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 Assumpt ions:  There are many molecules moving in random

direct ions at a variet y of speeds.

 The molecules are, on average, far apart from

each ot her. Their separat ion >> t heir diamet er.

 Molecules obey classical mechanics laws

regarding collisions, energy et c. And t hey int eract only t hrough collision, not t hrough at t ract ive forces (PE).

 Collisions wit h ot her molecules or t he cont ainer

wall are perfect ly elast ic. Time of collision << t ime bet ween collisions.

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 Equations:  pV = nRT  pV = nkT  PV/ T = const.  Boyle’s Law (V

, P), Charles’ Law (V , T), Gay- Lussac’s Law (P , T).

 KE = 3/ 2 kT = ½ mv2  V

rms = 3kT/ m

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 Linear Expansion:  Volume Expansion:  Y

• u’ re having trouble opening a glass j ar,

how can you make it easier?

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 Linear Expansion:  Volume Expansion:  Y

• u’ re having trouble opening a glass j ar,

how can you make it easier?

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AP Physics B

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 Density, pressure, volume…

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 Two systems are said to be in thermal

equilibrium if there is no net flow of heat between them when they are brought into thermal contact.

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 Two systems individually in thermal

equilibrium with a third system are in thermal equilibrium with each other.

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 Change in internal energy of a system is

equal to the heat transferred to/ from the system plus the work done on/ by the system:

ΔU = Q + W

 Comes from the law of conservation of

energy.

 S

ign convention: heat energy Q/ W is positive when the system gains heat and negative when the system loses heat.

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S uppose you had a piston filled with a specific amount of gas. As you add heat, the temperature rises and thus the volume of the gas

• expands. The gas then applies

a force on the piston wall pushing it a specific

• displacement. Thus it can be

said that a gas can do WORK.

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a)

Jogging along the beach one day you do 4.3 x 105 J of work and give off 3.8 x 105 J of heat. What is the change in your internal energy?

b)

S witching over to walking, you give off 1.2 x 105 J of heat and your internal energy decreases by 2.6 x 105 J. How much work have you done while walking?

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S ketch a PV diagram and find the work done by the gas during the following stages.

(a) A gas is expanded from a

volume of 1.0 L to 3.0 L at a constant pressure of 3.0 atm.

(b) The gas is then cooled at a

constant volume until the pressure falls to 2.0 atm

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c)

The gas is then compressed at a constant pressure of 2.0 atm from a volume of 3.0 L to 1.0 L.

d)

The gas is then heated until its pressure increases from 2.0 atm to 3.0 atm at a constant volume.

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What is the NET WORK?

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 Internal Energy is a funct ion of st at e – it

depends only on the state of a system, not

• n the method by which the system arrives

at a given state

 Quasi-static – a process that occurs slowly

enough that a uniform pressure and temperature exist throughout all regions of the system at all times. There is no friction nor dissipative

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To keep the temperature constant both the pressure and volume change to compensate. (Volume goes up, pressure goes down) “ BOYLES ’ LAW”

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Heat is added to the gas which increases the Internal Energy (U) Work is done by the gas as it changes in volume. The path of an isobaric process is a horizontal line called an isobar.

∆U = Q - W can be used

since the WORK is NEGATIVE in this case

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ADIABATIC- (GREEK- adiabatos- "impassable") In other words, NO HEAT can leave or enter the system.

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 Example

 A gas expands from an init ial volume of 0.40 m3 t o a final

volume of 0.62 m3 as t he pressure increases linearly from 110 kPa t o 230 kPa. Find t he work done by t he gas.

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Heat will only flow spontaneously from a

body of higher temperature to a body of lower temperature.

For the reverse to happen, work must be

done.

Example?

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Disorder in the universe can only increase.

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Heat flows from a HOT reservoir to a COLD reservoir QH = remove from, absorbs = hot QC= exhausts to, expels = cold

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In order to determine the

thermal efficiency of

an engine you have to look at how much ENERGY you get OUT vs how much you energy you take IN.

• Eff. = Wout = 1 -

Qout Qin Qin

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A heat engine with an efficiency of 24.0% performs 1250 J of

• work. Find (a) the heat absorbed from the hot reservoir, and

(b) the heat given off to the cold reservoir.

