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AP Physics Chp 15

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Thermodynamics – study of the relationship of heat and work System vs Surroundings Diathermal walls – allow heat to flow through Adiabatic walls – do not allow heat to flow

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Zeroth Law of Thermodynamics Two systems in thermal equilibrium with a third are also in equilibrium with each other

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First Law of Thermodynamics Internal energy changes based on the amount of heat and/or work done by/on the system. U = Q – W W = PV Q is positive when it goes in (endothermic) W is positive when the system does work

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What is the change in the internal energy if you supply 15 kJ to a 35 m 3 sample of helium at 101150 Pa and it is allowed to expand to 52 m 3 ?

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U = Q – W U = 15000 J – (101150 Pa)(52 m 3 – 35 m 3 ) U =

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If a process is slow enough then the P and T are uniform. When P is constant its called an isobaric process. W = PV Why is W negative when work is done on a system?

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Isochoric processes occur at constant volume This is the bomb calorimeter idea.

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At constant T its an isothermal process Adiabatic processes occur without the transfer of any heat

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One way to relate work for a system is to plot the P vs V graph and compare the area under the curve.

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How much work is done in compressing the gas from 4 m 3 to 3 m 3 ? Why is it more than 9 m 3 to 8 m 3 ?

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What would a graph for an isochoric process look like? Why does it show no work being done?

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What about isobaric, hows its graph look and is there any work?

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Isothermal process – Expansion or Compression Since T is constant the internal energy is constant so Q = W

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Any work done by the gas results in heat flowing out to the surroundings and vice versa.

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Adiabatic Processes – Expansion/Compression Since no heat is transferred the internal energy is related only to the work U = -W

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When the gas does work the T decreases and the internal energy of the gas has decreased

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If 2 moles of an ideal gas expands from 0.020 to 0.050 m 3 at a pressure of 101300 Pa, how much work is done? W = PV W = 101300Pa(0.050 m 3 -0.020 m 3 ) W = 3039Pa m 3 = J

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If the temperature is allowed/forced to remain constant how has the internal energy changed? 0 U = 3/2 nRT so with no change in T there is no change in internal energy

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How much heat was transferred? The same as the work. Q = W Q = 3039 J

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What is the temperature of the gas? 3039J = (2n)(8.31J/nK)T ln(0.050/0.020) T = 199.6 K

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Specific Heat Capacities Gases use a molar heat capacity at constant pressure and another for constant volume C p and C v

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Ideal Gases At constant pressure the heat is related to both the change in internal energy and work thus C p = 5/2R At constant volume its only the internal energy and C v = 3/2R So C p – C v = R

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Isobaric (P const) W = PV Isochoric (V const) W = 0 Isothermal (T const) W = nRT ln(V f /V o ) Adiabatic (no Q) W = 3/2nR(T o – T f )

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2 nd Law of Thermodynamics Heat flows spontaneously from a higher temperature to a lower temperature

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Heat engines use heat to perform work. – Heat comes from a hot reservoir – Part of the heat is used to perform work – The remainder is rejected to the cold reservoir Efficiencey e = W/Q H

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Efficiency can be multiplied by 100 to make it a percentage. Since Q H = W + Q C W = Q H – Q C e = 1 – Q C /Q H

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Carnot created a principle that says that a irreversible engine can not have a greater efficiency than a reversible one operating at the same temperatures. For a Carnot engine Q C /Q H = T C /T H e carnot = 1 – T C /T H

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If absolute zero could be maintained while depositing heat in then a 100% efficiency would be possible but its not.

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If my truck operates at a running temperature of 94 o C and the outside air is only -5 o C, what is the maximum efficiency for the engine?

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T H = 273 +94 = 367 K T C = 273 + -5 = 268 K e = 1 – T C /T H e = 1 – 268K / 367 K = 0.27 or 27%

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Refrigerators, Air Conditioners, Heat Pumps All of these take heat from the cold reservoir and put it into the hot reservoir by doing a certain amount of work. Its the reverse of the heat engine.

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Why cant you cool your house by running an air conditioner without having it exhaust outside? Coefficient of performance = Q C /W Heat pumps warm up a space by moving heat from the cold outside to the warm inside.

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Seems kind of weird that the cold outside has heat. If you use a Carnot heat pump to deliver 2500 J of heat to your house to achieve a temperature of 20 o C while it is -5 o C outside, how much work is required?

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W = Q H – Q C and Q C /Q H = T C /T H So Q C = Q H T C /T H and W = Q H – Q H T C /T H W = Q H (1-T C /T H ) W = 2500J (1- 268 K/293K) = 210 J

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Entropy Randomness or disorder gas>>>liquids>solids The entropy of the universe increases for irreversible process but stays constant for reversible

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Since carnot engines are reversible Q C /T C = Q H /T H Thus

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If we set the hot coffee pot at 372K on the table at 297K and they exchange 4700 J of heat, how much has the entropy of the universe changed?

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What happens to the energy in irreversible processes? Since the Suniv increases the increase is due to the energy being removed from being able to do any work

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W unavailable = T c Suniv So how much energy was lost to do work in the earlier example? Wunav = (295K)(3.3J/K) = 970 J

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