Physics 231 Topic 12: Temperature, Thermal Expansion, and Ideal Gases Alex Brown Nov

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Physics 231 Topic 12: Temperature, Thermal Expansion, and Ideal Gases Alex Brown Nov 18-23 2015 MSU Physics 231 Fall 2015 1

homework 3 rd midterm final Thursday 8-10 pm makeup Friday final 9-11 am MSU Physics 231 Fall 2015 2

Key Concepts: Temperature, Thermal Expansion, and Ideal Gases Temperature and Thermometers Thermal Energy & Temperature Thermal Expansion Coefficient of thermal expansion Ideal Gases State Variables Ideal gas law Kinetic Theory of Gases Kinetic & thermal energy Maxwell distribution Covers chapter 12 in Rex & Wolfson MSU Physics 231 Fall 2015 3

Conversions: T c = T k - 273.15 T f = (9/5)T c + 32 Helium boils at T k =4 MSU Physics 231 Fall 2015 4

Potential Energy Binding Forces Kinetic energy ~ T (temperature) 0 -E min R The curve depends on the material, e.g. E min is different for water and iron R 2 atom/molecules MSU Physics 231 Fall 2015 5

Potential Energy Solid (low T) Kinetic energy ~ T 0 R min R -E min The temperature (and thus kinetic energy) is so small that the atoms/molecules can only oscillate around a fixed position R min MSU Physics 231 Fall 2015 6

Potential Energy Liquid (medium T) Kinetic energy ~ T 0 R min R -E min On average, the atoms/molecules like to stick together but sometimes escape and can travel far. MSU Physics 231 Fall 2015 7

Gas (high T) Potential Energy R min Kinetic energy ~ T 0 -E min R The kinetic energy is much larger than E min and the atoms/molecules move around randomly. MSU Physics 231 Fall 2015 8

What happens if the temperature of a substance is increased? R min =R ave (T=0) Kinetic energy ~ T 0 -E min R T=0: Average distance between atoms/molecules: R min MSU Physics 231 Fall 2015 9

What happens if the temperature of a substance is increased? R min =R ave (T=0) Kinetic energy ~ T R ave (T>0) > R min 0 R -E min T>T o : The average distance between atoms/molecules is larger than R min : the substance expands MSU Physics 231 Fall 2015 10

length Thermal expansion L= L o T L surface A = A o T = 2 volume V = V o T = 3 : coefficient of linear expansion different for each material L 0 Some examples: = 24x10-6 1/K Aluminum = 1.2x10-4 1/K Alcohol T=T 0 T=T 0 +T MSU Physics 231 Fall 2015 11

MSU Physics 231 Fall 2015 12

A Heated Ring A metal ring is heated. What is true: a) The inside and outside radii become larger b) The inside radius becomes larger, the outside radius becomes smaller c) The inside radius becomes smaller, the outside radius becomes larger d) The inside and outside radii become smaller PHY MSU 231 13 Physics 231 Fall 2015 13

Demo: Bimetallic Strips top bottom Application: contact in a refrigerator top < bottom if the temperature increases, The strip curls upward, makes contact and switches on the cooling. MSU Physics 231 Fall 2015 14

Water: a special case Coef. of expansion is negative: If T drops the volume becomes larger Coef. Of expansion is positive: if T drops the volume becomes smaller Below this ice is formed (it floats on water) MSU Physics 231 Fall 2015 15

Ice (g/cm 3 ) 1 liquid Phase transformation 0.917 ice Ice takes a larger volume than water! A frozen bottle of water might explode MSU Physics 231 Fall 2015 16

Thermal equilibrium Thermal contact Low temperature Low kinetic energy Particles move slowly Transfer of kinetic energy High temperature High kinetic energy Particles move fast Thermal equilibrium: temperature is the same everywhere MSU Physics 231 Fall 2015 17

Zeroth law of thermodynamics If objects A and B are both in thermal equilibrium with an object C, than A and B are also in thermal equilibrium. There is no transfer of energy between A, B and C MSU Physics 231 Fall 2015 18

Ideal Gas: properties Collection of atoms/molecules that Exert no force upon each other The energy of a system of two atoms/molecules cannot be reduced by bringing them close to each other Take no volume The volume taken by the atoms/molecules is negligible compared to the volume they are sitting in MSU Physics 231 Fall 2015 19

Potential Energy R min 0 -E min R Ideal gas: we are neglecting the potential energy between The atoms/molecules Potential Energy 0 Kinetic energy R MSU Physics 231 Fall 2015 20

Properties of gases V = volume P = pressure T = temperature in K (Kelvin) n = number of moles Example balloon MSU Physics 231 Fall 2015 21

