Presentation on theme: "Chapter 13: Temperature and Ideal Gas"— Presentation transcript:
1Chapter 13: Temperature and Ideal Gas What is Temperature?Temperature ScalesThermal ExpansionMolecular Picture of a GasThe Ideal Gas LawKinetic Theory of Ideal GasesChemical Reaction RatesCollisions Between Molecules
2§13.1 TemperatureHeat is the flow of energy due to a temperature difference. Heat always flows from objects at high temperature to objects at low temperature.When two objects have the same temperature, they are in thermal equilibrium.
3The Zeroth Law of Thermodynamics: If two objects are each in thermal equilibrium with a third object, then the two objects are in thermal equilibrium with each other.
4§13.2 Temperature Scales (*) Values given at 1 atmosphere of pressure. Absolute or Kelvin scaleFahrenheit scaleCelsius scaleWater boils*K212 F100 CWater freezes*K32 F0 CAbsolute zero0 KFCCould incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 13, Questions 1, 2, and 3.(*) Values given at 1 atmosphere of pressure.
5The temperature scales are related by: Fahrenheit/CelsiusAbsolute/Celsius
6Example (text problem 13.3): (a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees?(b) At what temperature (if any) does the numerical value of Kelvin equal the numerical value of Fahrenheit degrees?
7§13.3 Thermal Expansion of Solids and Liquids Most objects expand when their temperature increases.Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 13, Questions 4, 5, 6, 7, and 17.
8An object’s length after its temperature has changed is is the coefficient of thermal expansionwhere T=T-T0 and L0 is the length of the object at a temperature T0.
9Example (text problem 13.84): An iron bridge girder (Y = 2.01011 N/m2) is constrained between two rock faces whose spacing doesn’t change. At 20.0 C the girder is relaxed. How large a stress develops in the iron if the sun heats the girder to 40.0 C?Using Hooke’s Law:
10How does the area of an object change when its temperature changes? L0L0+LThe blue square has an area of L02.With a temperature change T each side of the square will have a length change of L = TL0.
12The fractional change in volume due to a temperature change is: For solids =3
13§13.4 Molecular Picture of a Gas The number density of particles is N/V where N is the total number of particles contained in a volume V.If a sample contains a single element, the number of particles in the sample is N = M/m. N is the total mass of the sample (M) divided by the mass per particle (m).
14One mole of a substance contains the same number of particles as there are atoms in 12 grams of 12C. The number of atoms in 12 grams of 12C is Avogadro’s number.
15A carbon-12 atom by definition has a mass of exactly 12 atomic mass units (12 u). This is the conversion factor between the atomic mass unit and kg (1 u = 1.6610-27 kg). NA and the mole are defined so that a 1 gram sample of a substance with an atomic mass of 1 u contains exactly NA particles.Problem
16Example (text problem 13.39): Air at room temperature and atmospheric pressure has a mass density of 1.2 kg/m3. The average molecular mass of air is 29.0 u. How many air molecules are there in 1.0 cm3 of air?The total mass of air in the given volume is:
18§13.5 Absolute Temperature and the Ideal Gas Law Experiments done on dilute gases (a gas where interactions between molecules can be ignored) show that:For constant pressureCharles’ LawCould incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 13, Questions 8, 9, 10, 11, 12, 13, 14, and 15.For constant volumeGay-Lussac’s Law
19For constant temperature Boyle’s LawFor constant pressure and temperatureAvogadro’s Law
20Putting all of these statements together gives the ideal gas law (microscopic form): k = 1.3810-23 J/K is Boltzmann’s constantThe ideal gas law can also be written as (macroscopic form):R = NAk= 8.31 J/K/mole is the universal gas constant and n is the number of moles.
21Example (text problem 13. 41): A cylinder in a car engine takes Vi = 4 Example (text problem 13.41): A cylinder in a car engine takes Vi = 4.5010-2 m3 of air into the chamber at 30 C and at atmospheric pressure. The piston then compresses the air to one-ninth of the original volume and to 20.0 times the original pressure. What is the new temperature of the air?Here, Vf = Vi/9, Pf = 20.0Pi, and Ti = 30 C = 303 K.The ideal gas law holds for each set of parameters (before compression and after compression).
22The final temperature is Example continued:Take the ratio:The final temperature isThe final temperature is 673 K = 400 C.
23§13.6 Kinetic Theory of the Ideal Gas An ideal gas is a dilute gas where the particles act as point particles with no interactions except for elastic collisions.Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 13, Questions 16 and 18.
24Gas particles have random motions Gas particles have random motions. Each time a particle collides with the walls of its container there is a force exerted on the wall. The force per unit area on the wall is equal to the pressure in the gas.The pressure will depend on:The number of gas particlesFrequency of collisions with the wallsAmount of momentum transferred during each collision
25The pressure in the gas is Where <Ktr> is the average translational kinetic energy of the gas particles; it depends on the temperature of the gas.
26The average kinetic energy also depends on the rms speed of the gas where the rms speed isProblem
27The distribution of speeds in a gas is given by the Maxwell-Boltzmann Distribution.
28Example (text problem 13.60): What is the temperature of an ideal gas whose molecules have an average translational kinetic energy of 3.2010-20 J?
29On the Kelvin scale T = 25 C = 298 K. Element Mass (kg) Example (text problem 13.70): What are the rms speeds of helium atoms, and nitrogen, hydrogen, and oxygen molecules at 25 C?On the Kelvin scale T = 25 C = 298 K.ElementMass (kg)rms speed (m/s)He6.6410-271360H23.32 10-271930N24.64 10-26515O25.32 10-26482In atomic mass units, the masses used are 4u for He, 2u for H2, 14u for N2, and 16u for O2.
30§13.7 Temperature and Reaction Rates For a chemical reaction to proceed, the reactants must have a minimum amount of kinetic energy called activation energy (Ea).Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 13, Question 19.
31Ifthen only molecules in the high speed tail of Maxwell-Boltzmann distribution can react. When this is the situation, the reaction rates are an exponential function of T.
32Example (text problem 13.76): The reaction rate for the hydrolysis of benzoyl-l-arginine amide by trypsin at 10.0 C is times faster than at 5.0 C. Assuming that the reaction rate is exponential, what is the activation energy?where T1 = 10.0 C = 283 K and T2 = 5 C = 278 K; and r1 = r2.The ratio of the reaction rates is
33Solving for the activation energy gives: Example continued:Solving for the activation energy gives:
34§13.8 Collisions Between Gas Molecules On average, a gas particle will be able to travel a distanceCould incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 13, Question 20.before colliding with another particle. This is the mean free path. The quantity d2 is the cross-sectional area of the particle.
35After a collision, the molecules involved will have their direction of travel changed. Successive collisions produce a random walk trajectory.
36Substances will move from areas of high concentration to areas of lower concentration. This process is called diffusion.In a time t, the rms displacement in one direction is:D is the diffusion constant (see table 13.3).
37Example (text problem 13.81): Estimate the time it takes a sucrose molecule to move 5.00 mm in one direction by diffusion in water. Assume there is no current in the water.Solve for t
38Summary Definition of Temperature Temperature Scales (Celsius, Fahrenheit, Absolute)Thermal ExpansionOrigin of Pressure in a GasIdeal Gas LawExponential Reaction RatesMean Free Path