AP Physics II.C Kinetic Theory and Thermodynamics.

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AP Physics II.C Kinetic Theory and Thermodynamics

Transcript of AP Physics II.C Kinetic Theory and Thermodynamics.

Page 1: AP Physics II.C Kinetic Theory and Thermodynamics.

AP Physics II.C

Kinetic Theory and Thermodynamics

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14.1-2 The Ideal Gas Law

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Three Relationships for Gasses

• Pressure and volume (Boyle’s Law)

• Volume and temperature (Charles’ Law)

• Pressure and temperature (Gay-Lussac’s Laws)

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So combined . . . (are we leaving something out?)

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Mole – the number of grams of a substance numerically equal to the

molecular mass (and the mole of any substance has the same number of

particles as 1.0 g of carbon-12)

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Molecular mass of any molecule is the sum of the atomic masses (see the periodic table) of the

molecule

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Examples

• 1 mole of H2 = 2.0 g

• 1 mole of O2 = 32.0 g

• 1 mole of He = 4.0 g (note: inert gasses are not diatomic)

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So, back to gasses, PV = nRT is known as the Ideal Gas Law. R is the universal gas constant (8.315

J/mol·K), n is the number of moles and temperature is given in Kelvin.

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An ideal gas – one of low density (i.e. particles are far enough apart they have few interactions), low pressure (a little less than 1 atm.) and a temperature that is not near

the boiling point for that gas.

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Ex. Determine the volume of 1.00 mol of any ideal gas at STP (standard temperature and pressure – 273 K at1.00 atm).

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Ex. A flexible container of O2 gas at STP has a volume of 10.0cubic meters. What is the mass (in kg) of the gas?

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Ex. A helium party balloon assumed to be spherical has a radius of 18.0 cm. At 20.0º C its internal pressure is 1.05 atm. Find thenumber of moles of helium in the balloon. Find the mass of the helium.

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The Combined Gas Law

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Ex. The absolute pressure in an automobile tire is 301 kPa at 10.0 º C. After driving for a period of time, the temperature in The tire rises to 40.0 º C. What is the new pressure?

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The Ideal Gas Law and Avagadro’s Number

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Avagadro’s Principle – equal volumes of gas at the same

pressure and temperature contain equal number of molecules.

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A quick proof

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Ex. Estimate the number of molecules in one breath if you breathe 1.0 L of air.

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p. 421: 9, 11-12, 15, 17; 96B7

12. 1.10

96B7

a. 8.2 EE 4 Pa

b. 410 N

c. 5.5 EE -3 moles

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14.3 Kinetic Theory of Gasses

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Assumptions about an ideal gas

• There are a large number of molecules of mass m moving in random directions with a variety of speeds

• The molecules are far apart from each other (much greater than their diameter)

• The molecules interact only when the collide (their attractive forces are weak)

• Collisions are elastic

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A whole bunch of ‘em. The concept that matter is composed of atoms in

continuous motion is called the Kinetic Theory.

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A lengthy proof for the average kinetic energy of a single molecule that is probably not worth the time and effort but here goes anyway.

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So . . . What we’ve found is the average kinetic energy of gas

molecules is directly proportional to their absolute temperature

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And, the average speed for each molecule

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Ex. A tank contains two 2.0 mol of He gas at 20.0º C. Find theaverage kinetic energy per molecule and the average speed ofof the molecule.

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p. 422: 28, 30, 32, 35

28. 4.8 EE -21 J

30. 750 K

32. 3.02 EE 5 Pa

35. 9330 m

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15.1 Thermodynamics and their Surroundings

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Thermodynamics – the branch of physics which studies laws

relating heat and work

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System – collection of objects upon which we focus our

attention. Everything else is called the surroundings.

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Closed system (mostly what we are concerned with) – one in

which no mass enters or leaves (but energy may be exchanged

with the environment)

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State of the system – described by giving values for pressure,

volume, temperature and mass.

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15.3 First Law of Thermodynamics

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The internal energy of a system can change due to

• An addition or loss of heat (Q is positive if the system gains heat; Q is negative is the system loses heat)

• Work done on or by the system (work is positive if work is done on the system; work is negative if work is done by the system – this ain’t what your book says)

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The First Law of Thermodynamics as an equation

(a statement of conservation of energy)

ΔU = Q + W

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Ex. 2500 J of heat are added to a system and 1800 J of work is doneon the system. What is the change in internal energy? What is thechange in internal energy if 1800 J of work is done by the system?