Molecular mass m mass of one atom (or molecule) N A 6.0210 23 Avagodro's numbers M molar m N A molecular (molar) mass for example M molar (carbon) 12.0 g 0.0120 kg m (carbon) 0.012 6.0210 23 2.0010-26 kg MSU Physics 231 Fall 2015 22

Number of electrons Z Name X A molar mass in grams MSU Physics 231 Fall 2015 23

Weight of 1 mol of atoms 1 mol of atoms weighs A grams (A is the molar mass) Examples: 1 mol of Hydrogen weighs 1.0 g 1 mol of Carbon weighs 12.0 g 1 mol of Oxygen weighs 16.0 g 1 mol of Zinc weighs 65.4 g What about molecules? H 2 O 1 mol of water molecules: 2 x 1.0 g (due to Hydrogen) 1 x 16.0 g (due to Oxygen) Total: 18.0 g MSU Physics 231 Fall 2015 24

Number of atoms and moles N total number of atoms (or molecules) n N N A number of moles or N n N A so one mole contains N A atoms (or molecules) M m N total mass of all atoms (or molecules) M n M molar MSU Physics 231 Fall 2015 25

Example A cube of Silicon (molar mass 28.1 g) is 250 g. A) How much Silicon atoms are in the cube? Total number of moles n = M / M molar = 250/28.1 = 8.90 N = n N A = (8.9) (6.02x10 23 ) = 5.4x10 24 atoms B) What would be the mass for the same number of gold atoms (molar mass 197 g) M = n M molar = (8.90) (197 g) = 1750 g MSU Physics 231 Fall 2015 26

Question 1) 1 mol of CO 2 has a larger mass than 1 mol of CH 2 2) 1 mol of CO 2 contains more molecules than 1 mol of CH 2 a) 1) true 2) true b) 1) true 2) false c) 1) false 2) true d) 1) false 2) false MSU Physics 231 Fall 2015 27

Properties of gases V = volume P = pressure T = temperature in K (Kelvin) n = number of moles Example balloon MSU Physics 231 Fall 2015 28

Boyle s Law (fixed n and T) ½P 0 2V 0 P 0 V 0 2P 0 ½V 0 At constant temperature: P ~ 1/V implies that PV = constant MSU Physics 231 Fall 2015 29

Charles law (fixed n and P) 2V 0 2T 0 V 0 T 0 If you want to maintain a constant pressure, the temperature must be increased linearly with the volume V ~ T implies that (V/T) = constant MSU Physics 231 Fall 2015 30

Gay-Lussac s law (fixed n and V) P 0 T 0 2P 0 2T 0 If, at constant volume, the temperature is increased, the pressure will increase by the same factor P ~ T implies that (P/T) = constant MSU Physics 231 Fall 2015 31

Brown s law (fixed T and P) 2n 0 2V 0 n 0 V 0 If you double the number of particles the volume doubles n ~ V implies that (V/n) = constant MSU Physics 231 Fall 2015 32

Boyle & Charles & Gay-Lussac IDEAL GAS LAW PV nrt Does not depend on what type or atom or molecule n = number of moles R = universal gas constant 8.31 J/mol K If the number of moles is fixed PV T constant or PV 1 T 1 1 PV 2 T 2 2 MSU Physics 231 Fall 2015 33

Example An ideal gas occupies a volume of 1.0 cm 3 at 20 0 C at 1 atm. A) How many atoms are in the volume? PV = nrt, so n = PV/(TR) with R=8.31 J/mol K T=20 0 C=293K, P=1atm=1.013x10 5 Pa, V=1.0cm 3 =1x10-6 m 3 n=4.2x10-5 mol N = n N A = (4.2x10-5 ) N A =2.5x10 19 B) If the pressure is reduced to 1.0x10-11 Pa, while the temperature drops to 0 0 C, how many atoms remained in the volume? T = 0 0 C = 273K, P = 1.0x10-11 Pa, V = 1x10-6 m 3 n=4.4x10-21 mol N=2.6x10 3 particles (almost vacuum) MSU Physics 231 Fall 2015 34

And another! An air bubble has a volume of 1.50 cm 3 at 950 m depth (T=7 o C). What is its volume when it reaches the surface (T=20 o C). ( water =1.0x10 3 kg/m 3 )? P 950m =P 0 + water g h = 1.013 x 10 5 + (1.0x10 3 )(9.8)(950) = 94.2 x 10 5 Pa V PV 1 T 2 1 1 P1 P 2 PV 2 T 2 T T 2 1 2 V 1 ( 93.0)(1.046) V 1 V surface =146 cm 3 Expanded by a factor of 97 MSU Physics 231 Fall 2015 35

