10). thermodynamics (finished)

10). thermodynamics (finished)

• 1.
  x Thermodynamics
• 2.
  Thermodynamics Laws of Thermodynamics ZeroLaw First Law Second Law Third Law State 𝑃[𝑃𝑎], 𝑉[𝑚3 ], 𝑇[𝐾] 1st Law 𝑈 = 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 To change 𝑈1 to 𝑈2, there are 2 ways; (i) Change Q (ii) Change W ***When energy (𝑄) is supplied to a system, the system will do work (𝑊) In reality, the work done is lesser than the energy supplied. 𝑄 𝑖𝑠 𝑠𝑢𝑝𝑝𝑙𝑦 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑄 > 0 𝑊 𝑖𝑠 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑊 > 0 If the system releases the energy 𝑄 < 0 If work is done on the system 𝑊 < 0 𝑠𝑦𝑠𝑡𝑒𝑚 𝑄 𝑊 ∆𝑈 = 𝑄 − 𝑊 𝑈 = 𝐷𝑂𝐹 2 𝑁𝑘𝑇 Internal energy Work done Work done by gas; 𝑊 > 0 Work done on gas; W < 0 Since, ∆𝑊 = 𝐹∆𝑠 𝐹 = 𝑑𝑊 𝑑𝑠 Since, 𝑃 = 𝐹 𝐴 , 𝐹 = 𝑃𝐴 𝑑𝑊 = 𝑃 𝐴𝑑𝑠 𝑑𝑊 = 𝑃𝑑𝑉 𝑊 = 𝑑𝑊 = 𝑃𝑑𝑉 𝑉 𝑓 𝑉 𝑖 𝑊 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑠𝑦𝑠𝑡𝑒𝑚 𝑷-𝑽 diagram ***the area under the graph is work down 𝑊 = 𝑃𝑑𝑉 𝑉 𝑓 𝑉 𝑖 𝑃 𝑉 𝑉𝑓𝑉𝑖 𝑊 = 𝑃(𝑉𝑓 − 𝑉𝑖)
• 3.
  ∆𝑈 = 𝑄− 𝑊 Process Condition Isochoric Isovolumetric Constant volume 𝑊 = 0 Path dependent for 𝑸 & 𝑾 ; although the final and initial states (points) are the same, work done may not be equal. Path independent for ∆𝑼 ; it is state function ; a state function is a property of a system that depends only on the current state of the system, not on the way in which the system acquired that state (independent of path) ∆𝑈 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑝𝑎𝑡ℎ 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑊𝑖→𝑓→𝑖 = 𝑊𝑖→𝑓 + 𝑊𝑓→𝑖 𝑃 𝑉 𝑖 𝑓 𝑊𝑓→𝑖 < 0 𝑃 𝑉 𝑖 𝑓 𝑊𝑖→𝑓 > 0 𝑃 𝑉 𝑖 𝑓 𝑊𝑖→𝑓→𝑖 > 0 𝑊 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑔𝑎𝑠 = 𝑎𝑟𝑒𝑎 𝑖𝑛𝑠𝑖𝑑𝑒 𝑊 > 0 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑊 < 0 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑎𝑛𝑡𝑖 − 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑃 𝑉 𝑖 𝑓 𝑃 𝑉 ∆𝑈 = 𝑄
• 4.
  ∆𝑈 = 𝑄− 𝑊 Process Condition Isobaric Constant pressure ∆𝑃 = 0 Isothermal Constant temperature ∆𝑈 = 0 𝑃𝑉 = 𝑁𝑘𝑇 → 𝑃 = 𝑁𝑘𝑇 𝑉 𝑊 = 𝑃𝑑𝑉 𝑉 𝑓 𝑉𝑖 = 𝑁𝑘𝑇 𝑉 𝑑𝑉 𝑉 𝑓 𝑉𝑖 = NkT 1 𝑉 𝑑𝑉 𝑉 𝑓 𝑉𝑖 = 𝑁𝑘𝑇𝑙𝑛 𝑉𝑓 𝑉𝑖 = 𝑁𝑘𝑇𝑙𝑛 𝑃𝑖 𝑃𝑓 𝑃 𝑉 𝑊 = 𝑃∆𝑉 𝑃 𝑉 ∆𝑈 = 0, Q = W 𝑇1 𝑇2 𝑇3 𝑊 = 𝑁𝑘𝑇 𝑙𝑛 𝑉𝑓 𝑉𝑖 = 𝑁𝑘𝑇 𝑙𝑛 𝑃𝑖 𝑃𝑓 ∆𝑈 = 𝑄 − 𝑊 Process Condition Adiabatic Constant heat 𝑄 = 0 Heat capacity 𝐽 𝐾 Specific heat capacity 𝐽 𝑘𝑔 𝐾 Molar heat capacity 𝐽 𝑚𝑜𝑙 𝐾 Specific heat capacity at constant V, (𝒄 𝑽) ∆𝑄 = 𝑀𝑐∆𝑇 𝑐 = 1 𝑀 ∆𝑄 ∆𝑇 = 1 𝑀 𝜕𝑄 𝜕𝑇 At constant V, ∆𝑈 = 𝑄 𝑐 𝑉 = 1 𝑀 𝜕𝑄 𝜕𝑇 𝑉 = 1 𝑀 𝑑𝑈 𝑑𝑇 ***For an ideal gas, 𝑈 = 3 2 𝑁𝑘𝑇 ∴ 𝑐 𝑉 = 1 𝑀 𝑑𝑈 𝑑𝑇 = 1 𝑀 𝑑 𝑑𝑇 3 2 𝑁𝑘𝑇 = 3𝑁𝑘 2𝑀 = 3𝑁𝑘 2 𝑁𝑚 = 3𝑘 2𝑚 𝑀 = 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑚𝑜𝑙𝑒 Specific heat capacity at constant P, (𝒄 𝑷) ∆𝑄 = 𝑀𝑐 𝑉∆𝑇 𝑐 𝑃 = 1 𝑀 𝜕𝑄 𝜕𝑇 𝑃 = 1 𝑀 𝜕 𝜕𝑇 𝑈 + 𝑊 𝑃 = 1 𝑀 𝑑𝑈 𝑑𝑇 + 𝜕 𝜕𝑇 𝑃𝑉 𝑇3 > 𝑇2 > 𝑇1 𝑐 𝑉 = 3𝑘 2𝑚
• 5.
  ∆𝑈 = 𝑄− 𝑊 Process Condition Adiabatic Constant heat 𝑄 = 0 Specific heat capacity at constant P, (𝒄 𝑷) ∆𝑄 = 𝑀𝑐 𝑉∆𝑇 𝑐 𝑃 = 1 𝑀 𝑑𝑈 𝑑𝑇 + 𝜕 𝜕𝑇 𝑃𝑉 𝑐 𝑃 = 1 𝑀 𝑑𝑈 𝑑𝑇 + 𝜕 𝜕𝑇 𝑁𝑘𝑇 𝑐 𝑃 = 1 𝑀 𝑑𝑈 𝑑𝑇 + 𝑁𝑘 = 1 𝑀 𝑑𝑈 𝑑𝑇 + 𝑁𝑘 𝑁𝑚 Since 𝑐 𝑉 = 1 𝑀 𝑑𝑈 𝑑𝑇 , 𝒄 𝑷 = 𝒄 𝑽 + 𝒌 𝒎 For an ideal gas, 𝑐 𝑉 = 3𝑘 2𝑚 , 𝑐 𝑃 = 3𝑘 2𝑚 + 𝑘 𝑚 ***We define 𝜸 = 𝒄 𝑷 𝒄 𝑽 , (adiabatic index) γ = 𝑐 𝑉 + 𝑘 𝑚 𝑐 𝑉 = 1 + 𝑘 𝑚𝑐 𝑉 For an ideal gas, γ = 3𝑘 2𝑚 + 𝑘 𝑚 3𝑘 2𝑚 = 5 3 𝑐 𝑃 = 5𝑘 2𝑚 𝛾 = 𝑐 𝑃 𝑐 𝑉 ∆𝑈 = 𝑄 − 𝑊 Process Condition Adiabatic Constant heat 𝑄 = 0 Since 𝑄 = 0, ∆𝑈 = −𝑊 And since ∆𝑈 is a state function (path independent), ∆𝑈 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐= ∆𝑈𝑖𝑠𝑜𝑐ℎ𝑜𝑟𝑖𝑐 when the initial states and final states of the 2 processes are the same. ∴ 𝑑𝑈 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 = 𝑑𝑈𝑖𝑠𝑜𝑐ℎ𝑜𝑟𝑖𝑐 = 𝑄 = 𝑀𝑐 𝑉 𝑑𝑇 𝑑𝑈 = −𝑑𝑊 𝑀𝑐 𝑉 𝑑𝑇 = −𝑃𝑑𝑉 𝑑𝑇 = − 𝑃 𝑀𝐶 𝑣 𝑑𝑉 From an ideal gas equation, 𝑃𝑉 = 𝑁𝑘𝑇 𝑑 𝑃𝑉 = 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑁𝑘𝑑𝑇 ∴ 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑁𝑘 − 𝑃 𝑀𝐶 𝑣 𝑑𝑉 Since 𝑀 = 𝑁𝑚, 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = − 𝑘 𝑚𝐶 𝑣 𝑃𝑑𝑉 Since, γ = 𝑐 𝑉+ 𝑘 𝑚 𝑐 𝑉 = 1 + 𝑘 𝑚𝑐 𝑉 𝑘 𝑚𝑐 𝑉 = 𝛾 − 1 ∴ 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = − 𝛾 − 1 𝑃𝑑𝑉 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 1 − 𝛾 𝑃𝑑𝑉 𝑉𝑑𝑃 = −𝛾𝑃𝑑𝑉 𝑑𝑃 𝑃 = −𝛾 𝑑𝑉 𝑉 ∴ ln 𝑃 = −𝛾 ln 𝑉 + 𝑐 ln 𝑃 + 𝛾 ln 𝑉 = 𝑐
• 6.
  ∆𝑈 = 𝑄− 𝑊 Process Condition Adiabatic Constant heat 𝑄 = 0 ln 𝑃 + 𝛾 ln 𝑉 = 𝑐 ln 𝑃𝑉 𝛾 = 𝑐 Since 𝑐 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, ∴ 𝑃𝑉 𝛾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Since 𝑃𝑉 = 𝑁𝑘𝑇, 𝑃 = 𝑁𝑘𝑇 𝑉 𝑁𝑘𝑇 𝑉 𝑖 𝑉𝑖 𝛾 = 𝑁𝑘𝑇 𝑉 𝑓 𝑉𝑓 𝛾 ***the graph of adiabatic process is more steeper than the graph of isothermal process. 𝑃𝑖 𝑉𝑖 𝛾 = 𝑃𝑓 𝑉𝑓 𝛾 𝑇𝑖 𝑉𝑖 𝛾−1 = 𝑇𝑓 𝑉𝑓 𝛾−1 𝑃 𝑉 𝑖 𝑓 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 ∆𝑈 = 𝑄 − 𝑊 Process Condition Free expansion ∆𝑈 = 𝑄 = 𝑊 = 0 ***there is no graph plotted joining the initial and final point because there is no work done (area under graph). ***Note that Free expansion is the subset of Adiabatic process. Cyclical process ∆𝑈 = 0 ***the initial and final point is the same. 𝑇𝑖 = 𝑇𝑓 𝑃𝑖 𝑉𝑖 = 𝑃𝑓 𝑉𝑓 𝑃 𝑉 𝑖 𝑓 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑃 𝑉 𝑖 𝑓 𝑊𝑖→𝑓→𝑖 > 0
• 7.
  ∆𝑈 = 𝑄− 𝑊 Process Condition Cyclical process ∆𝑈 = 0 𝑊 > 0 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑊 < 0 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑎𝑛𝑡𝑖 − 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 Since 𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊 𝑎𝑛𝑑 𝑑𝑈 = 0, 𝑑𝑄 = 𝑑𝑊 2nd Law of Thermodynamics Reversible and Irreversible process Most of the processes in Thermodynamics are reversible. 2 mains irreversible processes (i). Transfer of heat energy between 2 systems with different temperatures. Such as; isobaric and isochoric (ii). Free expansion (not quasi-static) ***isothermal and adiabatic are reversible Entropy 𝑺 & Entropy change ∆𝑺 ;Entropy shows the property of the state which is disorder. For example, 𝑆 𝑔𝑎𝑠 > 𝑆𝑙𝑖𝑞𝑢𝑖𝑑 > 𝑆𝑠𝑜𝑙𝑖𝑑 For Entropy change, ∆𝑆 = 𝑆𝑓 − 𝑆𝑖 ∆𝑆 = 0 → 𝑆𝑖 = 𝑆𝑓 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Suppose there is an initial state, 𝑑𝑆 = 𝑑𝑄 𝑇 ***A tiny change in entropy is equal to the tiny input energy over a period of time. ***Entropy is a state function, meaning that Entropy change does not depend on the path it takes. Specifically for isothermal, 𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ∴ ∆𝑆 = 1 𝑇 𝑑𝑄 𝑓 𝑖 = 𝑄 𝑇 From the 1st Law of Thermodynamics, ∆𝑈 = 𝑄 − 𝑊 For isothermal ∆𝑈 = 0, 𝑄 = 𝑊 ∴ ∆𝑆 = 𝑄 𝑇 = 𝑊 𝑇 Since 𝑊 = 𝑁𝑘𝑇 𝑙𝑛 𝑉 𝑓 𝑉 𝑖 , ∆𝑆 = 𝑁𝑘𝑇 𝑙𝑛 𝑉𝑓 𝑉𝑖 𝑇 ∆𝑆 = 𝑆𝑓 − 𝑆𝑖 = 𝑑𝑄 𝑇 𝑓 𝑖 ∆𝑆 = 𝑁𝑘 𝑙𝑛 𝑉𝑓 𝑉𝑖
• 8.
  2nd Law ofThermodynamics For any closed system, 1. If 𝑠 of the system increases ∆𝑆 > 0 , it is irreversible process. 2. If 𝑠 of the system is constant ∆𝑆 = 0 , it is reversible process. In Thermodynamics, a closed system means that there is no transfer of heat energy between the system and surroundings. ***Only Adiabatic is a closed system because 𝑄 = 0. However, to increase the boundary of the application of the 2nd Law, we must consider, Closed system = system + surroundings so that there is no transfer of heat energy in and out of our closed system. For Reversible processes, ∆𝑺 = 𝟎 ; isothermal and adiabatic Isothermal, ∆𝑈 = 0, 𝑄 = 𝑊 ∆𝑆𝑐𝑙𝑜𝑠𝑒𝑑 𝑠𝑦𝑠𝑡𝑒𝑚≥ 0 ∆𝑆 = 𝑑𝑄 𝑇 𝑓 𝑖 Since 𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, ∆𝑆 = 𝑄 𝑇 = 𝑊 𝑇 From 𝑖 → 𝑓, 𝑊 < 0 (the gas is compressed) ∴ ∆𝑆 𝑔𝑎𝑠= 𝑊𝑖→𝑓 𝑇 = − 𝑄 𝑡 < 0 ***This is not a closed system since 𝑄 ≠ 0. Consider the surrounding. Since 𝑄 𝑔𝑎𝑠 < 0, heat energy is lost to the surrounding. 𝑄𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = −𝑄 𝑔𝑎𝑠 ∴ ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠= 𝑄 𝑇 ∆𝑆𝑡𝑜𝑡𝑎𝑙= 𝑄 𝑇 − 𝑄 𝑇 = 0 (𝑝𝑟𝑜𝑣𝑒𝑛) For Reversible processes, ∆𝑺 > 𝟎 ; Free expansion, isochoric, isobaric For Free expansion, ∆𝑈 = 𝑄 = 𝑊 = 0 Only gas itself is already a closed system since 𝑄 = 0 Since, ∆𝑠 is a state function ∆𝑆 𝐹𝑟𝑒𝑒 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛= ∆𝑠𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 ∆𝑆 = 𝑄 𝑇 > 0 (𝑝𝑟𝑜𝑣𝑒𝑛) 𝑃 𝑉 ∆𝑈 = 0, Q = W 𝑉 𝑖 𝑓 𝑖 ∆𝑆𝑐𝑙𝑜𝑠𝑒𝑑 𝑠𝑦𝑠𝑡𝑒𝑚≥ 0
• 9.
  Engine 𝑄 𝐻 →𝑊𝑜𝑢𝑡 𝑝𝑢𝑡 + 𝑄 𝐿 𝐻 = 𝐻𝑖𝑔ℎ, 𝐿 = 𝐿𝑜𝑤 Engine converts heat energy to work. However, not all heat energy will be entirely converted to work. (𝑄 𝐻 → 𝑊𝑜𝑢𝑡 𝑝𝑢𝑡 + 𝑄 𝐿) 𝑄 𝐻 > 𝑄 𝐿 Hot reservoir is the source heat energy 𝑄 𝐻 , has the temperature of 𝑇 𝐻. Cold reservoir is where it receives 𝑄 𝐿, has the temperature of 𝑇𝐿. ***𝑇 𝐻 = 𝑇𝐿; we assume that the hot and cold reservoir are very large. Small 𝑄 𝐻 and 𝑄 𝐿 do not make any noticeable change of their temperature. Working substance ; A substance, such as a fluid, used to effect a thermodynamic or other change in a system. ***The lost or gain of heat energy involves only the working substances and not the engine itself. Carnot Engine ; It is an ideal engine. ; We assume that all processes in Carnot Engine are reversible. ; All work is done by the working substance only and not the frictional force or viscosity. Carnot Cycle There are 4 steps. 𝑎 → 𝑏 → 𝑐 → 𝑑 → 𝑎 *** 𝑎 → 𝑏, 𝑐 → 𝑑 ∶ 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑏 → 𝑐, 𝑑 → 𝑎 ∶ 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝒂 → 𝒃 ∶ 𝑸 𝑯 𝒊𝒏 𝒄 → 𝒅: 𝑸 𝑳 𝒐𝒖𝒕 Since it is a clockwise cycle, 𝑊𝑎→𝑏→𝑐→𝑑→𝑎 > 0 Efficiency, η η = 𝑊𝑜𝑢𝑡 𝐸𝑖𝑛 = 𝑊 𝑄 𝐻 = 𝑄 𝐻 − 𝑄 𝐿 𝑄 𝐻 = 1 − 𝑄 𝐿 𝑄 𝐻 ℎ𝑜𝑡 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 𝐸𝑛𝑒𝑟𝑔𝑖𝑛𝑒 𝑐𝑜𝑙𝑑 𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 𝑄 𝐻 𝑄 𝐿 𝑇 𝐻 𝑇𝐿 𝑊 𝑃 𝑉 𝐶𝑎𝑟𝑛𝑜𝑡 𝐶𝑦𝑐𝑙𝑒 𝑎 𝑏 𝑑 𝑐 𝑇 𝐻 𝑇𝐿 𝑄 𝐿 𝑄 𝐻 = 𝑇𝐿 𝑇 𝐻
• 10.
  Carnot Cycle (step-by-step) 1).Isothermal expansion, 𝑎 → 𝑏 2). Adiabatic expansion, 𝑏 → 𝑐 3). Isothermal compression, 𝑐 → 𝑑 ∆𝑈 = 0, 𝑄 = 𝑊 𝑊𝑐→𝑑 = −𝑄 𝐿 = 𝑁𝑘𝑇𝐿 ln 𝑉𝑐 𝑉𝑑 < 0 4). Adiabatic compression, 𝑑 → 𝑎 𝑄 = 0, ∆𝑈 = −𝑊𝑑→𝑎 ∴ 𝑊𝑑→𝑎 = −∆𝑈 𝑑→𝑎< 0 ***−∆𝑈 𝑑→𝑎= −∆𝑈 𝑎𝑑= − 𝑈 𝑎 − 𝑈 𝑑 Carnot Cycle (step-by-step) 1). Isothermal expansion, 𝑎 → 𝑏 ∆𝑈 = 0, 𝑄 = 𝑊 𝑊𝑎→𝑏 = 𝑄 𝐻 = 𝑁𝑘𝑇 𝐻 ln 𝑉𝑏 𝑉𝑎 > 0 2). Adiabatic expansion, 𝑏 → 𝑐 𝑄 = 0, ∆𝑈 = −𝑊𝑏→𝑐 ∴ 𝑊𝑏→𝑐 = −∆𝑈 𝑏→𝑐> 0 ***−∆𝑈 𝑏→𝑐= −∆𝑈𝑐𝑏= − 𝑈𝑐 − 𝑈 𝑏 3). Isothermal compression, 𝑐 → 𝑑 4). Adiabatic compression, 𝑑 → 𝑎 𝑃 𝑉 𝑎 𝑏 𝑑 𝑐 𝑇 𝐻 𝑇𝐿 𝑃 𝑉 𝑎 𝑏 𝑑 𝑐 𝑃 𝑉 𝑎 𝑏 𝑑 𝑐 𝑃 𝑉 𝑎 𝑏 𝑑 𝑐
• 11.
