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**Chapter 10: Thermodynamics**

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**10-1 Relationship Between Heat and Work**

In a closed system there’s a direct relationship between heat and work. Heat and work both transfer energy to or from a system. Key idea: A system never has “heat” or “work”, it has internal energy which is affected by heat in/out or work done on/by the system Energy can be transferred to or from a substance as heat, changing its internal energy (+/-) The internal energy can be used to do work

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**Transfer of heat and work**

System: a set of particles or interacting particles considered to be a distinct physical entity Environment: the combination of conditions and influences outside a system that affects the behavior of the system Ex of closed systems: a gas confined in a cylinder by a piston, a calorimeter, a thermos…

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**Work Done on or By a Gas W = PΔV Is represented in the equation:**

W - in Joules (J) P = pressure in Pascal (Pa) 1 Pa = 1 N/m2 ΔV= volume change in (m3)

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Sample Problem Gas in a container is at a pressure of 1.6x105 Pa and a volume of 4.0 m3. What is the work by the gas if it expands at a constant pressure to twice its initial volume?

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Solution P= 1.6x105 Pa =1.6x105 N/m2 ∆V=Vf – Vi = 8.0 m m3 =4.0 m3 W= P ∆ V W = (1.6x105 N/m2)(4.0 m3) W= 6.4x105 J

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**Thermodynamic Processes**

Isovolumetric process: a thermodynamic process that takes place at a constant volume so that no work is done on or by the system, ex: a car with closed windows parked in a hot garage. Isothermal process: a thermodynamic process that takes place at a constant temperature, ex usually a slow process like a balloon expanding slowly during the day. Adiabatic process: a thermodynamic process during which heat energy is transferred to or from the system. ex: usually a fast process like filling a tank Isobaric process: a process that takes place at a constant pressure. ex: heating an open pot of water

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**10.2 The First Law of Thermodynamics**

The first law is a statement of conservation of energy that takes into account a system’s internal energy (U) as well as the energy transfer to/from the system by work and heat. It is expressed as:

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**Signs of Q and W For a System**

ΔQ = positive if heat is added to a system ΔQ = negative if heat is released from a system ΔW = positive if work is done by the system ΔW = negative if work is done on the system

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**First Law – Isovolumetric Process**

Δ U = Q – W ΔV = 0 Since W = P ΔV, W = 0 therefore, Δ U = Q Energy added to the system increases the system’s internal energy. Energy removed by heat decreases the system’s internal energy.

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**First Law – Isothermal Process**

ΔU = Q – W since ΔT = 0 , ΔU = 0 therefore Q = W Energy added to the system as heat is removed from the system as work done by the system.

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**First Law – Adiabatic Process**

Δ U = Q – W Q = 0, therefore Δ U = – W Work done on the system increases the system’s internal energy. Work done by the system decreases the system’s internal energy.

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**First Law – Isobaric Process**

Δ U = Q – W since W = PΔV Δ U = Q – PΔV example, a gas inside a cylinder with a fitted piston

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**First Law – Isolated System**

Δ U = Q – W since Q = W = 0 Δ U = 0 No heat or work interaction with surroundings so there’s no change in the system’s internal energy.

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Sample Problem A total of 135 J of work is done on a gaseous refrigerant as it undergoes compression. If the internal energy of the gas increases by 114 J during the process, what is the total amount of energy transferred as heat? work done on the system is – W

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**Solution W= -135 J (work done on the system is -) ∆ U= 114 J**

∆ U = Q - W Q = ∆ U + W Q= 114 J + (-135 J)= -21 J Q= -21 J In this problem, energy is removed from the gas as heat, which is indicated by the negative sign on the Q value ( Q < 0 ).

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Cyclic Processes A thermodynamic process in which a system returns to the same conditions under which it started (no change in system’s energy) ΔUnet = 0 and Qnet = Wnet Resembles an isothermal process in that all energy is transferred as work and heat. example, operation of a refrigerator

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The Heat Engine Any device that exploits a temperature difference to do mechanical work The net work done is equal to the difference in energy taken in as heat from a high-temp. reservoir (Qh) and the energy expelled as heat to the low temp. reservoir(Qc). Wnet = Qnet = Qh − Qc Basically a heat engine takes in heat from a hot place, uses some of it to do useful work and dumps the rest. The larger the temp difference, the more work done in a cycle.

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**10.3 The Second Law of Thermodynamics**

States that no cyclic process that converts heat entirely into work is possible. Includes the requirement that a heat engine give up some energy at a lower temperature in order to do work. So a heat engine cannot transfer all energy as heat to do work.

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**Second Law of Thermodynamics**

No cyclic process that converts heat entirely into work is possible

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**Efficiency of a Heat Engine**

The smaller the fraction of usable energy that an engine can provide, the lower its efficiency is. Efficiency = Can never be 1, or 100%.

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Sample Problem Find the efficiency of a gasoline engine that, during one cycle, receives 204 J of energy from combustion and loses 153 J as heat into the exhaust.

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Solution Qh= 204 J Qc= 153 J eff= 1- Qc Qh eff= J 204 J eff= J

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**Entropy Entropy: a measure of the randomness or disorder of a system**

A greater disorder means there is less energy to do work The motion of the particles of a system is not well ordered and therefore is less useful for doing work

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Once a system has reached a state of the greatest disorder, it will tend to remain in that state and have maximum entropy. The second law of thermodynamics states that the entropy of the universe increases in all natural processes.

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