heat engine information detail

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1 PHYS1001 Physics 1 REGULAR Module 2 Thermal Physics HEAT ENGINE ap06/p1/thermal/ptG_engines.ppt

Transcript of heat engine information detail

Page 1: heat engine information detail

1PHYS1001Physics 1 REGULARModule 2 Thermal Physics

HEAT ENGINE

ap06/p1/thermal/ptG_engines.ppt

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HEAT ENGINES AND REFRIGERATORS

Heat engines: Entropy (§20.1 p673 §20.2 p675 §20.5 p682)

Heat engines: Carnot cycle (§20.6 p684 §20.7 p690)

Internal Combustion Engines: Otto & Diesel Cycles (§20.3 p678)

Refrigerators (§20.4 p680)

References: University Physics 12th ed Young & Freedman

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Heat engine: device that transforms heat partly into work (mechanical energy) by a working substance undergoing a cyclic process.

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4HEAT ENGINE

Hot reservoir (heat source) TH

Cold reservoir (heat sink) TC

QH

QC

W useful mechanical work output

Dissipative losses – friction, turbulence

Cyclic process: W = |QH| - |QC|

Engine – working substance

petrol engine - hot exhaust gases + cooling system

petrol engine: fuel + air

W > 0|QH| = |W|+|QC|

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5All heat engines absorb heat QH from a source at a relatively high temperature (hot reservoir TH), perform some work W and reject some heat QC at a lower temperature (cold reservoir TC).

First Law for a cyclic process: U = 0 = Qnet – W

Qnet = |QH| - |QC| = W

Thermal efficiency, e represents the fraction of QH that is converted to useful work.

H

C

H QQ

QWe 1

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6Problem (Y & F Example 20.1)

A large truck is travelling at 88 km.h-1. The engine takes in 10 000 J of heat and delivers 2 000 J of mechanical energy per cycle. The heat is

obtained by burning petrol (heat of combustion Lcomb = 5.0107 J.kg-1).

The engine undergoes 25 cycles per second. Density of petrol = 700 kg.m-3

(a) What is the thermal efficiency of the heat engine?(b) How much heat is discarded each cycle?(c) How much petrol is burnt per hour (kg.h-1)?(d) What is the petrol consumption of the truck in L/100 km?

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Solution I S E E

v = 88 km.h-1 QH = 1.0104 J/cycle W = 2103 J/cycle

Lcomb = 5.0107 J.kg-1 f = 25 cycles.s-1

(a)e = ?e = W / QH = (2103) / (1.0104) = 0.20 = 20% sensible

(b)QC = ? J |QH| - |QC| = W |QC| = 8.0103 J/cycle

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8(c)QH = 1104 J/cycle QH /t = (25)(1104) J.s-1 = 2.5105 J.s-1 LC = 5.0107 J.kg-1

QH/t = (m/t) Lcomb m/t = (QH/t )/ Lcomb = (2.5105 ) / (5.0107) = 510-3 kg.s-1

m/t = (510-3)(60)(60) = 18 kg.h-1

(d)fuel consumption = ? L/100 kmv = 88 km.h-1 v = d/t t = ? hd = 100 km t = d / v = 100 / 88 h = 1.1364 h

mass used traveling 100 km m = (m/t)t = (18)(1.1364) = 20.455 kg

= m V 103 L = 1 m3

volume used in traveling 100 km V = m / = (20.455 / 700) m3 = 0.0292 m3 = 29.2 L

petrol consumption = 29.2 L/100 km

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Hot Reservoir

QH

Engine SurroundingsW

Can a heat engine be 100% efficient in converting heat into mechanical work ?

Why does this engine violate the Second Law ?

