termodinamika, entropy, dan energi dalam 2014

8/10/2019 Termodinamika, Entropy, dan Energi dalam 2014

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Internal Energy, Work, Enthalpy,

Ideal Gas

19 September 2013


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2

Processes 2 An isobaricprocess is one in which the pressure is constant.

An isochoricprocess is one in which the volume is constant. An isothermalprocess is one in which the temperature is

constant.

An adiabaticprocess is one in which no heat enters or leaves

the system; i.e. Q = 0. An isentropicprocess is one in which the entropy is constant.

It is a reversible adiabatic process.

If a system is left to itself after undergoing a non-quasistaticprocess, it will reach equilibrium after a time t much longerthan the longest relaxation time involved; i.e. t .

Metastable equilibrium occurs when one particularrelaxation time 0is much longer than the time t for which

the system is observed; i.e. 0 t .


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3

Three Types of Process

Adiabat

Isotherm

P

VHeat bath or reservoir

Isothermal process Adiabatic process

Adiabatic free expansionP

V

1

2

End points

System


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5

The Ideal Gas Law

The triple point of water is Ttr

= 273.16 K, Ptr

= 6.0 x 103atm.

Triple point of Water


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6

The Ideal Gas Law Fixed point (1954)

The triple point of water is Ttr

= 273.16 K, Ptr

= 6.0 x 103atm.

Ideal gas law PV = nRT or Pv = RT, where n is the no. of kmoles, v is the volume per kmole, T is the absolute temperature in K, and

the gas constantR = 8.314 x 103J/(K.kmol).

For a constant quantity of gas, P1V1/T1= P2V2/T2.

P P V

V

T increasing

T T

V increasing

P increasing


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Van der Waals Equation 1 The Van der Waals equation of state

(P + a/v2)(vb) = RT,

reproduces the behavior of a real gas more accurately thanthe ideal gas equation through the empirical parameters aand b, which represent the following phenomena.

i. The term a/v2

represents the attractive intermolecularforces, which reduce the pressure at the walls compared tothat within the body of the gas.

ii. The termb represents the volume occupied by a kilomoleof the gas, which is unavailable to other molecules.

As a and b become smaller, or as T becomes larger, theequation approaches ideal gas equation Pv = RT.

An inflection point, which occurs on the curve at the critical

temperatureTc, gives the critical point(Tc,Pc).


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Van der Waals Equation 2

Below the critical temperature Tc, the curves show maxima and minima. Aphysically reasonable result is obtained by replacing the portion xy, with a straightline chosen so that A1= A2. C is the critical point.

A vapor, which occurs below the critical temperature,differs from a gasin that itmay be liquefied by applying pressure at constant temperature.

P

V

P

V

C

C

Isotherms at

higher T

Tc

A1 A2

Tcxy

vapor

gas

lower T

Inflection point

Isotherms at

higher T

Isotherms at

higher T


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9

Thermodynamic Work 1 Sign convention

The work done by the system is defined to be positive.

With this definition, the work done on the systemtheexternal workof mechanicsis negative.

The work done in a reversible processthe configurationworkis given by the product of an intensive variable and itscomplementary external variable; e.g. dW = PdV.

Reversible isochoric process W = 0, since V = 0.

Reversible isobaric process W = P V = P(V2V1).

These results hold for all materials.

The work done is alwayspositive for expansion

andnegative for compression.


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Thermodynamic Work 2

Calculating the work done in a reversible isothermal process

requires the equation of state of the system to be known.

Reversible isothermal process for an ideal gas (PV = nRT)

W = PdV = nRT dV/V = nRT ln(V2/V1).

In both cases, the work done by the system equals the shadedarea under curve.

Isobaric process

W = P(V2V1)P

V1 V2

Isothermal

process

P

VV1 V2


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Thermodynamic Work 3 Reversible cyclic process

W = PdV equals thearea enclosed by the PVcurve.

W is positive if the area

is traversed in aclockwise sense (asshown), and negative iftraversed counter-clockwise.

P

V

The equality W = PdV applies to reversible processes only.The work done in an irreversible process is given by the

inequality W < PdV.


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Expansivity and Compressibiity An equation of statemay be written as

P = P(V,T), V = V(T,P) or T = T(P,V). Thus, for example,

dV = (V/T)PdT + (V/P)TdP.

In general (x/y)z (y/z)x (z/x)y=1,or (y/x)z =(y/z)x (z/x)y .

Two experimental quantities which may be used to

find the equation of state are the following:coefficient of volume expansion (1/V) (V/T)P;

isothermal compressibility (1/V) (V/P)T.

Thus dV = VdTV dP.


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Useful Theorem

Remember the

negative signs.


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Hysteresis Hysteresis curves are examples of processes which may be

quasistatic, but are not reversible.

Hysteresis is caused by internal friction, and is a well-knownfeature offerromagnetismandfirst-order phase transitions.

The specification of the state of a homogeneous system by asmall number of thermodynamic variables breaks down in the

presence of hysteresis, since the equilibrium state dependson the previous history of the system.

