Chapter 13: Temperature and Ideal Gas What is Temperature? Temperature Scales Thermal Expansion Molecular Picture of a Gas The Ideal Gas Law Kinetic Theory of Ideal Gases Chemical Reaction Rates Collisions Between Molecules 1
13.1 Temperature Heat 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.
The 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. 3
13. Temperature Scales Absolute or Kelvin scale Fahrenheit scale Celsius scale Water boils * 373.15 K 1 F 100 C Water freezes * 73.15 K 3 F 0 C Absolute zero 0 K -459.67 F -73.15 C (*) Values given at 1 atmosphere of pressure. 4
The temperature scales are related by: Fahrenheit/ Celsius T (.8 F/ C) + 3 F F 1 TC Absolute/ Celsius T T C + 73.15 5
Example (text problem 13.3): (a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees? T F 1.8T T C C + 3 40 C T C (b) At what temperature (if any) does the numerical value of Kelvin equal the numerical value of Fahrenheit degrees? T T F F 1.8T 1.8 1.8 C + 3 ( T 73) ( T 73) F 574 F + 3 + 3 6
13.3 Thermal Expansion of Solids and Liquids Most objects expand when their temperature increases. 7
An object s length after its temperature has changed is ( 1+ ) L 0 L αδt α is the coefficient of thermal expansion where ΔTT-T 0 and L 0 is the length of the object at a temperature T 0. 8
How does the area of an object change when its temperature changes? The blue square has an area of L 0. L 0 L 0 +ΔL With a temperature change ΔT each side of the square will have a length change of ΔL αδtl 0. 9
10 ( )( ) ( ) T A A T A TL L L T TL L TL L TL L Δ Δ Δ + Δ + Δ + Δ + Δ + Δ + α α α α α α α 1 A new area 0 0 0 0 0 0 0 0 0 0 0 The fractional change in area is:
The fractional change in volume due to a temperature change is: ΔV V 0 βδt For solids β3α 11
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). 1
One mole of a substance contains the same number of particles as there are atoms in 1 grams of 1 C. The number of atoms in 1 grams of 1 C is Avogadro s number. N 6.0 10 A 3 mol -1 13
A carbon-1 atom by definition has a mass of exactly 1 atomic mass units (1 u). This is the conversion factor between the atomic mass unit and kg (1 u 1.66 10-7 kg). N A and the mole are defined so that a 1 gram sample of a substance with an atomic mass of 1 u contains exactly N A particles. 14
Example (text problem 13.39): Air at room temperature and atmospheric pressure has a mass density of 1. kg/m 3. The average molecular mass of air is 9.0 u. How many air molecules are there in 1.0 cm 3 of air? number of particles 3 total mass of air in 1.0 cm average mass per air molecule The total mass of air in the given volume is: m ρv 1. kg 3 m 1. 10 1.0 cm 1 6 kg 3 1m 100 cm 3 15
Example continued: number of particles 3 total mass of air in 1.0 cm average mass per air molecule 1. 10 6 kg ( )( 7 9.0 u/particle 1.66 10 kg/u).5 10 19 particles 16
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 pressure V T Charles Law For constant volume P T Gay-Lussac s Law 17
For constant temperature P 1 V Boyle s Law For constant pressure and temperature V N Avogadro s Law 18
Putting all of these statements together gives the ideal gas law (microscopic form): PV NkT k 1.38 10-3 J/K is Boltzmann s constant The ideal gas law can also be written as (macroscopic form): PV nrt R N A k 8.31 J/K/mole is the universal gas constant and n is the number of moles. 19
Example (text problem 13.41): A cylinder in a car engine takes V i 4.50 10 - m 3 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 0.0 times the original pressure. What is the new temperature of the air? Here, V f V i /9, P f 0.0P i, and T i 30 C 303 K. P V i i P V f f NkT i NkT f The ideal gas law holds for each set of parameters (before compression and after compression). 0
1 Example continued: Take the ratio: i f i f i i f f T T NkT NkT PV P V The final temperature is ( ) K 673 303 K 9 0.0 i i i i i i f i f f V V P P T V V P P T The final temperature is 673 K 400 C.
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.
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 particles Frequency of collisions with the walls Amount of momentum transferred during each collision 3
The pressure in the gas is P 3 N V K tr Where <K tr > is the average translational kinetic energy of the gas particles; it depends on the temperature of the gas. K tr 3 kt 4
The average kinetic energy also depends on the rms speed of the gas where the rms speed is 1 K tr m v 1 mv rms K tr v rms 3 kt 1 3kT m mv rms 5
The distribution of speeds in a gas is given by the Maxwell- Boltzmann Distribution. 6
Example (text problem 13.60): What is the temperature of an ideal gas whose molecules have an average translational kinetic energy of 3.0 10-0 J? K tr T 3 K 3k tr kt 1550 K 7
Example (text problem 13.70): What are the rms speeds of helium atoms, and nitrogen, hydrogen, and oxygen molecules at 5 C? v rms 3kT m On the Kelvin scale T 5 C 98 K. Element Mass (kg) rms speed (m/s) He 6.64 10-7 1360 H 3.3 10-7 1930 N 4.64 10-6 515 O 5.3 10-6 48 8
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 (E a ). 9
If 3 E a >> kt then 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. reaction rates e E a kt 30
Example (text problem 13.76): The reaction rate for the hydrolysis of benzoyl-l-arginine amide by trypsin at 10.0 C is 1.878 times faster than at 5.0 C. Assuming that the reaction rate is exponential, what is the activation energy? r 1 r e e E a E a kt kt 1 where T 1 10.0 C 83 K and T 5 C 78 K; and r 1.878 r. The ratio of the reaction rates is r r 1 exp Ea kt 1 + Ea kt 31
Example continued: Solving for the activation energy gives: E a r 1 k ln r 1 1 T T1 ( 3 1.38 10 J/K) ln( 1.878) 1 78 K 1 83 K 1.37 10 19 J 3
13.8 Collisions Between Gas Molecules On average, a gas particle will be able to travel a distance Λ 1 ( N V ) πd / before colliding with another particle. This is the mean free path. The quantity πd is the cross-sectional area of the particle. 33
After a collision, the molecules involved will have their direction of travel changed. Successive collisions produce a random walk trajectory. 34
Substances 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: xrms Dt D is the diffusion constant (see table 13.3). 35
Example (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. xrms Dt Solve for t t x D ( 3 ) 5.00 10 m ( 10 10 m /s) rms 5.0 5000 s 36
Summary Definition of Temperature Temperature Scales (Celsius, Fahrenheit, Absolute) Thermal Expansion Origin of Pressure in a Gas Ideal Gas Law Exponential Reaction Rates Mean Free Path 37