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15.4 Thermal Processes

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The process occurs slowly enough that uniform pressure exists throughout all regions of the

system at all times.

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1. Isobaric (constant pressure) – basically a way to derive a

formula

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PV diagram for isobaric process (what is the area under the curve?)

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2. Isothermal Process (constant temperature)

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An expanding balloon

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PV = nRT and the combined gas law

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PV diagram for isothermal process

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Calculating work for an isothermal process

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An isothermal process and the First Law

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3. Adiabatic – no heat flows in or out of the system (Q is

constant) Examples: bicycle pump, diesel engine, stretching a rubber band, compressed gas

released from a container

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PV diagram

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An exponential equation (which you don’t have to remember)

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Calculating work – another non-boxer. Work is estimated from the area under the curve as it is for any

PV diagram.

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An adiabatic process and the First Law

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Note: two adiabatic processes are always connected by two isotherms (a Carnot engine –

more on this later)

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4. Isovolumetric (isochoric) – constant volume

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An isovolumetric process and the First Law

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PV diagram for an isochoric process

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Calculating work

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The First Law and an Isochoric Process

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Ex. An ideal gas undergoes two process. It is slowly compressed at a constant pressure of 2.0 atm from 10.0 L to 2.0 L. In this process, some heat flows out and the temperature drops. Then, heat is then added to the gas, holding the volume constant. The pressure is allowed to rise until the temperature reaches its original value. Calculate the a) total work done on the gas and b) the total heat flow into the gas.

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p. 452: 1, 4, 7-8, 13: Rev. p. 53: 35

4. a) -87 J b) ?

8. Yours

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15.7 The Second Law of Thermodynamics

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Second Law of Thermodynamics (Claussius statement) – heat flows spontaneously from a

substance with a higher temperature to one of a lower temperature, never in reverse.

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The First Law of Thermodynamics relates work and heat. The Second Law of Thermodynamics describes

the flow of heat. Both laws are needed to understand devices that

depend upon heat and work.

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15.8 Three Parts of a Heat Engine (and device that uses heat to perform

work)• Hot reservoir (place from which the engine

receives heat)

• Working substance (device on which the input heat performs work)

• Cold reservoir (remainder of input heat that is rejected at a temperature lower than the input heat)

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Efficiency of a heat engine

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A temperature difference between the hot reservoir and cold reservoir is

needed to operate a heat engine. Therefore, it is not possible to create a heat engine that is 100% efficient.

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Ex. An automobile engine is 20.0% efficient and produces 23 000 Jof work per second. How much heat is rejected by the engine persecond?

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A note on power

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15.9 Carnot’s Principle and the Carnot Engine

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Real processes happen quickly and include gas turbulence and

friction. Thus, the process can’t be repeated since the turbulence

would be different and the friction cannot be reversed.

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Reversible process – Carnot cycle is assumed to happen slowly

enough that the process can be considered a series of equilibrium states. The process can be done in

reverse with out changing the work done or heat exchange.

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2nd Law Again (Carnot Principle - max. eff. of a heat engine): no irreversible

engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same

temperatures.

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2nd Law One More Time (Kelvin-Plank statement) – no device is possible whose sole effect is to

transform a given amount of heat into work (no heat engine is 100%

efficient)

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Efficiency of a Carnot Engine

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Ex. An engine manufacturer makes the following claims: the heatinput per second of the engine is 9.0 kJ at 325 K. The output persecond is 4.0 kJ at 225 K. Are the claims valid?

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Concept Question Suppose a heat engine receives 1000 J of heatfrom a hot reservoir, delivers 1000 J of work and rejects no heatto the cold reservoir. Does this engine violate the 1st Law, 2nd Lawor both?

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15.10 Refrigerators air conditioners and Heat Pumps

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Heat can flow from hot to cold if work is done.

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Comparing a heat pump to a conventional heater

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PV diagram for a heat engine

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PV diagram for a heat pump

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Ex. An ideal (Carnot) heat pump is used to heat a house to a temperature of 294 K. How much work is done by the pump to deliver 3350 J of heat into the house when the outdoor temperature is 273 K?

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p. 454: 40-43, 47, 49-50, 52

40. 65 J

42. 1.82 EE 4 J

50. 1170 K

52. a) 735 K b) 271 K