Chapter 2. The First Law Physical Chemistry Fundamentals

Presentation on theme: "Chapter 2. The First Law Physical Chemistry Fundamentals"— Presentation transcript:

1 Chapter 2. The First Law Physical Chemistry Fundamentals
● Conservation of energy - energy can be neither created nor destroyed (experimented observation) - assess the energy changes that accompany physical and chem processes ● Concept of enthalpy - property for keeping track of the heat output of physical processes and chemical rxns at constant P ● Establish relations between different properties of a system ● Liquefaction of gases Energy Storage & Conversion Materials Lab.

2 Physical Chemistry Fundamentals
■ The basic concepts - System : the part of world in which we have a special interest ~rxn vessel, an engine, an Electrochemical cell, - Surrounding : the region outside the system - Boundary : system at surrounding을 구분 - Type of system : Characteristics of the boundary 1) Open system : Matter and Energy is transferred through boundary 2) Close system : Matter cannot pass but Energy can be transferred 3) Isolated system : Matter an Energy in not transferred Fig. 2.1 (a) An open system can exchange matter and energy with surroundings. (b) A closed system can exchange energy with its surroundings, but it cannot exchange matter, (c) An isolated system can exchange neither energy nor matter with its surroundings. Energy Storage & Conversion Materials Lab.

3 Physical Chemistry Fundamentals
■ 2.1 Work, heat and energy - Work : motion against an opposing force - Energy : the capacity to do work - Heating : the transfer of energy that make use of disorderly molecular motion - Work : the transfer of energy that make use of organized motion Operational definitions - Work : (Fundamental physical property in thermodynamics) Motion against force ex) Raising a weight against the pull of gravity, Expansion of a gas that pushes out a piston - Energy : Capacity to do work - Diathermic (투열벽 ) : Boundary permit energy transfer as heat - Adiabatic (단열벽 ) : Energy transfer가 일어나지 않는 경계 - Exothermic process : Process that releases energy as heat into its surrounding - Endothermic process : Energy in acquired from its surroundings as heat Energy Storage & Conversion Materials Lab.

4 에너지가 열로써 주위에 이동할 때 주위에 있는 원자들의
Physical Chemistry Fundamentals (b) The molecular interpretation of heat and work - Heating : the transfer of energy that make use of disorderly molecular motion ( 무질서한 분자운동을 이용하는 에너지의 이전 ) ※ Disorderly molecules motion ~ thermal motion 에너지가 열로써 주위에 이동할 때 주위에 있는 원자들의 무질서한 운동이 자극된다. Fig. 2.3 When energy is transferred to the surrounding as heat, the transfer stimulates random motion of the atoms in the surroundings. Transfer of energy from the surroundings to the system makes use of random motion (thermal motion) in the surroundings. - Work : the transfer of energy that make use of organized motion in surrounding ex) Spring : The atoms in a spring move in an orderly way Electrons : Electrons in an electric current move in an orderly direction 계가 일을 할 때 주위에 있는 원자나 전자들을 조직화된 방법으로 움직인다. Fig. 2.4 When a system does work, it stimulates orderly motion in the surroundings. For instance, the atoms shown here may be part of a weight that is being raised. The ordered motion of the atoms in a falling weight does work on the system. Work: Energy transfer making use of the organized motion of atoms in the surroundings Heat : Energy transfer thermal motion in the surroundings Energy Storage & Conversion Materials Lab.

5 Physical Chemistry Fundamentals ■ 2.2 The internal energy, u
The conservation of energy : First law of thermodynamics ( 경험 법칙 ) ① If a system is isolated from its surroundings, No change in internal energy take place ② 계와 주위 사이에서 에너지 受授(수수)가 일어날 때 계와 주위 에너지의 합은 항상 일정하다 ※ 受授(수수)=교환 − 𝑀𝑎𝑡ℎ𝑒𝑚𝑎𝑡𝑖𝑐𝑎𝑙 𝑠𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑙𝑎𝑤 : Δ𝑆𝑦𝑠𝑡𝑒𝑚 𝐸+Δ𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝐸=0 Δ𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝐸=±𝑞±𝑤 Δ𝑆𝑦𝑠𝑡𝑒𝑚 𝐸=Δu ∴ 𝜟𝐮=±𝒒±𝒘 부호의 정의 : system에다 일이나 열을 이전시킬 때 𝑤 >0, 𝑞 >0 ∴Δu=𝑞+𝑤 - Internal Energy um(𝑇)=um 𝑅𝑇 (monoatomic gas; translation only) um(𝑇)=um(0)+ 5 2 𝑅𝑇 (linear molecular; translation and rotation only) um(𝑇)=um(0)+3𝑅𝑇 (nonlinear molecular; translation and rotation only) 에너지 미소변화 : 𝑑𝑢=𝑑𝑞+𝑑𝑤 Energy Storage & Conversion Materials Lab.

6 Physical Chemistry Fundamentals
■ 2.3 Expansion Work ( 팽창 일 ) : The work arising from a change of vol the general expression for work ( 일반식 ) : 계가 팽창하면서 행하는 일 𝑑𝑤 = − 𝐹𝑑𝑧 𝐹=𝑃𝑒𝑥𝐴 𝑑𝑤=−𝑃𝑒𝑥𝐴𝑑𝑧=−𝑃𝑒𝑥𝑑𝑣 𝑊=- 𝑣𝑖 𝑣𝑓 𝑃𝑒𝑥𝑑𝑣 Fig. 2.6 When a piston od area A moves out through a distance dz, it sweeps out a volume dV = Adz. The external pressure pex is equivalent to a weight pressing on the piston, and the force opposing expansion is F = pex A. IfⅠ) Vf> Vi (계가 팽창) : 계가 주위에 일을 해줌(-) 𝑤=− Ⅱ) Vf< Vi (계가 수축) : 주위에서 계에 일을 해줌(+) 𝑤=+ Non-expansion work Energy Storage & Conversion Materials Lab.

