes up, the other goes down. P1V1 = P2V2 “Father of Modern Chemistry” Robert Boyle Chemist & Natural Philosopher Listmore, Ireland January 25, 1627 – December 30, 1690

Boyle’s Law: P1V1 = P2V2

Boyle’s Law: P1V1 = P2V2

Gas Laws, Absolute Temperature, and the Kelvin Temperature Scale When the temperature and mass of an ideal gas is held constant, pV is constant, Boyle’s law:

Calculations with Boyle’s Law

Calculation with Boyle’s Law Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mm Hg after its pressure is changed to 2200 mm Hg at constant T? 1. Set up a data table: Conditions 1 Conditions 2 P1 = 550 mm Hg P2 = 2200 mm Hg V1 = 8.0 L V2 = ?

Calculation with Boyle’s Law (Continued) 2. When pressure increases, volume decreases. Solve Boyle’s Law for V2: P1V1 = P2V2 V2 = V1 x P1 P2 V2 = 8.0 L x 550 mm Hg = 2.0 L 2200 mm Hg pressure ratio decreases volume

Gas Law Problems BOYLE’S LAW P V A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa P V WORK: P1V1 = P2V2 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL

Learning Check 1) 200. mm Hg 2) 400. mm Hg 3) 1200 mm Hg A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant). 1) 200. mm Hg 2) 400. mm Hg 3) 1200 mm Hg

Solution 1) 200. mm Hg Data table Conditions 1 Conditions 2 P1 = 600. mm Hg P2 = ??? V1 = 12.0 L V2 = 36.0 L P2 = P1 x V1 V2 600. mm Hg x 12.0 L = 200. mm Hg 36.0 L

PV Calculation (Boyle’s Law) A quantity of gas has a volume of 120 dm3 when confined under a pressure of 93.3 kPa at a temperature of 20 oC. At what pressure will the volume of the gas be 30 dm3 at 20 oC? P1 x V1 = P2 x V2 (93.3 kPa) x (120 dm3) = (P2) x (30 dm3) P2 = 373.2 kPa

Jacques-Alexandre Charles Mathematician, Physicist, Inventor Charles’ Law Volume of a gas varies directly with the absolute temperature at constant pressure. V = KT V1 / T1 = V2 / T2 Jacques-Alexandre Charles Mathematician, Physicist, Inventor Beaugency, France November 12, 1746 – April 7, 1823

Charles’s Law

Charles’ Law: V1/T1 = V2/T2

Charles’ Law: V1/T1 = V2/T2

Gas Laws, Absolute Temperature, and the Kelvin Temperature Scale When the pressure is held constant, V is proportional to T, Charles’ Law

EXAMPLE 6.15 Charles’s Law: Temperature-Volume Relationships A balloon indoors, where the temperature is 27 °C, has a volume of 2.00 L. What would its volume be (a) in a hot room where the temperature is 47 °C, and (b) outdoors, where the temperature is –23 ºC? (Assume no change in pressure in either case.)

Gas Law Problems CHARLES’ LAW T V A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK: P1V1T2 = P2V2T1 (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3

Charles’s Law: Temperature-Volume Relationships continued A sample of oxygen gas occupies a volume of 2.10 L at 25 °C. What volume will this sample occupy at 150 °C? (Assume no change in pressure.) b. A sample of hydrogen occupies 692 L at 602 °C. If the pressure is held constant, what volume will the gas occupy after being cooled to 23 °C? At what Celsius temperature will the initial volume of oxygen in Exercise 6.15A occupy 0.750 L? (Assume no change in pressure.)

