Thermodynamics: the first law Yongsik Lee 2004. 3.
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Transcript of Thermodynamics: the first law Yongsik Lee 2004. 3.
Thermodynamics: the first law
Yongsik Lee2004. 3.
Branches of Thermodynamics Thermodynamics
Classical thermodynamics Phenomenal, bulk properties
Statistical thermodynamics Connection between atomic and bulk thermodynamic pr
operties Branches
Thermochemistry Electrochemistry bioenergetics
The conservation of energy
Matter & Energy
물질 (matter) 공간을 차지하고 질량 (mass) 을 가진다
에너지 (energy) 일을 할 수 있는 능력 열 (heat) 과 일 (work) 의 형태를 가진다
E=mc2
물질과 에너지의 연관성
System and surroundings Consider matter & energy tr
ansfer Open system Closed system Isolated system
Wall Diathermic adiabatic
Work Gas push back the
surrounding atmosphere
Do wok on its surroundings
Energy transfer as work/heat
The measurement of work Transfer of energy that can cause motion against a
n opposing force Work = force x displacement
Physical work No force means no work
Work = pressure x volume change Expansion work Pressure = force / area Volume = area x displacement Work = F dx = P dV
에너지의 단위 에너지의 크기는 줄 (J) 이라는 단위로 표시 1 J 은 어떤 물체를 1 뉴턴 (N) 의 힘으로 1 m 만큼 움직이는 데
필요한 일의 양 1 J = 1 N-m = 1 kg m2 s-2
1 J 은 1 kg 의 책을 중력에 대해서 10 cm 만큼 들어올리는데 필요한 일의 양
인간의 심장이 한번 박동하는데 필요한 에너지 1840 년대 줄의 실험에 의해 물 1g 의 온도를 1℃ 올리는 데
필요한 열은 4.18J 의 일 (1cal) 에 해당한다는 사실이 밝혀졌다 . 1 cal = 4.184 J
따라서 J 은 일과 열 , 그리고 에너지를 나타내는 단위
다른 에너지 단위들 calorie
식품 분야에서 많이 쓰이는 에너지 단위 1 cal = 4.184 J 1 Cal = 1 kcal
Erg 1 J = 107 erg
BTU (British thermal unit) 일 파운드의 물을 화씨 일도 올리는데 필요한 에너지
Heat
열 (q) 높은 온도에서 낮은 온도의 물체로 흐른다 분자운동이 활발하면 열이 많다
열과 온도 계에 열을 가하면 보통 온도가 올라간다
Heat capacity C = q/ΔT
Work and Heat
계가 열을 받으면 부호는 +, 계가 열을 잃으면 - 가 된다 .
Expansion work W=-PexΔV W=- ∫PexdV Free expansion
When external pressure is zero
Means no work
Indicator Diagram
Reversible expansion Reversible
A change that can be reversed By an infinitesimal change in a variable Maximum work is possible
At Constant volume 일정부피에서 반응이
일어나면 , △V = 0, 그러므로 w = 0
Isothermal expansion
Perfect gas Isothermal (ΔT=0) process
Kinetic energy is constant
The measurement of heat Heat capacity
C = q/ΔT [C]=[J/K] Large C means small ΔT
heat capacity at constant pressure (Cp) at constant volume (Cv)
Heat capacity Molar heat capacity Specific heat capacity
일과 열은 결국 같은 종류가 아닐까 ?
일이나 열은 같은 효과를 얻을 수 있다 .
Joule 은 일과 열을 모두 ' 에너지 ' 라는 새로운 개념의 한 형태로 정의
에너지는 계의 상태가 변할 때 일이나 열의 형태로 나타낼 수 있는 추상적인 양
James P. Joule (1818-1889) born in Salford, Lancashir
e worked in a brewery Hero in Thermodynamics
질문 : 무게 25℃ 물이 1 g 있다 . 이 물의 온도를 올릴 수 있는 방법은 무엇인가 ?
두 가지 해결 방법일 ( 실험 2) 과 열 ( 실험 1) 은 같은 효과를 얻을 수 있다 .
Internal energy and enthalpy
Thermodynamic variables Intensive property Extensive property
dependent of the amount of the system Path function State function
Depends only on the current state of the system (independent of the path)
Molar property – per number Specific property – per mass
State function
Internal energy Measure of the energy
reserves of the system
ΔU=w+q Isothermal change
ΔT=0 ΔU=w+q=C ΔT=0 U is constant
Expansion at constant volume △U = q + w △U = q + 0 = qv
제 1 법칙 : 에너지 보존법칙 Crate energy from nothing?
