MSU Physics 231 Fall 2015 1 Physics 231 Topic 14: Laws of Thermodynamics Alex Brown Dec 7-11 2015.

MSU Physics 231 Fall 2015 1

Physics 231Topic 14: Laws of Thermodynamics

Alex BrownDec 7-11 2015


9th 10 pm attitude survey (1% for participation)

11th 10 pm last homework set

8th 10 pm correction for 3rd exam

10th 10 pm concept test timed (50 min) (1% for performance)

17th 8-10 pm final (Thursday) VMC E100


Clicker Question!

Ice is heated steadily and becomes liquid and then vapor.During this process:

a) the temperature rises continuously.b) when the ice turns into water, the temperature drops for a brief moment.c) the temperature is constant during the phase transformationsd) the temperature cannot exceed 100oC


Key Concepts: Laws of Thermodynamics

Laws of Thermodynamics

1st Law: U = Q + W

2nd Law: Heat flows from hotter cooler

Thermodynamic ProcessesAdiabatic (no heat flow)

Work done in different processes

Heat Engines & RefrigeratorsCarnot engine & efficiency

Entropy

Relationship to heat, energy.

Statistical interpretation

Covers chapter 14 in Rex & Wolfson


Engine based on a container of an idea gaswhere the P, V and T change (n is fixed)

1) Put in contact with a source of heat at high T during which heat energy flows in and piston is pushed up.

2) Put in contact with a source of heat at low T during which piston is pushed down and heat flows out.

3) Comes back to it original state (e.g. same value of P, V, and T)

4) End result is that we have turned heat energy into work

piston

P,V,T

area A

y

n fixed


Process visualized with a P-V diagram for the gas inside

isobaric line: pressure is constant volume changes

iso-volumetric line: volume is constant pressure changes

n fixed

V

P

i


P

V

lines with constant T

iso-thermal lines

PV = n R T (ideal gas equation from chapter 12)

P = n R T/V = c T/V (c = constant)

T1

T2

T3

T4

T1 < T2 < T3 < T4


A Piston EnginePiston is moved downwardslowly so that the gas remainsin thermal equilibrium:Volume decreases (obviously)Temperature increases Work is done on the gas

vin

voutvout > vin (speeds) work is done on the gasand temperature increases

Ti < Tf

Pf Vf Tf

piston

Pi Vi Ti

area A

y


Isobaric Compression

The pressure does not change while pushing down the piston (isobaric compression).

W = work done on the gas by pushing down on the piston

VVf Vi

f iP

P

piston

Pi Vi Ti

area A

y

Pf Vf Tf


Isobaric Compression

The pressure does not change while lowering the piston (isobaric compression).

W = work done on the gas

W = F d = - P A y (P=F/A)W = - P V = - P (Vf-Vi) (in Joule)

Sign of the work done on the gas:+ if V < 0- if V > 0

work is the area under the curve in a P-V diagram with V decreasing

VVf Vi

f iP

P

piston

Pi Vi Ti

area A

y

Pf Vf Tf


Non-isobaric Compression

In general, the pressure can change when lowering the piston.

The work (W) done by the piston on the gas when going from an initial state (i) to a final state (f) is the area under the line on the P-V diagram with V decreasing.

VVf Vi

f

Pi

P

i

Pf

piston

Pi Vi Ti

area A

y


Work Done on Gases:Getting the Signs Right!

If the arrow goes from right to left (volume becomes smaller) positive work is done by pushing the piston down on the gas (W > 0)

the internal (kinetic) energy of the gas goes up

V

P

i


Work Done on Gases:Getting the Signs Right!

If the arrow goes from left to right (volume becomes larger)W < 0 and Wg = -W > 0positive work (Wg) is done by the gas on the piston.the internal energy of the gas goes down

V

P

i


iso-volumetric process

v

P

Work done on/by gas: W = Wg = - PV = 0


Clicker Quiz!

A gas is enclosed in a cylinder with a moveable piston. Thefigures show 4 different PV diagrams. In which case is thework done by the gas largest?

Work: area under PV diagram Work done by the gas: volume must become larger, whichleaves (a) or (c). Area is larger for (a).


a) What is the pressure PA?

b) If the inside temperature is raised the lid moves up by 5 cm. How much work is done by the gas?