7 Physical Chemistry Fundamentals
(b) Free expansion ( 자유팽창 ) : expansion against zero 𝑃𝑒𝑥=0, 𝑑𝑤=0 ⇒ 𝑤=0 Ex) 진공 속으로 팽창 (c) Expansion against constant pressure ( 일정한 압력하에서의 팽창 ) 𝑊=−𝑃𝑒𝑥 𝑉𝑖 𝑉𝑓 𝑑𝑣 = −𝑃𝑒𝑥(𝑉𝑓=𝑉𝑖)=−𝑃𝑒𝑥∆𝑣 (d) Reversible Expansion (가역 팽창) : Infinitesimal ≒reversible , Pex ≒ P, Irreversible: Pex ≥ P : 계와 주위와 평행을 이루면서 팽창한다. Pex P 𝛿𝑧 − 외부압력을 내부압력보다 미소량 만큼 작게 해주면 기체는 약간 팽창한다 그러나 외부압력을 미소량 증가시키면 기체는 약간 압축한다 − 동일한 온도의 두 계가 열적 평형에 있다 어느 한 쪽 물체의 온도를 미소량 만큼 낮추면 그 곳으로 에너지가 흘러 들어가고 반대로 온도를 미소하게 높여 주면 에너지가 반대물체로 흘러 들어가므로 이 두 물체 사이의 에너지이동은 가역적이다. There is obviously a very close relationship between reversibility and equilibrium; systems at equilibrium are poised to undergo reversible change Energy Storage & Conversion Materials Lab.

8 Physical Chemistry Fundamentals
𝑃𝑒𝑥= 𝑃, 𝑑𝑤=−𝑃𝑒𝑥𝑑𝑉=−𝑃𝑑𝑉 𝑊=- 𝑣𝑖 𝑣𝑓 𝑃𝑑𝑉 𝑃=𝑓(𝑣) (e) Isothermal reversible expansion ( 등온가역팽창 ) 𝑃𝑉=𝑛𝑅𝑇 ⇒𝑃=𝑛𝑅𝑇/𝑉 ( 등온과정 : 𝑇=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ) 𝑤 =− 𝑉𝑖 𝑉𝑓 𝑝𝑑𝑉 = −𝑛𝑅𝑇 𝑉𝑖 𝑉𝑓 𝑑𝑉 𝑉 = − 𝑛𝑅𝑇 ln 𝑉𝑓 𝑉𝑖 𝐼𝑓) 𝑉𝑓 > 𝑉𝑖 (팽창 : 계가 주위에 일을 해줌, ) 𝑊= 음 𝐼𝑓) 𝑇↑ : 𝑊는 더 많은 일이 행하여진다. Fig. 2.8 The work done by a perfect gas when it expands reversibly and isothermally is equal to the area under the isotherm p = nRT/V. The work done on irreversible expansion against the same final pressure is equal to the rectangular area shown slightly darker. Fig. 2.7 The work done by a gas when it expands against a constant external pressure, pex, is equal to the shaded area. Wreversible > Wirreversible Energy Storage & Conversion Materials Lab.

9 Physical Chemistry Fundamentals
𝑊=− 𝑉𝑖 𝑉𝑓 𝑃𝑑𝑉 =−𝑛𝑅𝑇 ln 𝑉𝑓 𝑉𝑖 𝑃𝑖𝑉𝑖=𝑃𝑓𝑉𝑓 @ 𝑇=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑊=−𝑛𝑅𝑇 ln 𝑃𝑖 𝑃𝑓 =𝑛𝑅𝑇 ln 𝑃𝑓 𝑃𝑖 For van der waals gas : P= 𝑅𝑇 𝑉𝑚−𝑏 − 𝑎 𝑉𝑚2 = 𝑛𝑅𝑇 𝑉−𝑛𝑏 − 𝑛2𝑎 𝑉2 w=− 𝑉𝑖 𝑉𝑓 𝑃𝑑𝑉 =− 𝑉𝑖 𝑉𝑓 𝑛𝑅𝑇 𝑉−𝑛𝑏 − 𝑛2𝑎 𝑉2 𝑑𝑉 =− nRT ln 𝑉𝑓−𝑛𝑏 𝑉𝑖−𝑛𝑏 +an2 ( 1 𝑉𝑖 − 1 𝑉𝑓 ) Energy Storage & Conversion Materials Lab.

10 𝑞=𝑐 ΔT 𝑞=𝐼𝑡Δ𝜙 Physical Chemistry Fundamentals
■ 2.4 Heat transactions (열의 계량) ( 일반적 계의 내부 에너지 변화 ) Change in internal energy of system 𝒅𝒖 = 𝒅𝒒+𝒅𝒘𝒆𝒙𝒑+𝒅𝑾𝒆 (𝑑𝑊𝑒 : 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑤𝑜𝑟𝑘) 𝑊ℎ𝑒𝑟𝑒, 𝑑𝑊𝑒𝑥𝑝 : expansion work 𝑑𝑤𝑒 : 팽창 일 외에 추가되는 일 (non-expansion work, ex : electrical work) - 일정한 부피 하에서 추가적인 일이 없는 경우 : 𝑑𝑢=𝑑𝑞𝑉 ( v : 일정부피하에) ⟹ Δ𝑢=𝑞𝑉 𝑑𝑢=𝑑𝑞 Calorimeter : a device for measuring energy transferred as a heat (ΔU 측정) 𝑞=𝑐 ΔT 𝑞=𝐼𝑡Δ𝜙 ΔT→𝑞→Δu Fig. 2.9 A constant-volume bomb calorimeter. The calorimeter is the entire assembly shown here. To ensure adiabaticity, the calorimeter is immersed in a water bath with temperature continuously readjusted to that of the calorimeter at each stage of the combustion. Calorimeter constant ※ 1V = 1 JC-1 1J = 1 VC (AS) = 1 VAS Energy Storage & Conversion Materials Lab.

11 Cv= ( 𝜕𝑢 𝜕𝑇 )v = ( 𝜕𝑞𝑣 𝜕𝑇 )v Physical Chemistry Fundamentals
(b) Heat capacity ( 열용량 ) Cv= ( 𝜕𝑢 𝜕𝑇 )v = ( 𝜕𝑞𝑣 𝜕𝑇 )v Heat capacity at constant volume du = dqv Extensive property Cv,m= Cv/n : molar heat capacity Intensive property Cv,s= Cv/m : specific heat capacity (heat capacity per mass) Fig The internal energy of a system increases as the temperature is raised; The slope of the tangent to the curve at any temperature is the heat capacity at constant volume at that temperature. Note that the heat capacity is greater at B than at A. - Heat capacity는 Temp에 의존  온도에 따라 감소함 ( 매우 낮은 온도에서는 0으로 접근함 ) - R.T이상에서의 좁은 온도 범위에서는 거의 상수로 취급함 - Heat capacity가 크다는 것은 주어진 일정한 열량으로 온도를 조금 밖에 올리지 못함 (J/K,mol) 상변화시 : 열용량은 무한대 ○ Heat capacity vs. charge in internal energy at constant vol 𝑑𝑢 = 𝐶𝑉𝑑𝑇 ⇒ Δ𝑢 = 𝐶 𝑉𝑑𝑡=𝑞𝑉 ⇒ Δ𝑢=𝐶𝑉ΔT 계의 heat와 온도증가분의 비 ; 𝑞𝑉/ΔT⇒ Cv Energy Storage & Conversion Materials Lab.

12 𝑑𝑞 > 𝑑𝑢 ⇒ 𝑑𝑞=𝑑𝑢+𝑑𝑤 Physical Chemistry Fundamentals ■ 2.5 Enthalpy
등압하에서 계의 부피가 자유롭게 변할 때, 열로 공급된 에너지 일부가 주위로 밀어내는 필요한 일로 변환된다. 𝑑𝑞 > 𝑑𝑢 ⇒ 𝑑𝑞=𝑑𝑢+𝑑𝑤 Fig When a system is subjected to a constant pressure and is free change its volume, some of the energy supplied as heat may escape back into the surroundings as work. In such case, the change in internal energy is smaller than the energy supplied as heat. (a) The definition of energy 𝑯 = 𝑼 + 𝑷𝑽 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑛𝑡ℎ𝑎𝑙𝑝ℎ𝑦 (𝑠𝑡𝑎𝑡𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛) Change in enthalpy : equal to the energy supplied as heat at constant pressure - Infinitesimal change 𝐻+𝑑𝐻=𝑈+𝑑𝑈+ 𝑝+𝑑𝑝 𝑉+𝑑𝑉 =𝑈 +𝑑𝑈 +𝑝𝑉+𝑉𝑑𝑝 +𝑃𝑑𝑉+𝑑𝑃𝑑𝑉 𝐻+𝑑𝐻=𝐻+𝑑𝑈+𝑃𝑑𝑉+𝑉𝑑𝑃에 𝑑𝐻=𝑑𝑈+𝑃𝑑𝑉+𝑉𝑑𝑃 ≒0 Energy Storage & Conversion Materials Lab.

13 𝑑𝐻=𝑑𝑞−𝑃𝑑𝑉+𝑃𝑑𝑉+𝑉𝑑𝑃=𝑑𝑞+𝑉𝑑𝑃
Physical Chemistry Fundamentals (열역학제 1법칙 : 𝑑𝑈=𝑑𝑞+𝑑𝑤) 𝑑𝐻=𝑑𝑞+𝑑𝑤+𝑃𝑑𝑉+𝑉𝑑𝑃 (Expansion work reversibly : 𝑑𝑤 = −𝑃𝑑𝑉) 𝑑𝐻=𝑑𝑞−𝑃𝑑𝑉+𝑃𝑑𝑉+𝑉𝑑𝑃=𝑑𝑞+𝑉𝑑𝑃 (Heating occurs at constant pressure : 𝑑𝑃=0) ∴𝒅𝑯 =𝒅𝒒 (at constant pressure, no additional work) ↳𝜟𝐇=𝒒𝐩 “ The change in enthalpy in equal to the energy supplied as heat ” Energy Storage & Conversion Materials Lab.

14 Physical Chemistry Fundamentals
(b) The measurement of an enthalpy change - Isobatic calorimeter : a calorimeter at constant pressure - Adiabatic calorimeter for combustion : 물리, 화학 변화에 수반되는 온도 변화 측정. - For perfect gas 𝐻 =𝑈 + 𝑃𝑉 =𝑈 + 𝑛𝑅𝑇 Δ𝐻 = Δ𝑈 + Δ𝑛𝑔𝑅𝑇 𝑒𝑥) 2𝐻2 𝑔 +𝑂2 𝑔 ⟶2𝐻2𝑂 𝑙 Δ𝑛𝑔=−3𝑚𝑜𝑙 Δ𝐻−Δ𝑈= Δ𝑛𝑔𝑅𝑇 =−3×𝑅𝑇 ≅−7.5𝐾𝐽 - For solids and liquids PVm ≅0 (small molar vol) 𝐻𝑚=𝑈𝑚 + 𝑃𝑉𝑚 ∴ Δ𝐻 =ΔU for solids and liquids Fig A constant-pressure flame calorimeter consists of this component immersed in a stirred water bath. Combustion occurs as a known amount of reactant is passed through to fuel the flame, and rise of temperature is monitored. Energy Storage & Conversion Materials Lab.

15 Physical Chemistry Fundamentals
(c) The variation of enthalpy with temperature Cp,m : molar heat capacity, intensive property Fig The constant-pressure heat capacity at a particular temperature is the slope of the tangent to a curve of the enthalpy of a system plotted against temperature (at constant pressure). For gases, at a given temperature the slope of enthalpy versus temperature is steeper than that of internal energy versus temperature and Cp,m is larger than Cv,m. - For Infinitesimal change of temperature 𝑑𝐻 = 𝐶𝑝𝑑𝑇 at constant pressure ↳ΔH = CpΔT - Change in enthalpy = energy supplied as heat 𝜟H = qp, qp=Cp𝜟𝐓 - Temp. dependence of heat capacity ( CP ) 1) For monoatomic perfect gas : temp range is small Cp=constant 2) Convenient approximate empirical expression : Cp,m=a+bT+c/T2 Table 2.2* (a, b, c) Energy Storage & Conversion Materials Lab.