Gas Law Problems GAY-LUSSAC’S LAW P T A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? P T WORK: P1T2 = P2T1 (765 torr)T2 = (560. torr)(309K) T2 = 226 K = -47°C

Ideal Gas Laws Combining gives the ideal gas law: At STP p2=101300 Pa, Vμ = 22.24 l, V2 = nVμ, T2 = 273.15 K R=8.31 J/(mole x K) universal gas constant PV/T = nR or

The Ideal Gas Law P V = n R T (Equation of State) P V = N kB T n = number of moles R = ideal gas constant 8.31 J/(molK) P V = N kB T P = pressure in N/m2 (or Pascals) V = volume in m3 N = number of molecules T = absolute temperature in K k B = R/NA Boltzmann’s constant = 1.38 x 10-23 J/K Note: P V has units of N-m or J (energy!) 20

Ideal Gas Law ACT II PV = nRT You inflate the tires of your car so the pressure is 3.0 barr when the air inside the tires is at 20 degrees C. After driving on the highway for a while, the air inside the tires heats up to 38 C. Which number is closest to the new air pressure (assume V = const.)? 1) 1.6 barr 2) 3.2 barr 3) 5.7 barr Careful, you need to use the temperature in K P = P0 (38+273)/(20+273) 23

Ideal Gas Law: ACT 1 pV = nRT A piston has volume 20 ml, and pressure of 30 psi. If the volume is decreased to 10 ml, what is the new pressure? (Assume T is constant.) 1) 60 2) 30 3) 15 Change this to the pressure change with temperature, but give the temperature in C. V=20 P=30 V=10 P=?? When n and T are constant, pV is constant (Boyle’s Law) 26

Balloon ACT 1 What happens to the pressure of the air inside a hot-air balloon when the air is heated? (Assume V is constant) 1) Increases 2) Same 3) Decreases Balloon is still open to atmospheric pressure, so it stays at 1 atm 30

Balloon ACT 2 What happens to the buoyant force on the balloon when the air is heated? (Assume V remains constant) 1) Increases 2) Same 3) Decreases FB = r V g r is density of outside air! 32

Balloon ACT 3 What happens to the number of air molecules inside the balloon when the air is heated? (Assume V remains constant) 1) Increases 2) Same 3) Decreases PV = NkT P and V are constant. If T increases N decreases. 34

Gas Laws In terms of the ideal gas law, explain briefly how a hot air balloon works. Hot air has less mole density than cool air. So less hot air is required in order to achieve the same pressure as cool air. This makes the density of hot air less allowing it to float. When temperature increases the volume of the gas increases, thus reducing the density of the gas making it lighter that then surrounding air, which causes the balloon to rise. In regards to balloons, you should inhale helium again sometime, we all liked that. Note! this is not a pressure effect, it is a density effect. As T increases, the density decreases the balloon then floats due to Archimedes principle. The pressure remains constant! 36

Gas Law Problems COMBINED GAS LAW P T V V1 = 7.84 cm3 P1 = 71.8 kPa A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW GIVEN: V1 = 7.84 cm3 P1 = 71.8 kPa T1 = 25°C = 298 K V2 = ? P2 = 101.325 kPa T2 = 273 K P T V WORK: P1V1T2 = P2V2T1 (71.8 kPa)(7.84 cm3)(273 K) =(101.325 kPa) V2 (298 K) V2 = 5.09 cm3

Ideal Gas Law IDEAL GAS LAW P = ? atm n = 0.412 mol T = 16°C = 289 K Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm

Ideal Gas Law IDEAL GAS LAW V = ? n = 85 g T = 25°C = 298 K Find the volume of 85 g of O2 at 25°C and 104.5 kPa. IDEAL GAS LAW GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm3kPa/molK WORK: 85 g 1 mol = 2.7 mol 32.00 g = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K V = 64 dm3

Summary Ideal Gas Law PV = n R T Kinetic Theory of Monatomic Ideal Gas P = pressure in N/m2 (or Pascals) V = volume in m3 n = # moles R = 8.31 J/ (K mole) T = Temperature (K) Kinetic Theory of Monatomic Ideal Gas <Ktr> = 3/2 kB T 50

Summary Ideal gas law: Absolute zero is –273.15°C. Celsius–Kelvin conversion: