Thermodynamics: The First Law

자연과학대학 화학과 박영동 교수Classical thermodynamics

Thermodynamics: the first law

2.1 The conservation of energy

2.1.1 Systems and surroundings 2.1.2 Work and heat 2.1.3 The measurement of work 2.1.4 The measurement of heat 2.1.5 Heat influx during expansion

2.2 Internal energy and enthalpy

2.2.6 The internal energy 2.2.7 Internal energy as a state function 2.2.8 The enthalpy 2.2.9 The temperature variation of the enthalpy

Thermodynamic Systems, States and Pro-cesses

Objectives are to:• define thermodynamics systems and states of systems• explain how processes affect such systems• apply the above thermodynamic terms and ideas to the laws of

thermodynamics

Thermodynamic universe

the system is the region of interest; its region is defined by the boundary.

the rest of the world is its surroundings.

The surroundings are where observations are made on the system.

The universe consists of the system and surroundings.

sys-tem

surroundings

The Universe

Various Systems

system matter energy

open( 열린 계 ) allowed allowed

closed( 닫힌 계 )

forbidden allowed

isolated( 고립 계 )

forbidden forbiddenSome other thermodynamics termsstate, state functions,path, process,extensive properties,intensive properties.

closed

open

iso-lated

A State Function and Paths

The altitude is a state property, be-cause it depends only on the current state of the system. The change in the value of a state property is inde-pendent of the path between the two states.

The distance between the initial and final states depends on which path (as depicted by the blue and red lines) is used to travel between them. So it is not a state function.

pressure, temperature, vol-ume, ...

heat, work

-- Heat and work are forms of energy transfer and energy is conserved.

The First Law of Thermodynamics

U = Q + W

work doneon the sys-

tem

change intotal internal en-

ergy

heat added

to system

State Function Process Functions

Calculating the change in internal en-ergy

We see that the person’s internal energy falls by 704 kJ. Later, that energy will be restored by eating.

Suppose someone does 622 kJ of work on an exercise bicycle and loses 82 kJ of energy as heat. What is the change in internal energy of the person? Disregard any matter loss by perspiration.

Solution w = −622 kJ (622 kJ is lost by doing work on the bicycle),q = −82 kJ (82 kJ is lost by heating the surroundings).

Then the first law of thermodynamics gives us

U = q + w = (-82 kJ )+ (-622 kJ) = -704 kJ

Work, and the expansion(p-V) work

F F

dW

pex=F/A

exdW Fdy p Ady Increase in volume, dV

exdW p dV (p-V work)

Work = force • distance

force acting on the piston = pex × Areadistance when expand = dy

+y

Total Work Done

exdW p dV

f

i

V

exVW p dV

To evaluate the integral, we must know how the pressure depends (functionally) on the volume.

We will consider the following cases0. Constant volume work1. Free expansion2. Expansion against constant pressure3. Reversible isothermal expansion

0. Work for the constant volume process

0dV

0f

i

V

exVW p dV

For the constant volume process, there is no p-V work

Heat and Internal Energy

U = q + w = qV

For the constant volume process, there is no p-V work, so

If we add heat q to the system, the temperature of the system increases by T , and the internal energy increases by U.

VdUCdT

Constant volume heat capacity

VV

UCT

Internal Energy of a Gas

A pressurized gas bottle (V = 0.05 m3), contains helium gas (an ideal monatomic gas) at a pressure p = 1×107 Pa and temperature T = 300 K. What is the internal thermal energy of this gas?

pVkTNU23

23

J105.705.0105.1 537 mPa

molar constant volume heat capacity of monatomic gases

(at 1 atm, 25 °C)

Monatomic gas CV, m (J/(mol·K)) CV, m/RHe 12.5 1.50Ne 12.5 1.50Ar 12.5 1.50Kr 12.5 1.50Xe 12.5 1.50

Temperature and Energy distribution

The temperature is a parameter that indicates the extent to which the ex-ponentially decaying Boltzmann dis-tribution reaches up into the higher energy levels of a system. (a) When the temperature is low, only the lower energy states are occupied (as indi-cated by the green rectangles). (b) At higher temperatures, more higher states are occupied. In each case, the populations decay exponentially with increasing temperature, with the total population of all levels a constant.

A constant-volume bomb calorime-ter.

The constant-volume heat capacity is the slope of a curve showing how the internal energy varies with tempera-ture. The slope, and therefore the heat capacity, may be different at dif -ferent temperatures.

1. Work for free expansion case

0exp

0f

i

V

exVW p dV

For the free expansion process, there is no p-V work

U = q + w = q

2. Work for expansion against constant pres-sure

exp p f

i

V

exVW p dV p V

U = q + w = q - p V

qp = U + p V = (U + p V )

If we define Enthalpy as, H ≡ U + pV

qp = HEnthalpy change is the heat given to the sys-tem at constant pressure.

Pp

HCT

Cp, Cv for an ideal gasFor an ideal gas, U and H do not depend on volume or pressure.For an example, for an ideal monatomic gases, U = (3/2) nRT

( )

( ) ( )

P VdH dU d H UC CdT dT dT

d pV d nRT nRdT dT

, ,P m V mC C R

Heat capacity of CO2 and N2

The heat capacity with temperature as ex-pressed by the empiri-cal formula Cp,m/(J K-1 mol-1) = a+bT+c/T2. The circles show the mea-sured values at 298 K.

molar heat capacities,Cp,m/(J K-1 mol-1) = a+bT+c/T2

a b/(l0-3K-1) c/(105K2) @273K* @298K* @350K*

Monatomic gases 20.78 0 0 2.50 2.50 2.50

Other gasesBr2 37.32 0.5 -1.26 4.30 4.34 4.39 Cl2 37.03 0.67 -2.85 4.02 4.09 4.20 CO2 44.22 8.79 -8.62 4.22 4.47 4.84 F2 34.56 2.51 -3.51 3.67 3.77 3.92 H2 27.28 3.26 0.5 3.47 3.47 3.47 I2 37.4 0.59 -0.71 4.40 4.42 4.45 N2 28.58 3.77 -0.5 3.48 3.50 3.55

NH3 29.75 25.1 -1.55 4.15 4.27 4.48 O2 29.96 4.18 -1.67 3.47 3.53 3.62

H2O(l) 75.29 0 0 9.06 9.06 9.06 C(s, graphite) 16.86 4.77 -8.54 0.81 1.04 1.39 *calculated results are given in unit of R(8.314 J K-1 mol-1)

An exothermic process

When hydrochloric acid reacts with zinc, the hydrogen gas pro-duced must push back the sur-rounding atmosphere (repre-sented by the weight resting on the piston), and hence must do work on its surroundings. This is an example of energy leaving a system as work.

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

exothermic and endothermic processes

Work and Heat

Work is transfer of energy that causes or utilizes uniform motion of atoms in the surroundings. For example, when a weight is raised, all the atoms of the weight (shown magnified) move in unison in the same direction.

Heat is the transfer of energy that causes or utilizes chaotic motion in the surroundings. When energy leaves the system (the green region), it generates chaotic motion in the surroundings (shown magnified).

The expansion(p-V ) Work

When a piston of area A moves out through a distance h, it sweeps out a volume ΔV = Ah. The external pres-sure pex opposes the expansion with a force pexA.

3. Work for reversible isothermal expansion for an ideal gas

exnRTp pV

1

ln

f

i

f

i

V

V

V

V

f

i

nRTW dVV

nRT dVVV

nRTV

Reversible Isothermal Expansion Work for a perfect gas

The work of reversible, isothermal expansion of a perfect gas. Note that for a given change of volume and fixed amount of gas, the work is greater the higher the temperature.

Molecular Basis for Heat CapacityThe heat capacity depends on the availability of levels. (a) When the levels are close together, a given amount of energy arriving as heat can be accommodated with little ad-justment of the populations and hence the temperature that occurs in the Boltzmann distribution. This sys-tem has a high heat capacity. (b) When the levels are widely separated, the same incoming energy has to be accommodated by making use of higher energy levels, with a conse-quent greater change in the ‘reach’ of the Boltzmann distribution, and there-fore a greater change in temperature. This system therefore has a low heat capacity. In each case the green line is the distribution at low temperature and the red line that at higher tem-perature.

Temperature dependence of en-thalpy(H) and internal energy(U)

H U pV

The enthalpy of a system increases as its temperature is raised. Note that the enthalpy is always greater than the internal en-ergy of the system, and that the difference increases with tem-perature.

( )dH dU d pVdT dT dT

Temperature dependence of en-thalpy(H) and internal energy(U)

Note that the heat capacityies depend on temperature, and that Cp is greater than CV.

PdHCdT

VdUCdT

a DSC(differential scanning calorimeter)

A thermogram for the protein ubiquitin. The protein retains its native structure up to about 45°C and then undergoes an en-dothermic conformational change. (Adapted from B. Chowdhry and S. LeHarne, J. Chem. Educ. 74, 236 (1997).)

exex

dqCdT

exdq dq dq

dTsample refer-

ence

Cp and CVH U pV

p V V Vp p p

V Tp V T p

H U VC C C p CT T T

U U U V C VT T V T

: internal pressureTT

UV

1 : the expansion coefficient

p

VV T

( )p V V T Tp

HC C C V pV p VT

( , )

S V

T S V T

TT

U S VU UdU dV dS pdV TdSV S

U U U SV V S V

Sp TV

1

1p p

V T

T T

V VT V Tp

T V Vp V p

( )p V TV

pC C p V T VT

2( ) /p V T TC C p V TV

The Maxwell relations

( , );

( , );

( , );

( , );

U S V

H S p

G p T

A V T

;

;

;

;

dU TdS pdV

dH TdS V dp

dG V dp SdT

dA pdV SdT

; ;

; ;

; ;

; ;

V S

V S

pT

T V

U UT pS VU UT pS V

G GV Vp T

A Ap SV T

S V

pS

p T

V T

T pV S

T Vp S

V ST p

p ST V

4. Work for adiabatic expansion

U = q + w = w

U = CV T = w

qad = 0

If the heat capacity is inde-pendent of temperature,

CV dT = -pdV

4. Work for adiabatic expansion

For an adiabatic process for an ideal gas,

2 2 1 1

V VC CnR nRV T V T 2 2 1 1

c cV T V T where, VCc

nR

1

1

"

( ) '

'

'

c

c

c c

cc

T V const

pV V const

p V const

pV const

1 11 pV V V

V V V

CC C nR C nRc nRc nR c nR C C C

pV const

2 2

1 1

2 2

1 1

2 1 2 1

1 2 1 2

1 1

1 1 ln ln

ln ln

V V

VV

T VV V

T V

C CnR nR

CnRTC dT pdV dV dT dVV nR T V

C C T VdT dVnR T V nR T V

T V T VT V T V

For an ideal monatomic gas of n moles,calculate q, w, ΔU for each process.

w12 = -pb(Vb-Va)ΔU12 = CV(T2-T1) = CV(pbVb/nR - pbVa/nR) = CVpb(Vb - Va) /nRq12 = ΔU12 - w12

T2 = pbVb/nRT1 = T3 = pbVa/nR = paVb/nRpbVa = paVb

w2 3= 0ΔU23 = CV(T3-T2) = CV(pbVa/nR - pbVb/nR) = CVpb(Va - Vb) /nR = - ΔU12 q12 = ΔU23

w31= -nRT3 ln(Va/Vb) = -pbVa ln(Va/Vb) ΔU31 = 0q12 = - w31