16 Physical Chemistry Fundamentals
ex) ΔHm= ? For N2 25ºC→100ºC heating 𝑑𝐻 = 𝐶𝑝𝑑𝑇 𝐻1 𝐻2 𝑑𝐻= (𝑎+𝑏𝑇+ 𝑐 𝑇2 )𝑑𝑇 = ( 𝑥103 𝑇+ 1 𝑇2 (-0.5X10-5)dT=2.2KJmol-1 - Relationship between heat capacities of a perfect gas 𝐻 =𝑈 + 𝑃𝑉 =𝑈 + 𝑛𝑅𝑇 →𝑑𝐻 = 𝑑𝑈 + 𝑛𝑅𝑑𝑇 𝐶𝑝𝑑𝑇 = 𝐶𝑣𝑑𝑇 +𝑛𝑅𝑑𝑇 (÷𝑑𝑇) 𝐶𝑝−𝐶𝑣=𝑛𝑅 𝐶𝑝−𝐶𝑣,𝑚=𝑅 𝐹𝑜𝑟 1 𝑚𝑜𝑙𝑒 : 𝐶𝑝−𝐶𝑣=8 𝐽𝐾−1 𝑚𝑜𝑙−1 𝐶𝑣,𝑚=12 𝐽𝐾−1 𝑚𝑜𝑙−1 𝑓𝑜𝑟 𝑚𝑜𝑛𝑜𝑎𝑡𝑜𝑚𝑖𝑐 𝑔𝑎𝑠 Energy Storage & Conversion Materials Lab.

17 Physical Chemistry Fundamentals
■ 2.6 Adiabatic change 완전기체가 단열적으로 팽창할 경우 : (계가 외부에 일을 해주면 내부에너지가 감소하고 따라서 온도가 감소) - Adiabatic expansion of a perfect gas 1) Expansion at constant temp Δ𝑈=0 2) Temperature change at constant V 𝑞=0 ∵𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑑𝑈=𝑑𝑞+𝑑𝑤=𝑑𝑊𝑎𝑑 𝑑𝑈=𝐶𝑉𝑑𝑇 = −𝑃𝑑𝑉=𝑑𝑊𝑎𝑑 𝐶𝑉𝑑𝑇=− 𝑛𝑅𝑇 𝑉 𝑑𝑉→Cv𝑑𝑇/𝑇=−𝑛𝑅 𝑑𝑉 𝑉 𝐶𝑉 𝑇𝑖 𝑇𝑓 𝑑𝑇 𝑇 =−𝑛𝑅 𝑉𝑖 𝑉𝑓 𝑑𝑉 𝑉 ( 𝐶𝑉=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ) Cv ln 𝑇𝑓 𝑇𝑖 =−𝑛𝑅 ln 𝑉𝑓 𝑉𝑖 𝐶=𝐶𝑉/𝑛𝑅=𝐶𝑉,𝑚/𝑅 치환 ln 𝑇𝑓 𝑇𝑖 𝐶 = ln( 𝑉𝑖 𝑉𝑓 ) Fig To achieve a change of state from one temperature and volume to another temperature and volume, we may consider the overall change as composed of two steps. In the first step, the system expands at constant temperatures; there is no change in internal energy if the system consists of a perfect gas. In the second step, the temperate of the system reduced at constant volume. The overall change in internal energy is sum of the changes for the two steps. ∴ 𝑉𝑖 𝑉𝑓 = 𝑇𝑓 𝑇𝑖 𝑐→𝑇𝑓=𝑇𝑖( 𝑉𝑖 𝑉𝑓 )1/c 온도변화 계산 𝑜𝑟 𝑉𝑖𝑇𝑖𝑐 = 𝑉𝑓𝑇𝑓𝑐 Energy Storage & Conversion Materials Lab.

18 𝑇𝑖 𝑇𝑓 =( 𝑉𝑓 𝑉𝑖 )1/c 𝑃𝑖𝑉𝑖 𝑇𝑖 = 𝑃𝑓𝑉𝑓 𝑇𝑓 𝑃𝑖𝑉𝑖 𝑃𝑓𝑉𝑓 =( 𝑉𝑓 𝑉𝑖 )1/c
Physical Chemistry Fundamentals - Heat capacity ratio and adiabatics : 𝑇𝑖 𝑇𝑓 =( 𝑉𝑓 𝑉𝑖 )1/c 𝑃𝑖𝑉𝑖 𝑇𝑖 = 𝑃𝑓𝑉𝑓 𝑇𝑓 𝑃𝑖𝑉𝑖 𝑃𝑓𝑉𝑓 =( 𝑉𝑓 𝑉𝑖 )1/c 𝑷𝒊 𝑽 𝒊 𝟏+ 𝟏 𝒄 =𝑷𝒇 𝑽 𝒇 𝟏+ 𝟏 𝒄 1+ 1 𝐶 =1+ 𝑅 𝐶𝑣,𝑚 = 𝐶𝑣,𝑚+𝑅 𝐶𝑣,𝑚 = 𝐶𝑝,𝑚 𝐶𝑣,𝑚 =𝛾 (𝐶𝑝,𝑚 – 𝐶𝑣,𝑚= 𝑅) ∴ 𝑷𝒊 𝑽 𝒊 𝜸 =𝑷𝒇 𝑽 𝒇 𝜸 𝑤ℎ𝑒𝑟𝑒 𝐶𝑝,𝑚 𝐶𝑣,𝑚 >1 𝑜𝑟 𝑷𝑽𝜸=𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝑇𝑓=𝑇𝑖( 𝑉𝑖 𝑉𝑓 )1/c =𝑇𝑖( 𝑉𝑖 𝑉𝑓 )𝛾-1 or 𝑻𝒇 𝑻𝒊 = ( 𝑽𝒊 𝑽𝒇 )𝜸-1 (T vs V) 𝑃𝑖𝑉𝑖 𝑇𝑖 = 𝑃𝑓𝑉𝑓 𝑇𝑓 → 𝑉𝑖 𝑉𝑓 = 𝑇𝑖 𝑇𝑓 𝑃𝑓 𝑃𝑖 , 𝑇𝑓 𝑇𝑖 =( 𝑇𝑖 𝑇𝑓 )𝛾-1( 𝑃𝑓 𝑃𝑖 )𝛾-1 ∴ 𝑻𝒇 𝑻𝒊 =( 𝒑𝒇 𝒑𝒊 ) 𝜸−1 𝜸 (𝑇 𝑣𝑠 𝑃) Energy Storage & Conversion Materials Lab.

