Chapter 12: Laws of Thermodynamics Work in Thermodynamics Processes Work done on a gas Energy can be transferred to a system by heat and by work done

Chapter 12: Laws of Thermodynamics

Work in Thermodynamics Processes Work done on a gas

• Energy can be transferred to a system by heat and by work done on the system.

• In this chapter, all the systems we are concerned with are volumes of gas and they are in thermodynamic equilibrium: every part of the gas is at the same temperature and pressure.

• Consider a gas contained by a cylinder with a movable piston and in equilibrium. The gas occupies a volume V and exerts a uniform pressure P on the cylinder wall.

Suggested homework assignment: 4,21,33,39,48

Work in Thermodynamics Processes Work done on a gas (cont’d)

• The gas is compressed slowly enough so that the system remains essentially in thermodynamic equilibrium.

• As the piston is pushed downward by an external force F through a distance y, the work done on gas is:

yPAyFW

yAV

VPW under constant pressure

• If the gas is compressed, V is negative and the work done on the gas is positive. If the gas expands, the work done on the gas is negative.

Work in Thermodynamics Processes PV diagram

• PV diagram

graph) under the area(

)(

if VVPVPW isobaric process

• In general, the area under the graph in PV diagram is equal in magnitude to the work done on the gas, whether or not the process proceeds at constant pressure. Note that Pa x m3

= J.

First Law of Thermodynamics First law of thermodynamics

• The first law of thermodynamics is another energy conservation law that relates changes in internal energy – the energy associated with the position and movement of all the molecules of a system – to energy transfers due to heat and work.

• Two mechanisms to transfer energy between a system and its environment : a macroscopic displacement of an object by a force and by heat which occurs through random molecular collisions. Both mechanisms result in a change in internal energy U.

First law of thermodynamics

WQUUU if

Ui : initial internal energyUf : final internal energyQ : energy transferred to the system by heatW : work done on the system

First Law of Thermodynamics First law of thermodynamics (cont’d)

• The internal energy of any isolated system must remain constant.

Internal energy of a monatomic ideal gas

• We learned that the internal energy of a monatomic ideal gas is:

nRTU2

3

Molar specific heat at constant volume • Molar specific heat at constant volume Cv is the amount of heat Q needed to change the temperature by T per mole at constant volume and is defined as:

VV

nCT

Q

n : number of mole

• Therefore a change in U is proportional to a change in T:

TnRU 2

3

First Law of Thermodynamics Molar specific heat

• From the first law of thermodynamics :

WQUUU if • At constant volume, W=0. Therefore the first law becomes:

TnCQU V• However, for an ideal gas:

Therefore, whether its volume is constant or not,

TnRU 2

3

RCV 2

3

First Law of Thermodynamics Molar specific heat (cont’d)

• A gas with a larger specific heat requires more energy to realize a given temperature change. The size of molar specific heat depends on the structure of the gas molecule and how many ways it can store energy.

A monatomic gas (He etc.) can store energy as motion in three different directions (3-dimension). Degrees of freedom = 3 A diatomic gas (H2 etc.) can store energy as motion in three different directions (3-dimension), and can tumble and rotate in two different directions. Degrees of freedom = 5 Other molecules can store energy as vibrations, which add more degrees of freedom.

Each degree of freedom contribute to the molar specific heatby (1/2)R.

So, for example, the molar specific heat of a diatomic gas suchas oxygen O2 is : 5 x (1/2)R = (5/2)R.

First Law of Thermodynamics Molar specific heat (cont’d)

• See Table of molar specific heats

Isobaric process

• An isobaric process is a process where the pressure remains constant.

• Isobaric process and first law:

VPW

WQU

VPQU

VPUQ

First Law of Thermodynamics Isobaric process (cont’d)

• Ideal gas in isobaric process VPUQ

TnRU 2

3

TnRVPnRTPV

For a monatomic ideal gas :

For an isobaric process :

TnRTnRTnRQ 2

5

2

3

Define molar specific heat at constant pressure as: pp

nCT

Q

For an ideal gas in an isobaric process:

RCTnRTnCQ pp 2

5

2

5 RCC Vp

This is valid for all ideal gasses

First Law of Thermodynamics Adiabatic process• An adiabatic process is a process where no energy enters or leaves the system by heat – the system is thermally isolated.