QC= 0

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ENTROPY considerations:

S(engine) = 0 cyclic process

S(surrounding) = 0 no heat transfer to surroundings

S(hot reservoir) < 0 heat removed

S(total) < 0 violates Second Law

All heat engines (heat to work): efficiency, e < 1

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Hot reservoir

Cold reservoir

Engine surroundings

QH

QC

W

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S(hot reservoir) = - |QH|/ TH

S(cold reservoir) = + |QC|/ TC

S(engine) = 0 (cyclic process)

S(surroundings) = 0 (no heat transfer to surroundings)

S(total) = - |QH|/ TH + |QC|/ TC

Useful work can only be done if S(total) 0

| QH| / TH | QC | / TC

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13An ideal engine e.g. Carnot Cycle

S = 0 | QH / TH | = | QC / TC |

e = W / QH W = |QH| - |QC|

e = (|QH| - |QC| / QH)= 1 - |QC| / |QH|

e = 1 – TC / TH < 1This is the absolute max efficiency of a heat engine

Sadi Carnot (1824) Frenchman

NOTE: All heat (QH & QC) exchanges occurred isothermally in calculating the efficiency e = 1 – TC / TH

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CARNOT CYCLE

Heat engine with the maximum possible efficiency consistent with 2nd law.

All thermal processes in the cycle must be reversible - all heat transfer must occur isothermally because conversion of work to heat is irreversible.

When the temperature of the working substances changes, there must be zero heat exchange – adiabatic process.

Carnot cycle – consists of two reversible isotherms and two reversible adiabats

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4

3

1

2

QH

QC released tosurroundings

V

PCarnot Cycle

adiabatic isothermals

Diagram not to scale, adiabats are much steeper than shown

W

NOTE: All heat (QH & QC) exchanges occurred isothermally in calculating the efficiency e = 1 – TC / TH

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2 3Adiabaticcompression

1 2Isothermalcompression

QC QH

3 4Isothermalexpansion

4 1Adiabaticexpansion

All energy exchanges are reversible – there are no non-recoverable energy losses

V

p

4

3

12

1 and 2: “cold”3 and 4: “hot”

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17Proof: e = 1 - |QC| / |QH| = 1 - TC / TH

Isothermal change Q = n R T ln(Vf /Vi)

1 2: Isothermal compression heat QC rejected to sink at constant temperature TC.

|QC | = n R TC ln(V2 / V1)

3 4: Isothermal expansion heat QH supplied from source at constant temperature

|QH | = n R TH ln(V4 / V3)

e = 1 - |QC| / |QH| = 1 – TC ln(V2 / V1) / TH ln(V4 / V3) (Eq. 1)

V

p

4

3

12

1 and 2: “cold”3 and 4: “hot”

Adiabatic change Q = 0 TiVi-1 = TfVf

-1

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182 3: Adiabatic expansion

e = 1 - TC ln(V1 / V2) / TH ln(V4 / V3)

In practice, the Carnot cycle cannot be used for a heat engine because the slopes of the adiabatic and isothermal lines are very similar and the net work output (area enclosed by pV diagram) is too small to overcome friction & other losses in a real engine.

Efficiency of Carnot engine - max possible efficiency for a heat engine operating between TC and TH

1 12 2 3 3T V T V

1 11 1 4 4TV T V 4 1: Adiabatic expansion

1 2 3 4T T T T

11

1 4

2 3

1 4

2 3

V VV V

V VV V

1 C

H

TeT

(Eq. 1)

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19Carnot Engine

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 200 400 600 800 1000 1200 1400 1600 1800 2000

Temperature TH (oC)

effic

ienc

y

The strength & hardness of metalsdecreases rapidly above 750 oC

TC = 25 oCCCarnot

H1

Te

T

“Re-author” Y&F Example 20.3

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Diathermal wall: A highly thermally conducting wall.

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CAR ENGINE

The four–stroke OTTO cycle of a conventional petrol engine

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22The four–stroke OTTO cycle of a conventional petrol engine

intake compression ignition power exhaust stroke stroke stroke stroke

intake stroke: isobaric expansion compression stroke: adiabatic compression ignition: isochoric heating of gas power stroke: adiabatic expansion of gas exhaust stroke: isochoric cooling of gas / isobaric compression

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Intake1

Compression 2

Power 3

Exhaust 4

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5

4

3

1

2

Po

QH

QC released tosurroundings

V

POtto Cycle

adiabatic isothermals

W cooling of exhaust gases

IGNITIONfuel combustion

power stroke

compression stroke

intake stroke exhaust stroke V1 = r V2

r = compression ratioV2

1

11er

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5 1: inlet stroke volume increases as piston moves down creating a partial vacuum to aid air/fuel entering cylinder via the open inlet valve.