19 Summary 1) 𝑉𝑖 𝑉𝑓 = ( 𝑇𝑓 𝑇𝑖 ) c 2) 𝑇𝑓 𝑇𝑖 = ( 𝑉𝑖 𝑉𝑓 ) 𝛾−1
Physical Chemistry Fundamentals Summary 1) 𝑉𝑖 𝑉𝑓 = ( 𝑇𝑓 𝑇𝑖 ) c 2) 𝑇𝑓 𝑇𝑖 = ( 𝑉𝑖 𝑉𝑓 ) 𝛾−1 3) 𝑇𝑓 𝑇𝑖 = ( 𝑃𝑓 𝑃𝑖 ) 𝛾−1 𝛾 4) 𝑉𝑖 𝑉𝑓 = ( 𝑃𝑓 𝑃𝑖 ) 1 𝛾 𝑃∝ 1 𝑣 𝑃∝ 1 𝑣 𝛾 Fig An adiabatic depicts the variation of pressure with volume when a gas expands adiabatically. Note that the pressure declines more steeply for an adiabatic than it does for an isotherm because the temperature decreases in the former. Energy Storage & Conversion Materials Lab.

20 Physical Chemistry Fundamentals
Thermochemistry : study of energy transferred as heat during chemrxn (화학반응으로 생성되거나 소모되는 열에 관한 연구 ) − ΔU or ΔH →𝑞 𝑑𝑢𝑟𝑖𝑛𝑔 𝑟𝑥𝑛 − 𝐸𝑥𝑜𝑡ℎ𝑒𝑟𝑚𝑖𝑐 : Δ𝐻 <0 ( =𝑞𝑝 ) 𝐸𝑛𝑑𝑜𝑡ℎ𝑒𝑟𝑚𝑖𝑐 : Δ𝐻 >0 ( =𝑞𝑝 ) 2.7 Standard enthalpy change, Δ𝐻θ − Δ𝐻θ :Standard enthalpy change ; the change in enthalpy for a process in which the initial and final substances are in their standard states − Standard state 𝑎 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑡 𝑎 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝑡𝑒𝑚𝑝 𝑖𝑠 𝑖𝑡𝑠 𝒑𝒖𝒓𝒆 𝒇𝒐𝒓𝒎 𝑎𝑡 1 𝑏𝑎𝑟 - Standard enthalpy change for a reaction or a physical process is the difference between the products in their standard states and the reactants in their standard states. e.g.) Standard enthalpy of vaporization, Δ𝑣𝑎𝑝𝐻θ : the enthalpy change per mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar H2O (l) → H2O(g) ; Δ𝑣𝑎𝑝𝐻θ (373K) = kJ mol-1 Energy Storage & Conversion Materials Lab.

21 Physical Chemistry Fundamentals
Enthalpies of physical change Standard enthalpy of transition; standard enthalpy change during physical change Δ𝑡𝑟𝑠𝐻θ (Table 2.3, 표준전이 엔탈피) (물리적 상태변화 시 (기, 액, 고체의 상 변화) 를 동반하는 표준 엔탈피변화) Δ𝑣𝑎𝑝𝐻θ, Δ𝑓𝑢𝑠𝐻θ, Δ𝑠𝑢𝑏𝐻θ Enthalpy : state function e.g.) H2O (s) → H2O (g) Δ𝑠𝑢𝑏𝐻θ H2O (s) → H2O (l) Δ𝑓𝑢𝑠𝐻θ H2O (l) → H2O (g) Δ𝑣𝑎𝑝𝐻θ ∴ Δ𝑠𝑢𝑏𝐻θ = Δ𝑓𝑢𝑠𝐻θ + Δ𝑣𝑎𝑝𝐻θ Forward and revese processes : Δ𝐻θ (A→B) = - Δ𝐻θ (B→A) ex) Δ𝑣𝑎𝑝𝐻θ = 44 kJ mol-1, Δ𝑐𝑜𝑛𝐻θ = - 44 kJ mol-1 Enthalpies of transition: Table 2.4 Energy Storage & Conversion Materials Lab.

22 Physical Chemistry Fundamentals
(b) Enthalpies of chemical change ; chemical rxn 에 수반되는 엔탈피 Enthalpy change 1) Thermochemical eq’n: a combination of chemical eq’n and corresponding change in standard enthalpy CH4 (g)+2O2 𝑔 →𝐶𝑂2 𝑔 +2𝐻2𝑂 𝑙 , Δ𝐻θ = -890 kJ (표준상태의 반응물이 표준상태의 생성물이 될 때의 엔탈피 변화) Pure, separate reactants in their standard states → pure, separate products in their standard states 2) Chemical eq’n and standard rxn enthalpy, Δ𝑟𝐻θ CH4 (g)+2O2 𝑔 →𝐶𝑂2 𝑔 +2𝐻2𝑂 𝑙 , Δ𝑟𝐻θ = -890 kJ mol-1 standard rxn enthalpy For 2A + B → 3C + D Δ𝑟𝐻θ = {3Hmθ (C) + Hmθ (D)} – {2 3Hmθ (A) + Hmθ (B)} where Hmθ (J) : standard molar enthalpy of J Energy Storage & Conversion Materials Lab.

23 Physical Chemistry Fundamentals
In general, Δ𝑟𝐻θ = ∑ν Hmθ - ∑ν Hmθ = ∑νi Hmθ products reactants i Standard enthalpy of combustion: Δ𝑐𝐻θ standard rxn enthaply for the complete oxidation of an organic compound to CO2 gas and liquid H2O. (c) Hes’s Law; The standard enthalpy of an overall rxn is the sum of the standard enthalpies of the individual rxns into which a rxn may be divided. Individual rxn : maybe hypothetical rxns but chemical eq’ns should balance. Energy Storage & Conversion Materials Lab.