0Q

For an adiabatic process,

WU First law of thermodynamics

• It can also be shown that for an ideal gas in an adiabatic process :

V

p

C

CPV hereconstant w

First Law of Thermodynamics Isovolumetric process• An isovolumetric process, also called an isochoric process, is a process where the volume is constant.

• Since the volume does not change, there is no work done.

QU

For an ideal gas :

TnCQU V

First Law of Thermodynamics Isothermal process• An isothermal process is a process where the temperature is constant.

• For an ideal gas, since the internal energy depends only on the temperature:

0U

QW First law of thermodynamics

• It can be shown that the work done on an ideal gas is given by:

i

f

V

VnRTW ln

First Law of Thermodynamics Examples

• Example 12.4 : Expanding gasSuppose a system of monatomic ideal gas at 2.00x105 Pa and aninitial temperature of 293 K slowly expands at constant pressure froma volume of 1.00 L to 2.50 L.(a) Find the work done on the environment.

J 1000.3)m 1000.1m 10Pa)(2.50 1000.2( 2333-35 VPWenv

(b) Find the change in the internal energy of the gas.

K 733m 1000.1

m 102.50K) 239(

33

3-3

i

fif

i

f

ii

ff

V

VTT

T

T

VP

VP

mol 1021.8K) mol)(293J/(K 31.8(

)m 10Pa)(1.00 1000.2( 23-35

i

ii

RT

VPn

J 1050.42

3 2 TnRTnCU V


• Example 12.4 : Expanding gas (cont’d)

(c) Use the first law to obtain the energy transferred by heat.

J 1000.3 2

envWW

WUQWQUJ 107.50J) 1000.3(J 1050.4 222 Q

(d) Use the molar heat capacity at constant pressure to obtain Q.

J 1050.72

5 2 TnRTnCQ p

(e) How would the answers change for a diatomic gas?

J 1050.712

3 2

TnRTnCU V

J 1005.112

5 3

TnRTnCQ p


• Example 12.5 : Work and engine cylinderIn a car engine operating at 1.80x103 rev/min, the expansion of hot,high-pressure gas against a piston occurs in about 10 ms. becauseenergy transfer by heat typically takes a time on the order of minutesor hours, it is safe to assume that little energy leaves the hot gasduring expansion. Estimate the work done by the gas on the pistonduring this adiabatic expansion by assuming the engine cylindercontains 0.100 moles of an ideal monatomic gas which goes from1.20x103 K to 4.00x102 K typical engine temperatures, during theexpansion.

UUQUW 0

J 1097.9)(2

3 2 ifif TTnRUUU

J 1097.9 2 UWWpiston


• Example 12.6 : An adiabatic expansionA monatomic ideal gas at a pressure1.01x105 Pa expands adiabaticallyfrom an initial volume of 1.50 m3,doubling its volume.(a) Find the new pressure.

3

5

)2/3(

)2/5(

R

R

C

C

V

p

5511 m Pa 1099.1 VPC

Pa 1019.3 4222 PVPC

(b) Estimate the work done on the gas.


• Example 12.7 : An isovolumetric processHow much thermal energy must be added to 5.00 moles of monatomicideal gas at 3.00x102 K and with a constant volume of 1.50 L in orderto raise the temperature of the gas by 3.80x102 K?

J 1099.42

3 3 TnRTnCUQ V

• Example 12.8 : An isothermal expansionA balloon contains 5.00 moles of monatomic ideal gas. As energy isadded to the system by heat, the volume increases by 25% at aconstant temperature of 27.0oC. Find the work Wenv done by the gasin expanding the balloon, the thermal energy Q transferred to the gas,and the work W done on the gas.

)25.1 :(note J 102.78ln 3if

i

fenv VVWQ

V

VnRTW

Second Law of Thermodynamics Heat engines• A heat engine takes in energy by heat and partially converts it to other forms, such as electrical and mechanical energy.