All failed 한 계가 받아들인 열에너지는 그 계의 내부
에너지를 올리거나 , 계가 한 일로 간다 . dq = dU - dW The internal energy of an isolated system is
constant Not a single violation!
Fuel
Combustion of propane
C3H8(g) + 5 O2(g) → 3CO2(g) + 4H2O(l) at 298 K 1 atm
(1 atm = 101325 Pa), -2220 kJ = q
What is the work done by the system?For an ideal gas, propane;
pV = nRT (p = pex)n – no. of molesR – gas constantT = temperatureV – volumep = pressure
V= nRT/p or Vi = niRT/pex
6 moles of gas:Vi = (6 × 8.314 × 298)/ 101325 = 0.1467 m3
3 moles of gas:Vf = (3 × 8.314 × 298)/ 101325 = 0.0734 m3
work done = -pex × (Vf – Vi)
= -101325 (0.0734 – 0.1467) = +7432 Jat constant pressure Pex
work done = - pex (nfRT/pex – niRT/pex)
= (nf – ni) RT
Work done = - n∆ gasRT
i.e. work done = - (3 – 6) × 8.314 × 298 = + 7432.7 J
Can also calculate U∆∆U = q +w q = - 2220 kJ
w = 7432.7 J = 7.43 kJ
∆U = - 2220 + 7.43 = - 2212.6 kJ
NB:qp U ∆ why?
Only equal if no work is done i.e. V = 0∆
i.e. qv = U∆
Example : energy diagram
C3H8 + 5 O2 (Ui)
3CO2 + 4H2O(l) (Uf)
U
U
progress of reaction
reaction path
Reversible work For an infinitesimal change in volume, dV Work done on system = PdV
For ideal gas, PV = nRT
P = nRT/ V work = P(V) dV = nRT dV/ V
= nRT ln (Vf/Vi) because dx/x = ln x Work done by system= -nRT ln(Vf/Vi)
Vf
Vi
Vf
Vi
From heat to work = Engine Compare the work
C-D-A C-A
enthalpy
Definition H=U+PV
Properties State function ΔH=qp
Relation between Heat capacities For an ideal gas: H = U + p V For 1 mol: dH/dT = dU/dT + R Cp = Cv + R
Cp / Cv = γ (Greek gamma) At constant pressure
Some of the energy supplied as heat escapes as work when the system expands
The temperature does not rise as much
Variation of H with temperature
Suppose do reaction at 400 K, need to know H
f at 298 K for comparison with literature value. How?
As temperature ↑, Hm ↑
Hm ∝ T
Hm = Cp,m T
where Cp,m is the molar heat capacity at constant pressure.
H vs. Temperature qv = Cv ( T2 – T1) or Cv
T = U Cp = H / T Cv = U /T
For small changes Cp = dH / dT Cv = dU / dT
Chapter 3Thermochemistry
Yongsik Lee2004. 3
반응열 반응열
표준상태에서 이루어진 반응에서 발생한 열 STP (standard temperature and pressure)
반응열 측정 직접법 : 통열량계를 이용하여 직접 측정 , 또는
표에서 직접 계산 간접법 : 원소로부터 직접 합성이 안 되는 경우
Hess 의 법칙 이용
Bomb calorimeter
난 통 열량계는 잘 몰라요
통아저씨 이양승
Parr 통열량계
Using Bomb calorimeter submerges the reaction inside an insulated
container of water. An electrical heating device starts the
reaction inside a sealed reaction vessel and the temperature rise of the water which surrounds it is measured.
heat loss from a calorimeter (calibration) to keep the temperature of the water
surrounding the reaction vessel constant by heating or cooling it
Use well-known standard
Determine the heat of reaction
Diphenyl + Oxygen reaction combustion reaction
Using benzoic acid for reference Benzoic acid 0.9150 g (0.0217-0.0111 g) Fe wire
Bomb calorimeter with Fe wire Diphenyl 0.8214 g + 0.0031 g Fe wire
Heat capacity calculation
Heat of combustion calculation
enthalpy
Definition H=U+PV For a perfect gas
H=U + RTFor a solid/liquidH=U
Properties State function ΔH=qp
Physical change
Phase
Definition A specific state of matter Uniform throughout in composition and
physical state More specific than the state
Two solid phases of sulfur solid state Rhombic or monoclinic sulfur
Sulfur allotropes Sulfur has three allotropes:
Rhombic, Monoclinic, Plastic At room temperature, monoclinic sulfur is the most stable form. When he
ated, monoclinic sulfur melts to form a viscous liquid at 119 ℃ at the atmosphere pressure.