24 Physical Chemistry Fundamentals
EX) Standard enthalpy of combustion of propene? Δ𝑟𝐻θ (𝑘𝐽 𝑚𝑜𝑙−1) C3H6 (g) + H2 (g) → C3H8 (g) C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l) ,220 H2O (l) → H2 (g) + ½ O2 (g) (Table 2.7) C3H6 (g) + 9/2O2 (g) → 3CO2 + 3H2O (l) ,058 2.8 Standard enthalpies of formation, Δ𝑓𝐻θ : standard enthalpy of formation, Δ𝑓𝐻θ , of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference states Reference state of element : its most stable state at the specified temp and 1 bar; e.g.; at 298K, N2: gas, Hg : liq, carbon: graphite, tin: white metallic form Energy Storage & Conversion Materials Lab.

25 Physical Chemistry Fundamentals
Standard enthalpy of formation of liq benzene 6 C (graphite, S) + H2 (g) → C6H6 (l), Δ𝑓𝐻θ = 49 kJ mol-1 N2 (g) → N2 (g), Δ𝑓𝐻θ = 0 kJ mol-1 (standard enthalpy of formation of elements in their reference state are zero at all temp) The reaction enthalpy in reference in terms of enthalpies of formation Reactions: by decomposing the reactants into their elements and then forming those elements into their products Δ𝑟𝐻θ = ∑ν Δ𝑓𝐻θ - ∑ν Δ𝑓𝐻θ = ∑νJ Δ𝑓𝐻θ (J) Where, νJ : stoichiometric number Ex) 2NH3 (l) + 2NO (g) → H2O2 (l) + 4 N2 (g) products reactants J Energy Storage & Conversion Materials Lab.

26 Physical Chemistry Fundamentals
2.9 The temperature dependence of reaction enthalpies - standard reaction enthalpy를 여러 온도에서 측정하는 방법 ; may be calculated from heat capacities (more accurate) and reaction enthalpy From eq’n 2.23 a; H(T2) = H(T1) + 𝑇1 𝑇2 𝐶𝑝𝑑𝑇 for each substance Δ𝑟𝐻θ(T2) = Δ𝑟𝐻θ (T1) + 𝑇1 𝑇2 Δ𝑟𝐶𝑝θ𝑑𝑇 Kirchhoff’s law Where Δ𝑟𝐶𝑝θ : the difference of the molar heat capacities of products and reactants under standard conditions Δ𝑟𝐶𝑝θ = 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 ν𝐶𝑝,𝑚θ − 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 ν𝐶𝑝,𝑚θ = 𝐽 ν𝐶𝑝,𝑚θ(𝐽) Fig An illustration of the content of Kirchhoff’s law. When the temp is increased, the enthalpy of the products and the reactants both increase, but may do so to different extents. In each case, the change in enthalpy depends on the heat capacities of the substances. The change in reaction enthalpy reflects the difference in the changes of the enthalpies. Energy Storage & Conversion Materials Lab.

27 Physical Chemistry Fundamentals
■ Exact and inexact differential - 𝑆𝑡𝑎𝑡𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 : 𝑢, 𝐻 Δ𝑢= 𝑖,𝑝𝑎𝑡ℎ 𝑓 𝑑𝑞 : 𝐼𝑛𝑒𝑥𝑎𝑐𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑞, 𝑑𝑤로 표시 - 𝑃𝑎𝑡ℎ 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 : 𝑤, 𝑞 𝑞 = 𝑖, 𝑝𝑎𝑡ℎ 𝑓 𝑑𝑞 𝑖𝑛𝑒𝑥𝑎𝑐𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑞, 𝑑𝑤로 표시 (경로함수를 더 명확하게 표시) ※ U는 path에 관계없이 같은 w와 q에 의존함 Energy Storage & Conversion Materials Lab.

28 Physical Chemistry Fundamentals
■ changes in internal energy - For closed system : 𝑢=𝑓( 𝑉, 𝑇, 𝑃 ) 1 state of eq’n ⊐ → Two independent variables V and T, P and T, or P and V ↳u=f( V ,T ) (a) General conditions u; Exact differential 𝑈=𝑓(𝑉,𝑇) - 𝑉→𝑉+ 𝑑𝑉 ⇒ U→(𝑈)=U+dU 𝑈 ′ =U+ 𝜕𝑈 𝜕𝑣 TdV - 𝑇 →𝑇+𝑑𝑇 ⇒U→( U ′ =U+dU) U ′ =U+ 𝜕𝑈 𝜕𝑇 vdT - Both V and T change infinitesimally 𝑈 ′ =U+ 𝜕𝑈 𝜕𝑣 TdV+ 𝜕𝑈 𝜕𝑇 vdT ∴𝑑𝑈= ( 𝜕𝑈 𝜕𝑉 )TdV+ 𝜕𝑈 𝜕𝑇 vdT Energy Storage & Conversion Materials Lab.

29 Physical Chemistry Fundamentals
Cv= ( 𝜕𝑈 𝜕𝑇 )v 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡−𝑣𝑜𝑙 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝜋v= ( 𝜕𝑈 𝜕𝑇 )v : 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ( ∵압력과 같은 단위 가짐 ) ∴𝑑𝑈=𝜋T𝑑𝑉+𝐶𝑉𝑑𝑇 Energy Storage & Conversion Materials Lab.

30 Physical Chemistry Fundamentals
(b) The Joule experiment : ΔU ( change in internal energy ) 수조온도 변화 없었음 𝑤=0 (진공 속 팽창) , 𝑞=0 (ΔT=0) Joule’s conclusion; 가스가 등온적으로 팽창 시 U는 일정하며, 따라서 𝜋T=0 ∴ Δ𝑈=0, (ΔU=q+w) Fig A schematic diagram of the apparatus used by Joule in attempt to measure the change in internal energy when a gas expands isothermally. The heat adsorbed by the gas is proportional to the change in temperature of the bath. ↳그러나 실제 온도 변화 ∵장치의 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 大 (c) Changes in internal energy at constant pressure U=f(T) at constant pressure 𝑑𝑈=𝜋Td𝑉+𝐶𝑉𝑑𝑇 ÷dT at constant pressure ( 𝝏𝑼 𝝏𝑻 )p= 𝝅T ( 𝝏𝑽 𝝏𝑻 )p+𝑪v Let) 𝛼= 1 𝑉 ( 𝜕𝑉 𝜕𝑇 )p ; Volume expansion coefficient Energy Storage & Conversion Materials Lab.