• A heat engine, in general, carries some working substance through a cyclic process during which : (1) energy is transferred by heat from a source at a high temperature (2) work is done by the engine (3) energy is expelled by the engine by heat to a source at lower temperature.

• A steam engine, the working substance is water. The water in the engine is carried through a cycle in which it first evaporates into steam in a boiler and then expands against a piston. After the steam is condensed with cooling water, it returns to the boiler, and the process is repeated.

Second Law of Thermodynamics Heat engines (cont’d)

• The engine absorbs energy Qh from the hot reservoir, does work Weng, then gives up energy Qc to the cold reservoir. Because the working substance goes through a cycle, always returning to its initial thermodynamic state – the initial and final internal energy is the same, so U=0.

WQU 0

ch

engnet

QQ

WWQ

cheng QQW

The work Wenv done by a heat engineequals the net energy absorbed bythe engine.

Second Law of Thermodynamics Heat engines (cont’d)

• If the working substance is a gas, the work done by the engine for a cyclic process is the area enclosed by the curve representing the process on a PV diagram.

• The thermal efficiency of a heat engine is defined by :

h

c

h

ch

h

eng

Q

Q

Q

QQ

Q

We

1

Second Law of Thermodynamics Examples

• Example 12.10 : Efficiency of an engine

During one cycle, an engine extracts 2.00x103 J of energy from ahot reservoir and transfers 1.50x103 J to a cold reservoir.(a) Find the thermal efficiency of the engine.

25.0%or 250.01 h

c

Q

Qe

(b) How much work does this engine do in one cycle?

J 1000.5 2 cheng QQW

(b) How much power does the engine generate if it goes through four cycles in 2.50 s?

W1000.8s 50.2

J) 1000.5(00.4 22

t

WP


• Example 12.11 : Analyzing an engine cycle

A heat engine contains an ideal gasconfined to a cylinder by a movablepiston. The gas starts at A whereT=3.00x102 K and B->C is an iso-thermal process.(a)Find the number n of moles of the gas and the temperature at B.

mol 203.0A

AA

RT

VPn

K 1000.9 2nR

VPT BBB


• Example 12.11 : Analyzing an engine cycle (cont’d)

(b) Find U, Q, and W for isovolumetric process A->B.

J 1052.12

3 3

TRnTnCU VAB

00 ABWV

J 1052.1 3 ABAB UQ



(c) Find U, Q, and W for isothermal process B->C.

0 TnCU VAB

Pa 1067.1ln 3

B

CBC V

VnRTW

J 1067.1

03

BCBC

BCBCBC

WQ

WQU



(d) Find U, Q, and W for isobaric process C->A.

J 1053.2 3 CACACA WUQ

Pa) 101.01atm (1

J 1001.15

3

VPWCA

J 1052.12

3 3 nRTUCA



(e) Find the net change in internal energy Unet

0 CABCABnet UUUU

(f) Find the energy input, Qh; the energy rejected, Qc; the thermal efficiency; and the net work performed by the engine.

J 1053.2

J 1019.33

3

c

BCABh

Q

QQQ

207.01 h

c

Q

Qe

J 1060.6)( 2 CABCABeng WWWW

Second Law of Thermodynamics Refrigeration and heat pump

• A reversed heat engine is called refrigeration!

Energy is injected into the engine calledheat pump and that results in extractionof energy from the cold reservoir to thehot reservoir.

Examples are refrigerator and airconditioner.

• Coefficient of performance

For a heat pump in the cooling mode

W

Qcmode) (cooling COP

For a heat pump in the heating mode

W

Qhmode) (heating COP

Second Law of Thermodynamics Example 12.12 : Cooling the leftovers• 2.00 L of leftover soup at T=323 K is placed in a refrigerator. Assume the specific heat of soup is the same as that of water and the density 1.25x103 kg/m3. The refrigerator cools the soup to 283 K. (a) If the COP of the refrigerator is 5.00, find the energy needed, in the form of work, to cool the soup.

kg 50.2 Vm J 1019.4 5 TmcQQc

J 1038.800.5COP 4 WW

Qc

(b) If the compressor has a power rating 0.250 hp find the time needed to cool the food.