Quiz A volume with dimensions L x W x H is kept under pressure P at temperature T. If the temperature is raised by a factor of 2, and the height is made 5 times smaller, by what factor does the pressure change, i.e. what is P 2 /P 1? No gas leaks or is added. a) 0.4 b) 1 c) 2.5 d) 5 e) 10 Use the fact PV/T is constant if no gas is added/leaked P 1 V 1 / T 1 = P 2 V 2 / T 2 P 1 V 1 / T 1 = P 2 (V 1 /5) / (2T 1 ) P 2 = (5)(2)(P 1 ) = 10 P 1 a factor of 10. MSU Physics 231 Fall 2015 36

Standard temperature and pressure (STP) P 1atm 1.01310 5 Pa T 0 o C 273.15 o K MSU Physics 231 Fall 2015 37

Moles n number of moles N N A N total number of objects N A 6.02 10 23 Avagadro's number MSU Physics 231 Fall 2015 38

macroscopic to microscopic PV n RT macroscopic quantities PV N k B T N = number of atoms or molecules (microscopic) k B R N A.31 6.0210 8 23 1.3810 23 (J/K) (Boltzman's constant) n R N k B MSU Physics 231 Fall 2015 39

Quiz Given P 1 = 1 atm P 2 = 2 atm V 1 = 2 m 3 V 2 = 10 m 3 T 1 = 100 K N 1 = N A N 2 = 10 N A T 2 =? K A) 200 B) 500 C) 2000 D) 5000 E) 100 MSU Physics 231 Fall 2015 40

Example How many air molecules at in the room with a volume of 1000 m 3 (assume only molecular nitrogen is present N 2 )? PV = N k B T T = 293 P = 1.013x10 5 Pa V = 1000 m 3 N = 2.5x10 28 MSU Physics 231 Fall 2015 41

microscopic description: kinetic theory of gases 1) The number of objects is large (statistical model) 2) Their average separation is large 3) The objects follow Newton s laws 4) Any particular object can move in any direction with a distribution of velocities 5) The objects undergo elastic collision with each other 6) The objects make elastic collisions with the walls 7) All objects are of the same type MSU Physics 231 Fall 2015 42

Movie of gas in two dimensions MSU Physics 231 Fall 2015 43

mean free path d = average distance between collisions air at P = 1 atm d = 68 nm = 68 x 10-9 m high vacuum P = 10-5 Pa d = 1m in space P = 10-12 Pa d = 10 8 m MSU Physics 231 Fall 2015 44

The Maxwell Distribution However we can model the distribution of the velocities (& thus the kinetic energies) of the individual gas molecules. The result is the Maxwell Distribution. The root-mean-square (rms) velocity is v rms 2 v MSU Physics 231 Fall 2015 45

Energy of one object Objects inside the container have a distribution of velocities around an average so each object has an average kinetic energy given by K 1 mv 2 2 average translation kinetic energy average squared velocity mass of the object (atom or molecule) MSU Physics 231 Fall 2015 46

MSU Physics 231 Fall 2015 47

Relationship to ideal gas law The objects bounce off of each other and the walls of the container (elastic). One can derive the following result PV 2N K 3 combine to get with K 3 2 PV k B T N k B T How the average kinetic energy of one atom is related to temperature MSU Physics 231 Fall 2015 48

root-mean-square (rms) velocity for one atom or molecule combine K 3 2 k B T with K 1 mv 2 2 v rms v 2 3kBT m 3RT M molar MSU Physics 231 Fall 2015 49

Example What is the rms speed of air at 1 atm and room temperature (293 K)? Assume it consist of molecular Nitrogen only (N 2 )? v rms v 2 3kBT m 3RT M molar R = 8.31 J/mol K T = 293 K M molar = (2 x 14)x10-3 kg/mol v rms = 511 m/s = 1140 mph! MSU Physics 231 Fall 2015 50

Total thermal energy E th d 3 N K N d 2 k B T d 2 nrt d 2 PV (since Nk B nr) d is the number of degrees of freedom for the motion d = 3 for an atom (motion in x, y, z directions) like helium gas d = 5 for a diatomic molecule (motion in x, y, z and two ways to rotate) like nitrogen molecule N 2 or hydrogen molecule H 2 (Homework question for one degree of freedom use d = 1) MSU Physics 231 Fall 2015 51

Example What is the total thermal kinetic energy of the air molecules in the lecture room (assume only molecular nitrogen is present N 2 )? E th = (d/2) PV = 2.5x10 8 J d = 5 P = 1.013x10 5 Pa V = 1000 m 3 Using KE = (1/2) mv 2 this is equivalent to 1000 cars with m=1000 kg each moving with v = 22.3 m/s (50 mph) Can we use that energy to do work? MSU Physics 231 Fall 2015 52

Diffusion MSU Physics 231 Fall 2015 53