31 Physical Chemistry Fundamentals
Let) 𝛫𝑇= - 1 𝑉 ( 𝜕𝑉 𝜕𝑝 )T : Isothermal compressibility (등온압축 계수) 압력증가에 따라 dV < 0가 됨, 따라서 (–)를 곱하여 (+)로 만듦 ( 𝝏𝑼 𝝏𝑻 )p= 𝜶𝝅TV+𝑪v : general eqn’s for closed system For perfect gas ; 𝜋T = 0 ; ( 𝝏𝑼 𝝏𝑻 )p= 𝑪v Cp−Cv= ( 𝜕𝐻 𝜕𝑇 )p−( 𝜕𝑉 𝜕𝑇 )p 𝐻=𝑈+𝑃𝑉=𝑈+𝑛𝑅𝑇 → ( 𝜕𝐻 𝜕𝑇 )p= ( 𝜕𝑈 𝜕𝑇 )p +𝑛𝑅 Cp−Cv= ( 𝜕𝑈 𝜕𝑇 )p+nR−( 𝜕𝑈 𝜕𝑇 )p= 𝒏𝑹 Energy Storage & Conversion Materials Lab.

32 Physical Chemistry Fundamentals
- The general case ; Cp−Cv=( 𝜕𝐻 𝜕𝑇 )p−( 𝜕𝑈 𝜕𝑇 )V= ( 𝜕𝑈 𝜕𝑇 )p + ( 𝜕(𝑃𝑉) 𝜕𝑇 )p− ( 𝜕𝑈 𝜕𝑇 )v (( 𝜕𝑈 𝜕𝑇 )p= ( 𝜕𝑈 𝜕𝑇 )V) : ( 𝜕𝑈 𝜕𝑇 )p = 𝛼𝜋TV+Cv=𝛼𝜋TV+( 𝜕𝑈 𝜕𝑇 )v 즉 ( 𝜕𝑈 𝜕𝑇 )p−( 𝜕𝑈 𝜕𝑇 )v=𝛼𝜋TV ① [( 𝜕(𝑃𝑉) 𝜕𝑇 )p]=P ( 𝜕𝑉 𝜕𝑇 )P=𝛼PV, ② 𝛼= 1 𝑉 ( 𝜕𝑉 𝜕𝑇 )P ∴Cp−Cv=𝛼𝜋TV+ 𝛼PV= 𝛼(𝜋T+P)V 𝜋T=T( 𝜕𝑃 𝜕𝑇 )v−P From section 3-8 Cp−Cv=𝛼TV( 𝜕𝑃 𝜕𝑇 )v - Euler’s chain rule 이용 ( 𝜕𝑃 𝜕𝑇 )v( 𝜕𝑇 𝜕𝑉 )p( 𝜕𝑉 𝜕𝑃 )T=− 1, ( 𝜕𝑃 𝜕𝑇 )v=- 1 ( 𝜕𝑇 𝜕𝑉 )p( 𝜕𝑉 𝜕𝑃 )T =- ( 𝜕𝑉 𝜕𝑇 )p ( 𝜕𝑉 𝜕𝑃 )T = 𝛼 𝐾𝑇 ∴ Cp− Cv = 𝜶𝑻V 𝜶 𝑲𝑻 = 𝜶𝟐𝑻𝑽 𝑲𝑻 For perfect gas : 𝛼= 1 𝑇 , KT = 1 𝑃 ∴ Cp− Cv= 1 𝑇2 𝑇𝑉 1 𝑃 = 𝑃𝑉 𝑇 = nR Energy Storage & Conversion Materials Lab.

33 Physical Chemistry Fundamentals
− 𝐹𝑜𝑟 𝑠𝑜𝑙𝑖𝑑𝑠 𝑎𝑛𝑑 𝐿𝑖𝑞𝑢𝑖𝑑,𝛼 (= 1 𝑉 ( 𝜕𝑉 𝜕𝑇 )P) : small ∴Cp− Cv ≅ V ⇒ Cp ≅ Cv Not always, KT : small ⇒ 𝛼2 𝐾𝑇 : Large − 𝐹𝑜𝑟 𝑠𝑜𝑙𝑖𝑑𝑠 𝑎𝑛𝑑 𝐿𝑖𝑞𝑢𝑖𝑑 : KT (=− 1 𝑉 ( 𝜕𝑉 𝜕𝑃 )T) is small ⇒ 𝛼2 𝐾𝑇 large 𝐴 𝑔𝑟𝑒𝑎𝑡 𝑑𝑒𝑎𝑙 𝑜𝑓 𝑤𝑜𝑟𝑘 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑑𝑜𝑛𝑒 𝑡𝑜 𝑝𝑢𝑙𝑙 𝑎𝑡𝑜𝑚𝑠 𝑎𝑝𝑎𝑟𝑡 𝑓𝑟𝑜𝑚 𝑜𝑛𝑒 𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑎𝑠 𝑠𝑜𝑙𝑖𝑑 𝑒𝑥𝑝𝑎𝑛𝑑𝑠 𝑡ℎ𝑜𝑢𝑔ℎ 𝑜𝑛𝑙𝑦 𝑎 𝑙𝑖𝑡𝑡𝑙𝑒 𝑤𝑜𝑟𝑘 𝑛𝑒𝑒𝑑 𝑏𝑒 𝑑𝑜𝑛𝑒 𝑡𝑜 𝑝𝑢𝑠ℎ 𝑏𝑎𝑐𝑘 𝑡ℎ𝑒 𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑒 ex) For H2O at 25ºC ( Cp,m = 75.3 JK-1 mol-1 ) ( Cv,m = 74.8 JK-1 mol-1 ) In some cases, difference in Cp,m-Cv,m is 30% ■ 2.12 The Joule-Thomson effect H : state function, dH : exact differential dH = ( 𝜕𝐻 𝜕𝑃 )TdP + ( 𝜕𝐻 𝜕𝑇 )PdT H=f(P,T) - Euler’s chain rule 이용 ( 𝜕𝐻 𝜕𝑃 )T( 𝜕𝑃 𝜕𝑇 )H( 𝜕𝑇 𝜕𝐻 )P=-1 ( 𝜕𝐻 𝜕𝑃 )T=- 1 ( 𝜕𝑃 𝜕𝑇 )H( 𝜕𝑇 𝜕𝐻 )P = -( 𝜕𝑇 𝜕𝑃 )H( 𝜕𝐻 𝜕𝑇 )P 𝐷𝑒𝑓𝑖𝑛𝑒 𝝁≡( 𝝏𝑻 𝝏𝑷 )H : Joule-Thomson coefficient ∴ ( 𝜕𝐻 𝜕𝑃 )T=−𝜇Cp 𝒅𝑯=−𝝁𝑪𝒑𝒅𝑷+𝑪𝒑𝒅𝑻 Energy Storage & Conversion Materials Lab.