W187hp) W/1746)(hp 250.0( P

s 448P

Wt

Second Law of Thermodynamics Second law of thermodynamics

• There are limits to the efficiency of heat engines.• An ideal engine which would convert all the input energy into work does not exist.

• The Kelvin-Planck formulation of the second law of thermodynamics:

No heat engine operating in a cycle can absorb energy from areservoir and use it entirely for the performance of an equalamount of work.

• This means that the efficiency e=Weng/|Qh| of engines must always be less than one. Some energy Qc must always be lost to the environment. • It is theoretically impossible to construct a heat engine with an efficiency 100%.

We cannot get a greater amount of energy out of a cyclic processthat we put in.

Second Law of Thermodynamics Reversible and irreversible processes• No engine can operate with 100% efficiency, but different designs yield different efficiencies.

• One design called Carnot cycle (engine) delivers the maximum possible efficiency.

• A reversible process is a process in which every state along the same path is an equilibrium state. In a reversible process, the system can return to its initial condition (state) by going along the same path in reverse direction.• An irreversible process is a process which does not satisfy the condition for a reversible process.

• Most natural processes are irreversible, but some of them are almost reversible. If a real process occurs so slowly that the system is virtually always in equilibrium, the process can be considered reversible.

Second Law of Thermodynamics Carnot engine

• Consider a heat engine operating in an ideal, reversible cycle called a Carnot cycle.

Second Law of Thermodynamics Carnot engine (cont’d)

• Consider a heat engine operating in an ideal, reversible cycle called a Carnot cycle.

A->B : Isothermal expansion at Th. Qh from hot reservoir. WAB done by gas.B->C : Adiabatic expansion Th->Tc. No heat goes out or comes in. WBC done by gas.C->D : Isothermal compression at Tc. Qc to cold reservoir. WCD done on gas.D->A : Adiabatic compression Tc->Th. No heat goes out or comes in. WDA done on gas.

Second Law of Thermodynamics Carnot engine (cont’d)

• Ratio of heat input to output vs. ratio of temperatures

h

c

h

c

T

T

Q

Q

• Thermal efficiency of a Carnot enegine

h

c

h

cC T

T

Q

Qe 11 Kin , ch TT

All Carnot engines operating reversibly between the same twotemperatures have the same efficiency.

• All real engines operate irreversibly, due to friction and brevity of their cycles, and are therefore less efficient that the Carnot engine.

Second Law of Thermodynamics Example 12.13 : Steam engine

• A steam engine has a boiler that operates at 5.00x102 K. The energy from the boiler changes water to steam which drives the piston. The temperature of the exhaust is that of the outside air, 300 K. (a) What is the engine’s efficiency if it is an ideal engine?

(40%) 400.01 h

cC T

Te

(b) If the 3.50x103 J of energy is supplied from the boiler, find the work done by the engine on its environment.

J 1010.2 3h

chc

h

c

h

c

T

TQQ

T

T

Q

Q

J 1040.1 3 cheng QQW

Entropy Definition of entropy

• Let Qr be the energy absorbed or expelled during a reversible, constant temperature process between two equilibrium states. Then the change in entropy during any constant temperature process connecting the two equilibrium states id defined by:

T

QS r SI unit: joules/kelvin (J/K)

• A similar formula holds even when the temperature is not constant.• Although calculation of S during a transition between two equilibrium states requires finding a reversible path that connects the states, the entropy change calculated on that reversible path is taken to be S for the actual path. This is valid logic as the change in entropy S depends only on the initial and final states and not on the path taken.

• The entropy of the Universe increases in all natural processes.

Entropy Meaning of entropy• In nature a disorderly arrangement is much more probable than an orderly one if the laws of nature are allowed act without interference.

• Using statistical mechanics it can be concluded that isolated systems tend toward great disorder, and entropy is a measure of that disorder.

WkS B ln kB : Boltzman constantW : a number proportional to the probability that system has a particular configuration.