  Summary 1). Isothermal expansion,𝑎 → 𝑏 𝑊𝑎→𝑏 = 𝑄 𝐻 2). Adiabatic expansion, 𝑏 → 𝑐 𝑊𝑏→𝑐 = −∆𝑈 𝑏→𝑐= −𝑈𝑐 + 𝑈 𝑏 3). Isothermal compression, 𝑐 → 𝑑 𝑊𝑐→𝑑 = −𝑄 𝐿 4). Adiabatic compression, 𝑑 → 𝑎 𝑊𝑑→𝑎 = −∆𝑈 𝑑→𝑎= −𝑈 𝑎 + 𝑈 𝑑 ∴ 𝑊𝑎→𝑏→𝑐→𝑑→𝑎 = 𝑄 𝐻 −𝑈𝑐 + 𝑈𝑏 −𝑄 𝐿 −𝑈 𝑎 + 𝑈 𝑑 Since for isothermal, ∆𝑈 = 0 𝑎 → 𝑏 ∶ 𝑈 𝑎 = 𝑈 𝑏 𝑐 → 𝑑 ∶ 𝑈𝑐 = 𝑈 𝑑 Efficiency, η η = 1 − 𝑄 𝐿 𝑄 𝐻 Consider ∆𝑠 𝐶𝑎𝑟𝑛𝑜𝑡 𝐶𝑦𝑐𝑙𝑒 ∆𝑆 𝐶𝑎𝑟𝑛𝑜𝑡 𝐶𝑦𝑐𝑙𝑒= ∆𝑆 𝑎→𝑏→𝑐→𝑑→𝑎= 0 ∆𝑆 = 𝑑𝑄 𝑇 𝑏 𝑎 + 𝑑𝑄 𝑇 𝑐 𝑏 + 𝑑𝑄 𝑇 𝑑 𝑐 + 𝑑𝑄 𝑇 𝑎 𝑑 ∆𝑆 = 𝑑𝑄 𝑇 𝑏 𝑎 + 0 + 𝑑𝑄 𝑇 𝑑 𝑐 + 0 ∆𝑆 = 𝑑𝑄 𝑇 𝑏 𝑎 + 0 + 𝑑𝑄 𝑇 𝑑 𝑐 + 0 ∆𝑆 = 𝑄 𝑎→𝑏 𝑇 𝐻 + 𝑄 𝑐→𝑑 𝑇𝐿 = 0 𝑄 𝐻 𝑇 𝐻 + −𝑄 𝐿 𝑇𝐿 = 0 𝑊 = 𝑄 𝐻 − 𝑄 𝐿 𝑄 𝐻 𝑇 𝐻 + −𝑄 𝐿 𝑇𝐿 = 0 𝑄 𝐻 𝑇 𝐻 = 𝑄 𝐿 𝑇𝐿 𝑇𝐿 𝑇 𝐻 = 𝑄 𝐿 𝑄 𝐻 (𝑝𝑟𝑜𝑣𝑒𝑛) ***For engine, it is impossible for 𝑄 𝐿 = 0 𝑜𝑟 η = 100% η = 1 − 𝑄 𝐿 𝑄 𝐻 = 1 − 𝑇𝐿 𝑇 𝐻