5

4

3

1

2

Po

V2 V1

QH

QC released tosurroundings

V

P Otto Cycle

adiabatic isothermals

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1 2: compression stroke inlet valve closes piston moves up compressing the air/fuel mixture adiabatically.

5

4

3

1

2

Po

V2 V1

QH

QC released tosurroundings

V

P Otto Cycle

adiabatic isothermals

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2 3: ignition – spark plug fires igniting mixture - constant volume combustion.

5

4

3

1

2

Po

V2 V1

QH

QC released tosurroundings

V

P Otto Cycle

adiabatic isothermals

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3 4: expansion or power stroke – heated gas expands adiabatically as the piston is pushed down doing work (Vmax = r Vmin). Compression ratio, r

5

4

3

1

2

Po

V2 V1

QH

QC released tosurroundings

V

P Otto Cycle

adiabatic isothermals

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4 1 start of Exhaust stroke – outlet valve opens and mixture expelled at constant volume then 1 5

5

4

3

1

2

Po

V2 V1

QH

QC released tosurroundings

V

P Otto Cycle

adiabatic isothermals

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1 5: Exhaust stroke – piston moves up producing a compression at constant pressure, Po (atmospheric pressure).

5

4

3

1

2

Po

V2 V1

QH

QC released tosurroundings

V

P Otto Cycle

adiabatic isothermals

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Otto cycle – standard petrol engine (4 stroke)

Idealized model of the thermodynamic processes in a typical car engine.

For compression ratio, r ~ 8 and = 1.4 (air) TH (peak) ~ 1800 °C TC (base) ~ 50 °C e ~ 56% (ideal engine) e ~ 35% (real engine).

Efficiency increases with larger r engine operates at higher temperatures pre–ignition knocking sound and engine can be damaged.

Octane rating – measure of anti-knocking – premium petrol r ~ 12.

In practice, the same air does not enter the engine again, but since an equivalent amount of air does enter, we may consider the process as cyclic.

1

11er

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32Comparison of theoretical and actual pV diagrams for the four-stroke Otto Cycle engine.p

V

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33Diesel Cycle

5 to 1: intake stroke isobaric expansion

1 to 2: compression stroke abiabatic compression

2 to 3: ignition isobaric heating

3 to 4: power stroke adiabatic expansion

4 to 1: exhaust stroke (start) isochoric cooling1 to 5: exhaust stroke (finish) isobaric compression

5

4

3

1

2

Po

V2 V1

QH

QC

released tosurroundings

V

Padiabatic isothermals

Rudolf Diesel

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5

4

3

1

2

Po

V2 V1

QH

QC released tosurroundings

V

P Diesel Cycle

adiabatic isothermals

power stroke

cooling of exhaust gases

compression stroke

fuel ignition

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36DIESEL CYCLENo fuel in the cylinder at beginning of compression stroke. At the end of the adiabatic compression high temperatures are reached and then fuel is injected fast enough to keep the pressure constant. The injected fuel because of the high temperatures ignites spontaneously without the need for spark plugs.

Diesel engines operate at higher temperatures than petrol engines, hence more efficient.

For r ~ 18 and = 1.4 (air) TH (peak) ~ 2000 °C TC (base) ~ 50 °C e ~ 68% (ideal engine) real efficiency ~ 40 %

Petrol engine ~ 56% (ideal engine) real efficiency ~ 35 %

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37Diesel engines

• Heavier (higher compression ratios), lower power to weight ratio.

• Harder to start.

• More efficient than petrol engines (higher compression ratios).

• No pre-ignition of fuel since no fuel in cylinder during most of the compression

• They need no carburettor or ignition system, but the fuel-injection system requires expensive high-precision machining.

• Use cheaper fuels less refined heavy oils – fuel does need to be vaporized in carburettor, fuel less volatile hence safer from fire or explosion.

• Diesel cycle – can control amount of injected fuel per cycle – less fuel used at low speeds.

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38http://people.bath.ac.uk/ccsshb/12cyl/

These engines were designed primarily for very large container ships.