M=50 kg A=100 cm2 = 0.010 m2 mass and area of the lid

PA

Patm


a) What is the pressure PA?

b) If the inside temperature is raised the lid moves up by 5 cm. How much work is done by the gas?

M=50 kg A=100 cm2 = 0.010 m2 mass and area of the lid

PA

Patm

a) PA = Patm + Mg/A = 1.50 x 105

b) Wg = PA V = 75.0 J


One mole of an ideal gas initially at 0° C undergoes an expansion at constant pressure of one atmosphere tofour times its original volume.

a) What is the new temperature?

b) What is the work done by the gas?

For ideal gas PV=nRT


One mole of an ideal gas initially at 0° C undergoes an expansion at constant pressure of one atmosphere tofour times its original volume.

a) What is the new temperature?

b) What is the work done by the gas?

For ideal gas PV=nRT

a) Use PV = nRT to get Tf = (Vf/Vi) Ti = 1092 K

b) W = -PV – P(4Vi-Vi) = -3PVi = -3P(nRTi/P)

Wg = -W = 3nRTi = 6806 J


First Law of Thermodynamics

By performing work on an objectthe internal energy can increased

By transferring heat to an objectthe internal energy can increased

The change in internal energy depends on the work done on the object and the amount of heat transferred to the object.

Internal energy (KE+PE) where KE is the kinetic energy associated with translational, rotational, vibrational motion of atoms


First Law of Thermodynamics

U = Uf - Ui = Q + W

U = change in internal energy

Q = energy transfer through heat (+ if heat is transferred to the system)

W = energy transfer through work (+ if work is done on the system)

This law is a general rule for conservation of energy


Applications to ideal gas in a closed container (number of moles, n, is fixed)

PV = n R T (chapter 12)

U = (d/2) n R T (chapter 12) (d=3 monatomic)

(d=5 diatomic)

So U = (d/2) P V (useful for P-V diagram)

(d/2) n R = constant So U = (d/2) n R T

Example for P-V diagram (in class)


First Law: Isobaric ProcessA gas in a cylinder is kept at 1.0x105 Pa. The cylinder is brought in contact with a cold reservoir and 500 J of heat is extracted from the gas. Meanwhile the piston has sunk and the volume decreased by 100cm3. What is the change in internal energy?

Q = -500 J

V = -100 cm3 = -1.0x10-4 m3

W = - P V = 10 J

U = Q + W = - 500 + 10 = - 490


First Law: General Case

V(m3)

P (Pa)

1 4

3

6In ideal gas (d=3) is compressed A) What is the change in internal energyB) What is the work done on the gas?C) How much heat has been transferred to the gas?

A) U = (3/2)PV U = 3/2(PfVf - PiVi) = 3/2[6x1 - 3x4] = -9 J

i

f

B) Work: area under the P-V graph: (9 + 4.5) = 13.5 (positive since work is done on the gas)

C) U = Q+W so Q = U-W = -9 - 13.5 = -22.5 J Heat has been extracted from the gas.


Types of Processes

A: Iso-volumetric V=0B: Adiabatic Q=0C: Isothermal T=0D: Isobaric P=0

PP


Iso-volumetric Process (V = 0)

V = 0W = 0 (area under the curve is zero)4) U = Q = (d/2) n R T 5) P/T = constant

When P = + (like in the figure)T = + (5) U = + (4) Q = + (4) (heat added)

When P = -T = -U = - Q = - (heat extracted)

1) PV = n R T

2) U = W + Q

3) U = (d/2) n R T


Isobaric Process (P = 0)1) PV = n R T

2) U = W + Q

3) U = (d/2) n R T

P = 04) W = - PV = - n R T5) Q = U - W = [(d+2)/2] n R T6) V/T = constant

When V = - (like in the figure)T = - (6)W = + (4) (work done on gas)U = - (3)Q = - (5) (heat extracted)

When V = +T = +W = - (work done by gas)U = + Q = + (heat added)


molar heat capacities

Constant volume Q = (d/2) n R T = Cv n T

where Cv = (d/2) R molar heat capacity at constant volume

Constant pressure Q = [(d+2)/2] n R T = CP n T where CP = [(d+2)/2] R molar heat capacity at constant pressure