Signal amplitudeof high

temperature phase.

Signal amplitude

of low

temperature phase.


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15

First Law of Thermodynamics

Microscopic picture

The internal energy U is made up of the translational androtational KE, and intermolecular PE of the gas molecules ofthe system.

For anideal monatomic gas, U is the total translational KE,known as the thermal energy, since it is proportional to T.

Q

WU

U = QW ,whereU is the increase of internal energy of the system, Q is

the heat entering the system, and W is the work done by the

system.

U depends onW because gas molecules

rebound off the piston moving to right with a

lower speed, thus reducing the KE of the gas.


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Exact and Inexact Differentials The differential form of the 1stLaw is

dU = dQdW,where dU is an exact differential, because U is a statevariable,and both dQ and dW are inexact differentials,since Q and W are not state variables.

Exact differential dF(x,y)dF is an exact differential if F(x,y) is a function of thevariables x and y. Thus

dF = A(x,y) dx + B(x,y) dy,

whereA(x,y) = (F/x)yand B(x,y) = (F/y)x.

Inexact differential dF(x,y)

IfdF = A(x,y) dx + B(x,y) dy is an inexact differential, there

is no function F(x,y) from which dF can be derived.


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Tests for an Exact Differential

Note: for a state function F = F(V,T,N),

dF = (F/V)T,NdV + (F/T)N,VdT + (F/N)V,TdN,

and 2

F/VT = 2

F/TV, etc.


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Heat Capacities The heat capacity at constant parameter i is given by

Ci= (dQ/dT)i .Note that one cannot use the partial form (Q/T)i ,since dQ is an inexact differential.

Heat capacity at Constant Volume CV dQ = dU + PdV, so thatCV= (dQ/dT)V= (U/T)V.

Heat capacity at Constant Pressure CP

CP= (dQ/dT)P= (U/T)P + P(V/T)P. The enthalpy H is defined as H = U + PV, so that

(H/T)P= (U/T)P + P(V/T)P.

Thus, CP= (H/T)P .


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Adiabatic process in an Ideal Gas 1 Ratio of specific heats = cP/cV= CP/CV.

For a reversible process, dU = dQrPdV. For an adiabatic process, dQr= 0, so thatdU =P dV.

For an ideal gas, U = U(T), so thatCV= dU/dT.

Also, PV = nRTandH = U + PV, so thatH =H(T).Thus, H = H(T) andCP= dH/dT.

Thus,CPCV= dH/dTdU/dT = d(PV)/dT = nR.

CPCV= nR is known as Mayers Equation, which holdsfor an ideal gas only.

For 1 kmole,cPcV= R, where cPand cVare specificheats.


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Adiabatic process in an Ideal Gas 2

SincedQ= 0for an adiabatic process,

dU =P dV anddU = CVdT,so that dT =(P/CV) dV .

For an ideal gas,PV = nRT,

so thatP dV +V dP = nR dT = (nRP/CV) dV.

HenceV dP + P (1 +nR/CV) dV = 0.

Thus,CVdP/P + (CV+ nR) dV/V = 0.

For an ideal gas,CPCV= nR.

so that CVdP/P + CPdV/V = 0,or dP/P + dV/V = 0.

Integration gives ln P + ln V = constant, so thatPV = constant.


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Adiabatic process in an Ideal Gas 2 Work done in a reversible adiabatic process

Method 1: direct integration For a reversible adiabatic process,PV =K.

Since the process is reversible, W = PdV,

so thatW = K V dV = [K/( 1)] V(1) |

=[1/( 1)] PV |

W =[1/( 1)] [P2V2 P1V1].

For an ideal monatomic gas, = 5/3, so that

W =(3/2)] [P2

V2

P1

V1

].

V2

V1

P1V1

P2V2


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Adiabatic process in an Ideal Gas 3 Work done in a reversible adiabatic process

Method 2: from 1stLaw For a reversible process,W = QrU

so that W = U, since Qr= 0 for an adiabatic process.

For an ideal gas, U = CVT = ncVT = ncV(T2T1).Thus,W =ncV(T2T1).

For an ideal gas PV = nRT,

so thatW = (cV/R)[P2V2 P1V1]. ButR = cPcV (Mayers relationship for an ideal gas),

so that W = [cV/(cPcV)][P2V2 P1V1]

i.e. W = [1/( 1)] [P2

V2

P1

V1

].


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Reversible Processes for an Ideal Gas

Adiabaticprocess IsothermalprocessIsobaricprocess Isochoricprocess

PV= K

= CP

/CV

T constant P constant V constant

W = [1/( 1)]

.[P2V2 P1V1]

W = nRT ln(V2 /V1) W = P V W = 0

U = CVT U = 0 U = CVT U = CVT

PV = nRT, U = ncVT, cPcV = R, = cP/cV.

Monatomic ideal gascV= (3/2)R, = 5/3.

termodinamika, entropy, dan energi dalam 2014

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