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S

• metimes it is useful to express the

energy usage of an engine as a RA TE. For example: The RATE at which heat is absorbed! The RATE at which heat is expelled. The RATE at which WORK is DONE

POWER t W t Q t Q

C H



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Our goal is to figure out just how efficient such a heat engine can be: what’s the most work we can possibly get for a given amount of fuel? The efficiency question was first posed—and solved—by Sadi Carnot in 1820, not long after steam engines had become efficient enough to begin replacing water wheels, at that time the main power sources for industry. Not surprisingly, perhaps, Carnot visualized the heat engine as a kind of water wheel in which heat (the “fluid”) dropped from a high temperature to a low temperature, losing “potential energy” which the engine turned into work done, just like a water wheel.

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Carnot a believed that there was an absolute zero of temperature, from which he figured out that on being cooled to absolute zero, the fluid would give up all its heat energy. Therefore, if it falls only half way to absolute zero from its beginning temperature, it will give up half its heat, and an engine taking in heat at T and shedding it at ½T will be utilizing half the possible heat, and be 50%

• efficient. Picture a

water wheel that takes in water at the top of a waterfall, but lets it out halfway down. S

• , the efficiency of an

ideal engine operating between two temperatures will be equal to the fraction of the temperature drop towards absolute zero that the heat undergoes. Carnot Eff. = TH – TC TH

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Carnot temperatures must be expressed in KEL VIN!!!!!!

The Carnot model has 4 parts

• An Isothermal Expansion
• An Adiabatic Expansion
• An Isothermal Compression
• An Adiabatic Compression

The PV diagram in a way shows us that the ratio of the heats are symbolic to the ratio of the 2 temperatures

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If the heat engine from the example before is operating at a maximum efficiency, and its cold reservoir is at a temperature of 295 K, what is the temperature of the hot reservoir?

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A particular engine has a power output of 5000 W and an efficiency of 25% . If the engine expels 8000 J of heat in each cycle, find (a) the heat absorbed in each cycle and (b) the time for each cycle

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The efficiency of a Carnot engine is 30% . The engine absorbs 800 J of heat per cycle from a hot temperature reservoir at 500 K. Determine (a) the heat expelled per cycle and (b) the temperature of the cold reservoir

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The relationship for a Carnot engine, can be rearranged to

. The quantity Q/T is the change in entropy, ΔS:

The temperature, again, must be in Kelvins, the subscript R refers to reversible process. Units for entropy = J/ K Entropy is a function of state (like internal energy) – only the state

• f the system determines the entropy

H C H C

T T Q Q 

H H C C

T Q T Q 

R

T Q ΔS       

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Calculate the change in entropy when a 0.125 kg chunk of ice melt at 0ºC. Assume the melting occurs reversibly.

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 The entropy of a Carnot Engine:

 as the engine operates, the entropy of the hot reservoir

decreases, since heat QH leaves. the change in the entropy of the hot reservoir is (minus indicates decrease in S )

 the change in the entropy of the cold reservoir is  Thus, the total change in entropy is

(equals zero because )

 Thus, ΔS = 0 for a Carnot Engine. This is also true for any

reversible process: the total entropy of the universe does not change

Reversible processes do not change the total entropy of the universe. (The ent ropy of one part of t he universe may change, but if so, t he ent ropy of anot her part must change in t he opposit e way by t he same amount .) Irreversible processes increase the entropy of the universe.

ΔS > 0

H H H

T Q ΔS  

C C C

T Q ΔS  

T Q T Q S ΔS

H H C C H C

     

H H C C

T Q T Q 

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A hot reservoir at the temperature 576 K transfers 1050 J of heat irreversibly to a cold reservoir at the temperature 305

• K. Find the change in entropy of the universe.

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The Second Law of Thermodynamics in terms of Entropy: The total entropy of the universe does not change when a reversible process occurs and increases when an irreversible process occurs.

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AP Physics B

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By definit ion, a fluid is a subst ance t hat has no fixed shape and yields easily t o ext ernal pressure.

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Typically, liquids are considered to be incompressible. That is once you place a liquid in a sealed container you can DO WORK on the FLUID as if it were an obj ect. The PRESSURE you apply is transmitted throughout the liquid and over the entire length of the fluid itself.