For all U = (d/2) n R T = Cv n T


1) PV = n R T

2) U = W + Q

3) U = (d/2) n R T

Isothermal Process (T = 0)

T = 0 work done on gas is the U = 0 area under the curve:Q = -WPV = constant

When V = - (like in the figure)P = + (like in the figure)W = + (work done on gas, from area)Q = - (heat extracted, Q = -W)

When V = +P = -W = - (work done by gas) Q = + (heat added)

W=−nRT ln(𝑉 𝑓

𝑉 𝑖)


1) PV = n R T

2) U = W + Q

3) U = (d/2) n R T

Adiabtic Process (Q = 0)

Q = 0 (system is isolated)W = U (work goes into internal energy)

P (V) = constant = Cp/Cv = (d+2)/d

> 1

When V = - (like in the figure)P = + (like in the figure)T = + (see figure)U = + (3) W = + (work done on gas, area)

When V = +P = -T = -U = - W = - (work done by gas)

dashed lines are isotherms


Ideal gas (monatomic d = 3) (diatomic d = 5) Cv = (d/2) R Cp = [(d+2)/2] R

Process U Q WIsobaric nCv T nCp T -P V

Adiabatic nCv T 0 U

Isovolumetric

nCv T U 0

Isothermal nCv T=0 -W -nRTln(Vf/Vi)

General nCv T U-W (PV Area)negative if V expands


First Law: Adiabatic processA piston is pushed down rapidly. Because thetransfer of heat through the walls takes along time, no heat can escape. During the moving of the piston, the temperature has risen 1000C. If the container contained 10 mol of an ideal gas, how much work has been done during the compression? (d=3)

U = (3/2) nRT Q = 0 and U = Q + W

piston

P,V,T

area A

y

W = U = (3/2) nR T = (3/2)(10)(8.31)(100) = 1.25x104 J


Clicker Quiz!A vertical cylinder with a movable cap is cooled. The process corresponding to this is:

a) CBb) ABc) ACd) CAe) Not shown

After the cooling of the gas and the lid has come to rest, the pressure is the same as before the cooling process.


Adiabatic process

An molecular hydrogen gas goes from P1 = 9.26 atm and V1 = 0.0118 m3 to P2 and V2 via an adiabatic process. If P2 = 2.66 atm, what is V2 ?

H2 (d=5) and adiabatic:

PV =Constant with = Cp/Cv = (d+2)/d= 7/5

P1 (V1)1.4 = P2 (V2)

1.4

(V2)1.4 = (P1 /P2)(V1)

1.4 = 0.0069

V2 = 0.00690.714 = 0.029 (1/1.4) = 0.714)


Cyclic processes (monatomic with d=3)

In a cyclic process,

The system returns to its original state.

Therefore, the internal energy must be the same after completion of the cycle [U = (3/2) PV and U=0]

P (Pa)

V (m3)10 50

5.0

1.0

A

BC


P (Pa)

V (m3)10 50

5.0

1.0

Cyclic Process: Step by Step (1)Process A to B Negative work is done on the gas: (the gas is doing positive work).W= - Area under P-V diagram

= - [ (50-10)(1.0-0.0) +½(50-10)(5.0-1.0) ]= - 40 - 80 = - 120 J (work done on gas)Wg = 120 J (work done by gas)

U = 3/2 (PBVB - PAVA) = 1.5[(1)(50) - (5)(10)] = 0The internal energy has not changed

U=Q+W so Q = U-W = 120 JHeat that was added to the system was used to do the work!

A

BC


P (Pa)

V (m3)10 50

5.0

1.0

Process B-CW = Area under P-V diagram

= - [(50-10)(1.0-0.0)]W=40 JWork was done on the gas

U = 3/2(PcVc-PbVb) = 1.5[(1)(10) - (1)(50)] = - 60 JThe internal energy has decreased by 60 J

U=Q+W so Q = U-W = - 60 - 40 J = - 100 J100 J of energy has been transferred out of the system.

A

BC

Cyclic Process: Step by Step (2)


P (Pa)

V (m3)10 50

5.0

1.0

Process C-AW=-Area under P-V diagramW=0 JNo work was done on/by the gas.