• The second law of thermodynamics is really a statement of what is most probable rather than of what must be.

• The entropy of the Universe always increases

Entropy Examples

• Example 12.14 : Melting a piece of lead

(a)Find the change in entropy of 300 g of lead when it melts at 327oC. Lead has a latent heat of fusion of 2.45x104 J/kg.

J 1035.7 3 fmLQ K 1000.6273 2 CTT

J/K 3.12T

QS

(b) Suppose the same amount of energy is used to melt part of a piece of silver, which is already at its melting point of 961oC. Find the change in the entropy of the sliver.

K 1023.1273 2 CTT

J/K 96.5T

QS

Entropy Examples• Example 12.15 : Ice, steam, and the entropy of the Universe

A block of ice at 273 K is put in thermal contact with a containerat 373 K, converting 25.0 g of ice to water at 273 K while condensingsome of the steam to water at 373 K.(a) Find the change in entropy of the ice.

J 1033.8 3 fmLQ

J/K 5.30ice

iceice T

QS

(b) Find the change in entropy of the steam.

J/K 3.22

steam

ice

steam

steamsteam T

Q

T

QS

Thermal energy lost by thesteam is equal to the thermalenergy gained by the ice.

(c) Find the change in entropy of the Universe.

J/K 2.8 steamiceUni SSS

Entropy Examples

• Example 12.16 : A falling boulder

A chunk of rock of mass 1.00x103 kg at 293 K falls from a cliff ofheight 125 m into a large lake, also at 293 K. Find the change inentropy of the lake, assuming that all of the rock’s kinetic energyupon entering the lake converts to thermal energy absorbed bythe lake.

J/K 1020.4 3T

PE

T

QS

J 1023.1 6mghPE The rock’s kinetic energy at the timeof entrance to the lake.

Human Metabolism Application of thermodynamics to living organisms• Animals do work and give off energy by heat, and this lead us to believe the first law of thermodynamics can be applied to living organisms.

• Let’s apply the first law in terms of the time rates of change of U, Q, and W.

t

W

t

Q

t

U

On average, energy Q flows out of the body, and

work is done by the body on its surroundings:Q/t and W/t are negative. U/t is negative

• Without supply of energy, the internal energy and the body temperature would decrease. But in reality, all animals acquire internal energy (chemical potential energy) by eating and breathing.

• Overall the energy from oxidation of food ultimately supplies the work done by the body and energy lost from the body by heat. From this point of view, U/t is the rate at which internal energy is added to our bodies by food, which balances the rate of energy loss by heat and work.

Human Metabolism Measuring the metabolic rate • The metabolic rate U/t is the rate at which chemical potential energy in food and oxygen are transformed into internal energy to balance the body losses of internal energy by work and heat.

• The metabolic rate U/t is directly proportional to the rate of oxygen consumption by volume.

t

V

t

U O

28.4For an average diet, the consumption of oneliter of oxygen releases 4.8 kcal or 20 kJ ofenergy.

L/s

Human Metabolism Metabolic rate, activity, and weight gain • Table below summarizes the measured rate of oxygen consumption in mL/(min kg) and the calculated metabolic rate for 65-kg male engaged in various activities.

Human Metabolism Physical fitness and efficiency of the human body as a machine • One measure of a person’s physical fitness is his or her maximum capacity to use or consume oxygen. Table below gives some idea how well a person fit. • The body’s efficiency e is defined as the ratio of the mechanical power supplied by a human to the metabolic rate:

tUtW

e

Human Metabolism Example 12.17 : Fighting fat

• In the course of 24 hours, a65-kg person spends 8 h at a desk puttering around the house, 1h jogging 5 miles, 5 h in moderate activity, and 8 h sleeping. What is the change in her internal energy during this period?

kcal 5000

h) kcal/h)(8 70(h) kcal/h)(5 400(

h) kcal/mi)(1 mi/h)(120 5(h) kcal/h)(10 200(

ii tPU

Documents

Chapter 12: Laws of Thermodynamics Work in Thermodynamics Processes Work done on a gas Energy can be transferred to a system by heat and by work done