President University Erwin Sitompul Thermal Physics 8/1

Lecture 8Thermal Physics

Dr.-Ing. Erwin SitompulPresident University

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President University Erwin Sitompul Thermal Physics 8/2

Change in EntropyChapter 20 Second Law of Thermodynamics

Entropy as a State Function We have assumed that entropy, like pressure, energy, and

temperature, is a property of the state of a system and is independent of how that state is reached.

That entropy is indeed a state function can be deduced only by experiment.

To make the process reversible, it is done slowly in a series of small steps, with the gas in an equilibrium state at the end of each step.

For each small step, the energy transferred as heat to or from the gas is dQ, the work done by the gas is dW and the change in internal energy is dEint.

These are related by the first law of thermodynamics in differential form:

intdE dQ dW

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Change in EntropyChapter 20 Second Law of Thermodynamics

Because the steps are reversible, with the gas in equilibrium states, we can use dW = pΔV and dEint = nCVdT.

Solving for dQ leads to:VdQ pdV nC dT

Using the ideal gas law, we replace p = nRT/V. After some rearrangements,

VdQ dV dTnR nCT V T

Integrating each term of the above equation between an arbitrary initial state i and an arbitrary final state f,

f f f

Vi i i

dQ dV dTnR nCT V T

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Change in EntropyChapter 20 Second Law of Thermodynamics

The left term is the entropy change ΔS (= Sf – Si) as the definition. Substituting this and integrating the quantities on the right:

ln lnf ff i V

i i

V TS S S nR nC

V T

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CheckpointAn ideal gas has temperature T1 at the initial state i shown in the p-V diagram here. The gas has a higher temperature T2 at final states a and b, which it can reach along the paths shown. Is the entropy change along the path to state a larger than, smaller than, or the same as that along the path to state b?

path to a < path to bPath to b requires more Q

Chapter 20 Second Law of Thermodynamics

ln lnf ff i V

i i

V TS S S nR nC

V T

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ProblemSuppose 1.0 mol of nitrogen gas is confined to the left side of the container as shown here. You open the stop-cock, and the volume of he gas doubles. What is the entropy change of the gas for this irreversible process?Treat the gas as ideal.

f iT T free expansion, isothermic

ln f

i

VQ nRT

V

Chapter 20 Second Law of Thermodynamics

int 0E Q W QST

ln f inR V V

(1.0)(8.31)(ln 2)

5.76 J K

irrev rev 5.76 J KS S

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ProblemThe figure shows two identical copper blocks of mass m = 1.5 kg; block L at temperature TiL = 60°C and block R at temperature TiR =

20°C. The blocks are in a thermally insulated box and are separated by an insulating shutter.When we lift the shutter, the blocks eventually come to the equilibrium temperature Tf = 40°C What is the net entropy change of the two block system during this irreversible process? The specific heat of copper is 386 J/kg·K.

Chapter 20 Second Law of Thermodynamics

● Irreversible process from initial state to final state

● Reversible process from initial state to final state

● Using thermal reservoir, gradual temperature change dT and gradual heat transfer dQ

President University Erwin Sitompul Thermal Physics 8/8

ProblemChapter 20 Second Law of Thermodynamics

f

Li

dQST

f

iL

T

T

mcdTT

f

iL

T

T

dTmcT

ln f

iL

Tmc

T

(40 273)(1.5)(386) ln(60 273)

35.86 J K

f

Ri

dQST

f

iR

T

T

mcdTT

f

iR

T

T

dTmcT

ln f

iR

Tmc

T

(40 273)(1.5)(386) ln(20 273)

38.23 J K

L RS S S 35.86 38.23 2.37 J K

● Step 1

● Step 2

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The Second Law of ThermodynamicsChapter 20 Second Law of Thermodynamics

A closed system may consists of a number of subsystems, which may interact between one and another.

Between the subsystems, there can be a number of processes. If the processes in a closed system are irreversible, the entropy of

the system will always increase. If the processes are reversible, the entropy will remain constant.

Closed System

Gas

Reservoir

Surroundingair

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The Second Law of ThermodynamicsChapter 20 Second Law of Thermodynamics

As depicted below, isothermal expansion of an ideal gas brings the gas from a to b, from initial state i to final state f. The heat is transferred at constant temperature from the reservoir to the gas.

Now, if we want to reverse the direction of the process, the heat must be extracted from the gas and thus its entropy is decreased.

gas

QS

T

However, if we have to include the reservoir also as part of the system, to have a closed system. The entropy of the reservoir will be:

reservoirQ

ST

The entropy change of the closed system is the sum of these two quantities = 0.

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The Second Law of ThermodynamicsChapter 20 Second Law of Thermodynamics

If a process occurs in a closed system, the entropy of the system increases for irreversible process and remains constant for reversible processes. It never decreases.

Although entropy may decrease in part of a closed system, there will always be an equal or lager entropy increase in another part of the system.

Thus, the entropy of the system as a whole never decreases. This fact is one form of the second law of thermodynamics:

0S Second Law of Thermodynamics

In the real world almost all processes are irreversible due to friction, turbulence, and other factors. So, the entropy of real closed systems undergoing real process always increases.

Process in which the system’s entropy remains constant are always idealizations.

President University Erwin Sitompul Thermal Physics 8/12

Class Group Assignments1. A 10-g ice dice at –10°C is placed in a huge water tank whose temperature is 15°C.

Calculate the change in entropy of the ice cube as it becomes water and reach the lake temperature.

2. 1 mol of a certain ideal gas with initial temperature of 25°C and pressure of 2 atm is heated at constant pressure until the volume is increased by 200%. Assume reversible process. Given CP = 25.895 + 32.999×10–3T – 30.46×10–7T2 J/mol·K, determine ΔS:(a) 0 (d) 26 J/K (b) –26 J/K (e) 47 J/K (c) –68 J/K

3. An copper stick conducts 7.50 cal/s from a heat source maintained at 240°C to a large body of water at 27°C. Determine the rate at which entropy changes per unit time in a closed system that covers this process.(a) +1.03 J/K·s (d) +0.065 J/K·s(b)+0.043 J/K·s (e) +0.031 J/K·s(c) +0.017 J/K·s

(a) 0.8 J/K (b) 2.0 J/K (c) 3.1 J/K (d) 3.8 J/K (e) 15.2 J/K

Chapter 20 Second Law of Thermodynamics

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Homework 81.(20-17)2.(20-18)

Deadline: Wednesday, 18 June 2014. Validation test will be conducted in the beginning of the class.

Chapter 20 Second Law of Thermodynamics