U = 3/2(PcVc-PbVb)= = 1.5[ (5)(10) - (1)(10) ] = 60 JThe internal energy has increased by 60 J

U=Q+W so Q = U-W = 60-0 J = 60 J60 J of energy has been transferred into the system.

A

BC

Cyclic Process: Step by Step (3)


Summary of the processQuantityProcess

Work on gas (W)

Heat(Q) U

A-B -120 J 120 J 0 J

B-C 40 J -100 J -60 J

C-A 0 J 60 J 60 J

SUM(net)

-80 J 80 J 0

A-B B-C C-A

P (Pa)

V (m3)10 50

5.0

1.0

A

BC


What did we do?

The gas performed net work (80 J) (Wg = -W)while net heat was supplied (80 J):We have built an engine that converts heat energy into work!

P (Pa)

V (m3)10 50

5.0

1.0

A

BC

When the path on the P-V diagram is clockwise work is done by the gas (engine) – heat engine

The work done by the gas is equal to the area of the loop Wg = (5-1)(50-10)/2 = 80

QuantityProcess

Work on gas (W)

Heat(Q) U

A-B -120 J 120 J 0 J

B-C 40 J -100 J -60 J

C-A 0 J 60 J 60 J

SUM(net)

-80 J 80 J 0


P (Pa)

V (m3)10 50

5.0

1.0

A

BC

QuantityProcess

Work on gas (W)

Heat(Q) U

A-B -120 J 120 J 0 J

B-C 40 J -100 J -60 J

C-A 0 J 60 J 60 J

SUM (net)

-80 J 80 J 0

Qh = 180 heat input from hot sourceQc = 100 heat output to cold source (wasted heat)Wg = -W = Qh – Qc = 80 work output by gas (engine)

efficiency e = Wg /Qh = 80/180 = 0.4444


Generalized Heat Engine

Heat reservoir Th

Cold reservoir Tc

engine Work

Qh (heat input)

Qc (heat output)

Wg = Qh - Qc

efficiency: Wg/Qh

e = 1 - Qc/Qh

Wg

The efficiency is determinedby how much of the heat yousupply to the engine is turnedinto work instead of being lostas waste.

Water turned to steam

The steam moves a piston

Work is done

The steam is condensed


Reverse Direction: The Fridge


heat reservoir Th

cold reservoir Tc

engine work

Qh

Qc

W

heat is expelled to outside

a piston compresses the coolantwork is done

the fridge is cooled Coefficient of performanceCOP = |Qc|/WQc: amount of heat removedW: work input W= Qh - Qc

Heat Pump (fridge)


On the P-V diagram the heat pump (fridge) is given by a path that goes counter clockwise.

The area inside the loop is the amountof work done on the gas to remove heat from the cold source.

P (Pa)

V (m3)10 50

5.0

1.0

A

BC


Clicker Quiz!

V (m3)

P (Pa)

1 3

3x105

1x105

Consider this clockwise cyclicprocess. Which of the following is true?

a) This is a heat engine and the work done by the gas is +4x105 b) This is a heat engine and the work done by the gas is +6x105 c) This is a heat engine and the work done by the gas is –4x105 d) This is a fridge and the work done on the gas is +4x105 Je) This is a fridge and the work done on the gas is +6x105 J

Clockwise: work done by the gas, so heat engineWork by gas=area enclosed = (3-1) x (3x105-1x105) = 4x105 J


What is the most efficient engine we can make

given a hot and a cold reservoir?

What is the best path to take on the P-V diagram?


Carnot engineAB isothermal expansion

DA adiabatic compression

W+, T+

Th

CD isothermal compressionTcW+, Q-

BC adiabatic expansion

W-, T-W-, Q+

Q=0

T=0Q=0

T=0


Carnot cycle

Work done by engine: Weng

Weng = Qh - Qc

Efficiency: ecarnot = 1-(Tc/Th)

e = 1-(Qc/Qh) also holds sincethis holds for any engine

inverse Carnot cycle

A heat pump or a fridge!By doing work we cantransport heat


Carnot engine

e =1 - (Qc/Qh ) always

ecarnot =1 - (Tc/Th ) carnot only!!