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 Can exert pressure in any direct ion.  Pressure always act s perpendicular t o t he surface.

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P

at is a direct result of

the weight of the air above us.

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S uppose a Fluid (such as a liquid) is at RES T , we call this HYDROSTATIC PRESSURE.

Notice that the arrows on TOP of the objects are smaller than at the

• BOTTOM. This is because pressure is greatly affected by the DEPTH of

the object. Since the bottom of each object is deeper than the top the pressure is greater at the bottom.

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S uppose we had an obj ect submerged in wat er wit h t he t op part t ouching t he at mosphere. If we were t o draw an FBD for t his obj ect we would have t hree forces

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But recall, pressure is force per unit area.

Note: The initial pressure in this case is atmospheric pressure, which is a CONSTANT. Po=1x105 N/m2 FINAL EQUATION:

DP = rgh

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a) Calculat e t he absolut e pressure at an ocean dept h of 1000 m. Assume t hat t he densit y of wat er is 1000 kg/ m3 and t hat P

• = 1.01 x 105 Pa

(N/ m2). b) Calculat e t he t ot al force exert ed on t he out side

• f a 30.0 cm diamet er circular submarine window

at t his dept h.

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Therefore: P A = PB = PC = PD (because they all have the same depth)

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 Mercury Barometer:

measures atmospheric pressure

 Open Tube

Manometer: measures pressure in a container

Po = 0 P = Patm Patm = 0 + ρgh Patm = ρgh P = Patm + ρgh Example: blood pressure cuff

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If you take a liquid and place it in a system that is CLOS ED like plumbing for example or a car’s brake line, the PRES S URE is the same everywhere. S ince this is true, if you apply a force at

• ne part of the system the pressure is

the same at the other end of the

• system. The force, on the other hand

MAY or MAY NOT equal the initial force

• applied. It depends on the AREA.

Y

• u can take advantage of the fact that

the pressure is the same in a closed system as it has MANY applications. The idea behind this is called P AS CAL’ S PRINCIPLE

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To inspect a 14,000 N car, it is raised with a hydraulic lift. If the radius of the small piston is 4.0 cm, and the radius of the large piston is 17cm, find the force that must be exerted on the small piston to lift the car.

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When an object is immersed in a fluid, such as a liquid, it is buoyed upwards by a force called the buoyant force.

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" An object is buoyed up by a force equal to the weight of the fluid displaced."

In the figure, we see that the difference between the weight in AIR and the weight in WATER is 3 lbs. This is the buoyant force that acts upward to cancel out part of the force. If you were to weight the water displaced it also would weigh 3 lbs.

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*V = A(h2 – h1)

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A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight in air to be 7.84 N. S he then weighs the crown while it is immersed in water (density of water is 1000 kg/ m3) and now the scale reads 6.86 N. Is the crown made

• f pure gold if the density of gold is 19.3 x 103 kg/ m3?

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A piece of wood with a density o 706 kg/ m3 is tied with a string to the bottom

• f a water-filled flask. The

wood is completely immersed, and has a volume of 8.00 x 10-6 m3. What is the tension in the string?

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 S

t eady – velocity of the fluid particles at any point is constant as time passes

 Compressible – density of the

fluid changes as pressure changes (usually gases)

 Viscous – “ a large viscosity” –

doesn’ t readily flow: the viscosity hinders the neighboring layers of fluid from sliding past each other.

 Rot at ional – a part of the fluid

has rotational as well as translational motion. Place a paddle wheel in the fluid, if it rotates, flow is rotational.

 Unst eady – velocity at a

point in the fluid changes as time passes (ex: Turbulent flow: extremely unsteady flow)

 Incompressible – density

• f the fluid remains

constant as pressure changes (usually liquids)

 Nonviscous – “ a low

viscosity” – flows readily – layers are not hindered from sliding past each

• ther.

 Irrot at ional – fluid has

• nly translational motion.

The paddle wheel will not turn.

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A A v v L L Consider a pipe with a fluid moving within it.