At higher pressure, the monoclinic sulfur → the rhombic sulfur.
Both crystalline forms have the S8 crown shaped molecule and the plastic sulfur has a chain structure of unspecified number of atoms Sn
Enthalpy of phase transition Phase transition
Conversion of one phase to another phase
State function ΔsubH = ΔfusH + ΔvapH ΔfusH = - ΔfreezeH Path independent
Enthalpy of vaporization (ΔvapH)
Atomic and molecular change Succession of ionization
First ionization enthalpy Second ionization enthalpy
More energy is needed to separate an electron from a positively charged ion
Electron gain enthalpy Reverse of ionization Cl(g) + e-(g) → Cl-(g)
ΔH = -349 kJ
Born-Haber cycle : 1 mol of NaCl
Bond Enthalpy
C-H bond enthalpy in CH4
CH4 (g) → C (g) + 4 H (g) , at 298K.
Need: Hf of CH 4 (g) =- 75 kJ mol-1
Hf of H (g) = 218 kJ mol-1
Hf of C (g) = 713 kJ mol-1
Hdiss = nHf (products) - nHf
( reactants)
= 713 + ( 4x 218) – (- 75) = 1660 kJ mol-1 Since have 4 bonds : C-H = 1660/4 = 415 kJ mol-1
Mean bond enthalpies Definition
Average of bond enthalpies Table 3.4 ΔH(AB) selected bond enthalpies Table 3.5 ΔHB mean bond enthalpies
Example H-O-H H2O(g) → 2H(g) + O(g) ΔH=+927 kJ H2O(g) → HO(g) + H(g) ΔH=+479 kJ HO(g) → H(g) + O(g) ΔH=+428 kJ Mean bond enthalpy ΔHB (O-H) = 463 kJ/mol
Chemical change
Standard enthalpy changes
ΔrH° Standard reaction enthalpy
Reactants in their standard states Products in their standard states
Standard state Pure substances at exactly 1 bar Temperature is not defined
Hess’ Law
the standard enthalpy of a reaction is the sum of the standard enthalpies of the reaction into which the overall reaction may be divided.
ΔrH°(overall reaction) = ΣΔrH°(each step) C (g) + ½ O2(g) → CO (g)
Hcomb =? at 298K
C (g) + ½ O2(g) → CO(g) Hcomb=?
From thermochemical data: C(g) +O2(g) → CO 2(g)
Hcomb =-393.5 kJmol-1
CO (g) +1/2 O2 (g) →CO 2(g) H
comb = -283.0 kJ mol-1
C (g) + O2 (g) – CO (g) – 1/2 O2 (g) → CO2 (g) – CO2 (g) C (g) + ½ O 2 (g) → CO (g) , H
comb= -393.5 – (-283.0) = - 110.5 kJ mol-1
standard enthalpies of formation
ΔfH° The standard enthalpy (per mole of the substan
ce) for its formation from its elements in their reference states
Reference state The most stable form under the prevailing condi
tions At specified temperature and 1 bar
화학반응 = 결합의 재배치 화학결합
결합을 끊는데 에너지가 필요하다 . 분자에 에너지를 가해주어야 한다 . 흡열반응
화학반응 화학결합을 끊고 새로운 화학결합을 만든다 약한 결합을 끊고 강한 결합을 만들면 에너지가
남는다 . ( 반응열의 발생 )
Standard enthalpies of formation Standard reaction enthalpy, ΔrHº ΔrHº =Σ ν Hº(products)- Σ νHº(reactants)
화학결합과 에너지 (kJ/mol)
Hydrazine synthesis reaction
N2 + 2H2 → N2H4
Energy absorbed for bond breaking +946 + 2 x (+436) = +1818 kJ
Energy released by bond making (-389) x 4 + (-163) = -1719 kJ
Energy change during the reaction (+1818) + (-1719) = +99 kJ
Water Generation reaction
Kirchoff’s equation : H(T)
Cp,m = Hm/ T (= [J mol-1/ K])
(= [J K-1 mol-1]) HT2°= HT1°+ Cp ( T2 - T1)
Cp = n Cp(products)- nCp(reactants)
For a wide temperature range: Cp ∫ dT between T1 and T2. Hence : qp = Cp( T2- T1) or H = Cp T
exercises
2-4, 2-5, 2-9, 2-17 3-5, 3-15
References
http://www.whfreeman.com/ECHEM/
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