In general: e < ecarnot

The Carnot engine is the most efficient way to operatean engine based on hot/cold reservoirs because theprocess is reversible: it can be reversed without lossor dissipation of energyUnfortunately, a perfect Carnot engine cannot be built.


ExampleThe efficiency of a Carnot engine is 30%. The engine absorbs800 J of heat energy per cycle from a hot reservoir at500 K. Determine a) the energy expelled per cycle and b) the temperature of the cold reservoirc) how much work does the engine do per cycle?

a) Generally for an engine: efficiency: e = 1 – (Qc/Qh) Qc = Qh(1-e) = 800(1-0.3) = 560 J

b) for a Carnot engine: efficiency: e = 1 - (Tc/Th ) Tc = Th(1-e) = 500(1-0.3) = 350

c) W = Qh – Qc = 800 – 560 = 240 J


The 2nd law of thermodynamics1st law: U=Q+W In a cyclic process (U=0) Q=-W: we cannot do more workthan the amount of energy (heat) that we put inside

2nd law in equivalent forms:

- Heat flows spontaneously ONLY from hot to cold masses

- Heat flow is accompanied by an increase in the entropy (disorder) of the universe

- Natural processes evolve toward a state of maximum entropy


Entropy

Lower Entropy

Higher Entropy


Reversing Entropy

We can only reverse the increase in entropy if we do work on the system

Do work to compress the gas back to a smaller volume


EntropyThe CHANGE in entropy (S):

Adiabatic process Q=0 and S = 0

If heat flows out (Q < 0) then S < 0 entropy decreases

If heat flows in (Q > 0) then S > 0 entropy increases

For a Carnot engine, there is no change in entropyover one complete cycle

(J/K unit)


Entropy and Work

Entropy represents an inefficiency wherein energy is “lost” and cannot be used to do work.

Shot = -Qhot/Thot = -24000J / 400K = -60 J/KScold = Qcold/Tcold = +24000J / 300K = +80 J/K

Shot + Scold = -60 J/K + 80 J/K = +20 J/K

Entropy increases!

Cold mass: Gained heat, can do more work.Hot mass: Lost heat, can do less work.

Cold mass gained less potential to do work than host mass lost.Net loss in the ability to do work.


Review: calorimetryIf we connect two objects with different temperatureenergy will transferred from the hotter to the coolerone until their temperatures are the same. If the system is isolated:

Energy flow into cold part = Energy flow out of hot part

mc cc ( Tf - Tc) = mh ch (Th - Tf)

the final temperature is: Tf =

mc cc Tc + mh ch Th

mc cc + mh ch


Phase Change

GAS(high T)

liquid (medium T)

Solid (low T)Q=cgasmT

Q=cliquidmT

Q=csolidmT

Gas liquid

liquid solid

Q=mLv Q=mLf


Heat transfer via conduction

Conduction occurs if there is a temperature difference betweentwo parts of a conducting medium

Rate of energy transfer P

P = Q/t (unit Watt = J/s)

P = k A (Th-Tc)/x = k A T/x

k: thermal conductivity Unit: J/(m s oC)

Metals k~300 J/(m s oC)Gases k~0.1 J/(m s oC)Nonmetals k~1 J/(m s oC)

Th Tc

x

A


Multiple Layers

iii

ch

kL

TTA

t

QP

)/(

)(

Th Tc

A

L1 L2 (x)

k1 k2


Net Power Radiated (photons)

An object emits AND receives radiation,energy radiated per second = net power radiated (J/s)

PNET = A e (T4-T04)

= Power radiated – Power absorbed

where

T: temperature of object (K)T0: temperature of surroundings (K) = 5.6696x10-8 W/m2K4

A = surface areae = object dependent constant emissivity (0-1) for a black body e=1 (all incident radiation is absorbed)


Wavelength where the radiant energy is maximum

where b=2.90×10−3 m KWiens displacement constant

MSU Physics 231 Fall 2015 1 Physics 231 Topic 14: Laws of Thermodynamics Alex Brown Dec 7-11 2015.

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Transcript of MSU Physics 231 Fall 2015 1 Physics 231 Topic 14: Laws of Thermodynamics Alex Brown Dec 7-11 2015.