34 Physical Chemistry Fundamentals
(a) Observation of the Joule-Thomson effect - Joule-Thomson coefficient : liquafaction of gases와 관련 - 전과정 : Isothermally (등엔탈피 과정) 앞쪽의 온도가 낮고 고압 쪽에 온도가 높다 이 온도차는 압력에 비례함 ⇒ 이러한 (“단열팽창”)으로 인한 냉각은 Joule –Thomson effect 라 함 Isenthalpic expansion : ( 𝜕𝑇 𝜕𝑃 )H=𝜇 기체가 등엔탈피 (단열적)적으로 팽창할 때 압력변화에 따른 온도 변화 비 Fig The apparatus used for measuring the Joule –Thomson effect. The gas expands through the porous barrier, which acts a throttle, and the whole apparatus is thermally insulated. As explained in the text, this arrangement corresponds to an isenthalpic expansion (expansion at constant enthalpy). Whether the expansion results in heating or cooling of the gas depends on the conditions. Energy Storage & Conversion Materials Lab.

35 Physical Chemistry Fundamentals
Adiabatic changes : 𝑞=0, ∴ Δ𝑈=𝑤 1) Compressed isothermally : 등온 압축 Pi → Pi, Vi → 0, Ti → Ti 𝑊1=𝑃𝑖𝑉𝑖( =−𝑃𝑖(0 − 𝑉𝑖) ) 2) Expand isothermally : 등온 팽창 𝑊2=− 𝑃𝑓𝑉𝑓 ( =− 𝑃𝑓 (𝑉𝑓− 0) ) ∴ 𝑊=𝑊1+𝑊2=𝑃𝑖𝑉𝑖− 𝑃𝑓𝑉𝑓=Δ𝑈=𝑈𝑓 −𝑈𝑖 ∴ 𝑈𝑓+𝑃𝑓𝑉𝑓=𝑈𝑖+𝑃𝑖𝑉𝑖 Fig The thermodynamic basis of Joule-Thomson expansion. The pistons represent the upstream and downstream gases, which maintain constant pressures either sides of the throttle. The transition from the top diagram to bottom diagram, which represents the passage of a given amount of gas through the throttle, occurs without change of enthalpy. ∴ 𝐻𝑓=𝐻𝑖 ; 등엔탈피과정 Energy Storage & Conversion Materials Lab.

36 Physical Chemistry Fundamentals
modern method of measuring 𝜇: 간접적으로 𝜇 측정 𝝁T=( 𝝏𝑯 𝝏𝑷 )T=−𝝁Cp Euler’s chain rule 이용 ( 𝜕𝐻 𝜕𝑃 )T( 𝜕𝑃 𝜕𝑇 )H( 𝜕𝑇 𝜕𝐻 )P=-1 ( 𝜕𝐻 𝜕𝑃 )T=- 1 ( 𝜕𝑃 𝜕𝑇 )H( 𝜕𝑇 𝜕𝐻 )P =-( 𝜕𝑇 𝜕𝑃 )H( 𝜕𝐻 𝜕𝑇 )P= −𝜇Cp 팽창에 수반되는 냉각을 상쇄시키기 위해 Heater로 가해주는 열량 측정 : 이 열은 기체의 ∆H값과 같으며 ∆P를 측정하므로 ∆H/∆P값을 구하고 ∆p →0으로보내 𝜇T = ( 𝜕𝐻 𝜕𝑃 )T를 구한다. Fig A schematic diagram of the apparatus used for measuring the isothermal Joule-Thomson coefficient. The electrical heating required to offset the cooling arising from expansion is interpreted as ∆H and used to calculate ( 𝝏𝑯 𝝏𝑷 )T, which is then converted to 𝜇 as explained in the text. ○ Real gas 𝜇= f (identity of gas, pressure, molecular interaction) 𝜇>0 :기체 팽창시 냉각됨 ( dP<0→dT<0 ) Coolilng 𝜇<0 :기체 팽창시 가열 됨 ( dP<0→𝑑𝑇>0 ) Heating ( 𝜕𝑃 𝜕𝑇 )H Energy Storage & Conversion Materials Lab.

37 Physical Chemistry Fundamentals
한 온도에서 팽창 시에 가열효과 (𝜇<0)를나타내는 기체는 그 온도를 inversion Temp, TI이하로 낮추면 냉각효과 (𝜇>0)를나타낸다. Fig The sign of the Joule-Thomson coefficient, , depends on the conditions. Inside the boundary, the blue area, it is positive and outside is negative. The temperature corresponding to the boundary at a given pressure is the “inversion temp.” of the gas at that pressure. Fig The principle of the Linde refrigerator is shown in this diagram. The gas is recirculated, and, so long as it is beneath its inversion temperature, it cools on expansion through the throttle. The cooled gas cools the high-pressure gas, which cools still further as it expands. Eventually liquefied gas drips from the throttle. Fig The inversion temperatures for the three real gases, nitrogen, hydrogen, and helium. - For perfect gas : 𝜇=0 ;𝐽𝑜𝑢𝑙𝑒−𝑇ℎ𝑜𝑚𝑠𝑜𝑛 팽창을 해도 온도가 변하지 않는다. - Real gas : P → 0 ( 실제기체상태 방정식 → 완전기체 상태 방정식 ) 𝜇≠0 : “분자간 인력”의 영향 Energy Storage & Conversion Materials Lab.