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39Problem

For the theoretical Otto cycle, calculate:

(a) max cycle temperature

(b) work per kilogram of fuel

(c) Efficiency

(d) Max efficiency Carnot engine working between same temperatures

Engine characteristics: cp = 1.005 kJ.kg-1.K-1 cV = 0.718 kJ.kg-1.K-1

compression ratio = 8:1

Inlet conditions p = 97.5 kPa and T = 50 oC

Heat supplied = 950 kJ.kg-1

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40SolutionIdentify / Setupcp = 1.005 kJ.kg-1.K-1 cV = 0.718 kJ.kg-1.K-1

V1/ V2 = 8p1 = 97.5103 PaT1 = 50 oC = 323 KQH = 950 kJ.kg-1

TH = T3 = ? KW = ? kJ.kg-1

e = ? eCarnot = ?

Adiabatic change Q = 0 T V -1 = constant = cp / cV = 1.005 / 0.718 = 1.4

Qp = m cp T QV = m cV T W = |QH| - |QC|

e = W / |QH| eCarnot = 1 – TC / TH

5

4

3

1

2

Po

V2 V1

QH

QC released tosurroundings

V

P Otto Cycle

adiabatic isothermals

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41Adiabatic compression ExecuteT2 / T1 = (V1/V2)-1

T2 = T1(V1 / V2)-1 = (323)(8)0.4 K = 742 KConstant volume heatingQH = m cV (T3 – T2)

T3 = (QH/m)/cV +T2 = (950/0.718 + 742) K = 2065 K = 1792 oCAdiabatic expansionT4 / T3 = (V3 /V4)-1

T4 = T3(1/8)0.4 = 899 KConstant volume heat rejectionQC = m cV (T4 – T3) QC/m = (0.718)(899 - 323) kJ.kg-1 = 414 kJ.kg-1

W = |QH| - |QC| = (950 – 414) kJ.kg-1 = 536 kJ.kg-1

e = W / |QH| = 536 / 950 = 0.56 (real engine < 0.4)

eCarnot = 1 – TC / TH = 1 – 323 / 2067 = 0.84

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5

4

3

1

2

Po

V2 V1

QH

QC

released tosurroundings

V

PDiesel Cycle

adiabatic isothermals

Calculate the above quantities for the diesel cycle with a compression ratio = 20

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43Adiabatic compression V1 / V2 = 20T2 / T1 = (V1/V2)-1 T2 = T1(V1 / V2)-1 = (323)(20)0.4 K = 1071 K

Isobaric heatingQH = m cp (T3-T2) QH/m = cp (T3 –T2)T3 = T2 + (1/cp)(QH/m) = 1071 +(1/1.005)(950) K = 2016 K

Isobaric heating / Adiabatic expansionV2 / T2 = V3 / T3 V3 / V2 = T3 / T2 V4 / V2 = 20 V2 = V4 / 20V3 / V4 = (1/20)(T3/T2) = (1/20)(2016/1071) = 0.0941

T3V3-1 = T4V4

-1 T4 = T3 (V3/V4)-1 = (2016)(0.0941)0.4 = 783 K

Isochoric coolingQC = m cv (T4 – T1) QC/m = cv (T4 – T1) = (0.718)(783 – 323) K = 330.3 kJ.kg-1

W = QH – QC = (950 – 330) kJ = 620 kJ.kg-1

e = W / QH = 620 / 950 = 0.65 e = 1- QC/QH = 1 – 330/950 = 0.65

The work output and efficiency are considerably higher than for Otto Cycle

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44Example Consider two engines, the details of which are given in the following diagrams. For both engines, calculate the heat flow to the cold reservoir and the changes in entropy of the hot reservoir, cold reservoir and engine. Which engine violates the Second Law? What is the efficiency of the working engine?

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45SolutionFirst Law: U = Qnet – W

Engine: cyclic process U = 0

Qnet = W |QH| - |QC| |QC | = |QH| - W

Engine 1: |QC| = 1000 - 200 = 800 J

Engine 2: |QC| = 1000 - 300 = 700 J

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46S(total) = - |QH|/ TH + |QC|/ TC

Engine 1: S = (- 2.5 + 2.7) J.K-1 = + 0.2 J.K-1 > 0 Second Law validated

Engine 2: S = (- 2.5 + 2.3) J.K-1 = - 0.2 J.K-1 < 0 Second Law not validated

Engine 1 is the working engine

efficiency, e = (work out / energy input) 100

= (200 / 1000)(100) = 20 %

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47Semester 1, 2007 Examination question (5 marks)

A hybrid petrol-engine car has a higher efficiency than a petrol-only car because it recovers some of the energy that would normally be lost as heat to the surrounding environment during breaking.