Mass flow rate is: m = Vr = ALr = Avr Dt Dt Dt (m = rV)

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A1 A2 v1 v2 L1=v1t L2=v2t

The first thing you MUST understand is that MASS is NOT CREATED OR DESTROYED! IT IS CONSERVED. The MASS that flows into a region = The MASS that flows out of a region.

Using the Mass Flow rate equation and the idea that a certain mass of water is constant as it moves to a new pipe section: We have the Fluid Flow Continuity equation

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The speed of blood in the aorta is 50 cm/ s and this vessel has a radius of 1.0 cm. If the capillaries have a total cross sectional area of 3000 cm2, what is the speed of the blood in them? (Equation of continuity – the last equation we derived last class.)

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What happens to the roof

• f the “w ind tunnel”?

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The S wiss Physicist Daniel Bernoulli, was interested in how the velocity changes as the fluid moves through a pipe of different area. He especially wanted to incorporate pressure into his idea as well. Conceptually, his principle is stated as: "an increase in velocity of a stream of fluid results in a decrease of pressure in the fluid”

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Assumptions:

• Laminar flow.
• Steady flow.
• Incompressible fluid.

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Work is done by a section of water applying a force on a second section in front of it over a displacement. According to Newton’s 3rd law, the second section of water applies an equal and opposite force back on the first. Thus is does negative work as the water still moves FORWARD. Pressure*Area is substituted for Force. X = L F1 on 2

• F2 on 1

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A1 A2 v1 v2 L1=v1t L2=v2t y2 ground Work is also done by GRAVITY as the water travels a vertical displacement UPWARD. As the water moves UP the force due to gravity is DOWN. So the work is NEGATIVE. The fluid in section 1 flows towards section 2 a distance L1 and in doing so pushes the fluid in section 2 a distance L2. y1

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The fluid in the first section is pushed by the fluid to the left of it, and work is done. W1 = F1L1 = P1A1L1 (S ince P = F/ A)

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The fluid in the first section is pushed by the fluid to the left of it, and work is done. W1 = F1L1 = P1A1L1 (S ince P = F/ A) The fluid in the second section is held back by the fluid to the right of it, and negative work is done. W2 = -F2L2 = -P2A2L2

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The fluid in t he first sect ion is pushed by t he fluid t o t he left of it , and work is done. W1 = F1L1 = P1A1L1 (S ince P = F/ A) The fluid in t he second sect ion is held back by t he fluid t o t he right of it , and negat ive work is done. W2 = -F2L2 = -P2A2L2 And negative work is also done by gravity, resisting the motion: W3 = -mg (y2 – y1)

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Total work done: W = W1 + W2 + W3

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Total work done: W = W1 + W2 + W3 And this work is equal to the change in kinetic energy of the system, so: ½ mv2

2 – ½ mv1 2 = P1A1L1 - P2A2L2 - mgy2 + mgy1

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Total work done: W = W1 + W2 + W3 And this work is equal to the change in kinetic energy of the system, so: ½ mv2

2 – ½ mv1 2 = P1A1L1 - P2A2L2 - mgy2 + mgy1

Realize that m = rV = rAL

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Total work done: W = W1 + W2 + W3 And this work is equal to the change in kinetic energy of the system, so: ½ mv2

2 – ½ mv1 2 = P1A1L1 - P2A2L2 - mgy2 + mgy1

Realize that m = rV = rAL Plug this value of m in and cancel the AL in each term, using the fact that A1L1 = A2L2 to get: ½rv2

2 – ½ rv1 2 = P1 – P2 – rgy2 + rgy1

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½rv2

2 – ½ rv1 2 = P1 – P2 – rgy2 + rgy1

Rearrange to get: P1 + ½ rv1

2 + rgy1 = P2 + ½rv2 2 + rgy2

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½rv2

2 – ½ rv1 2 = P1 – P2 – rgy2 + rgy1

Rearrange to get: P1 + ½ rv1

2 + rgy1 = P2 + ½rv2 2 + rgy2

Or more simply:

P + ½ rv2 + rgy = const.

Notice, if v = 0, this becomes the hydrostatic equation.

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Water circulates throughout the house in a hot-water heating system. If the water is pumped at a speed of 0.50 m/ s through a 4.0 cm diameter pipe in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a 2.6 cm-diameter pipe on the second floor 5.0 m above?

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