(a)If the efficiency of a typical petrol-only car engine is 20%, what efficiency could be achieved if the amount of heat loss during breaking is halved?

(b)Is it possible to recover all the energy lost as heat during braking and convert it into mechanical energy? Explain your answer.

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48SolutionIdentify / Setup efficiency 1H C C

H H H

W Q Q Qe

Q Q Q

Second Law of Thermodynamics 100% of heat can not be transformed into mechanical energy e < 1

Execute(a)

1 0.2 0.80C C

H H

Q Qe

Q Q

Reduce heat loss by half 0.40 0.6C

H

Qe

Q

(b) Would require |QC| = 0, this would be a violation of the Second Law

1 1 0 0C CC

H H

Q Qe Q

Q Q

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What is a heat pump ?

Better buy this quick: 500 % efficiency

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W

QC

QH

evaporatorgas absorbs

heat from interior of frig.

cold

hot

compressorheats gas by compression

condensergas to liquid (high

pressure)

expansion valuerapid expansion: liquid to gas, sudden large drop in temp (~35 oC)

|QH| = |Qc| + |W|

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The compressor compresses the gas (e.g. ammonia). The compressed gas heats up as it is pressurized (orange). The gas represents the working substance eg ammonia and the compressor driven by an electric motor does work W.

The condenser coils at the back of the refrigerator let the hot ammonia gas dissipate its heat QH. The ammonia gas condenses into ammonia liquid (dark blue) at high pressure gas (gas liquid).

The high-pressure ammonia liquid flows through the expansion valve The liquid ammonia immediately boils and vaporizes (light blue), its temperature dropping to about –35 °C by the expansion. This makes the inside of the refrigerator cold by absorption of heat QC .

The cold ammonia gas enters the compressor and the cycle repeats.

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52Refrigeration Cycle

Heat engine operating in reverse – it takes heat from a cold place and gives it off at a warmer place, this requires a net input of work.

|QH| = |QC| + |W|

Best refrigerator – one that removes the greatest amount of heat |QC| from inside the refrigerator for the least expenditure of work |W| coefficient of performance, K (higher K value, better the refrigerator)

C

H C

CQ QK

W Q Q

K = what we want / what we pay forK = extraction of max heat from cold reservoir / least amount of work

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53p

V

QH

QC

W

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Refrigerator

Walls of room

TCW < 0

QC

QH

QC = QH - W

refrigerator

Inside refrigerator|QH| = |Qc| + |W|

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Air Conditioner

Walls of roomTC

W < 0

QC

QH

QH = W + QC

Outside

|QH| = |Qc| + |W|

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Heat Pump

Walls of room

TC

W < 0

QC

QH

QH = W + QCOutside |QH| = |Qc| + |W|

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57PHYS1001 2009 Exam Question 10

In the figure above, cylinder A is separated from cylinder B by an adiabatic piston which is freely movable. In cylinder A there is 0.010 mole of an ideal monatomic gas with an initial temperature 300 K and a volume of 1.0x10-4 m3. In cylinder B, there is 0.010 mole of the same ideal monatomic gas with the same initial temperature and the same volume as the gas in cylinder A. Suppose that heat is allowed to flow slowly to the gas in cylinder A, and that the gas in cylinder B undergoes the thermodynamic process of adiabatic compression. Heat flows into cylinder A until finally the gas in cylinder B is compressed to a volume of 0.5x10-5 m3. Assume that CV = 12.47 J.mol-1.K-1 and the ratio of heat capacities is = 1.67. (a) Calculate the final temperature and pressure of the ideal monatomic gas in cylinder B after the adiabatic compression. (b) How much work does the gas in cylinder B do during the compression? Explain the meaning of the sign of the work. (c) What is the final temperature of the gas in the cylinder A? (Hint: the pressure exerted by the gas in cylinder A on the piston is equal to that exerted by the gas in cylinder B on the piston.) (d) How much heat flows to the gas in the cylinder A during the process?

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