AP Physics Chp 15. Thermodynamics – study of the relationship of heat and work System vs...
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Transcript of AP Physics Chp 15. Thermodynamics – study of the relationship of heat and work System vs...
AP Physics Chp 15
• 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
• Zeroth Law of Thermodynamics
• Two systems in thermal equilibrium with a third are also in equilibrium with each other
• 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
• What is the change in the internal energy if you supply 15 kJ to a 35 m3 sample of helium at 101150 Pa and it is allowed to expand to 52 m3?
• ∆U = Q – W
• ∆U = 15000 J – (101150 Pa)(52 m3 – 35 m3)
• ∆U =
• If a process is slow enough then the P and T are uniform.
• When P is constant it’s called an isobaric process.
• W = P∆V Why is W negative when work is done on a system?
• Isochoric processes occur at constant volume
• This is the bomb calorimeter idea.
• At constant T its an isothermal process
• Adiabatic processes occur without the transfer of any heat
• One way to relate work for a system is to plot the P vs V graph and compare the area under the “curve”.
How much work is done in compressing the gas from 4 m3 to 3 m3?
Why is it more than 9 m3 to 8 m3?
• What would a graph for an isochoric process look like? Why does it show no work being done?
• What about isobaric, how’s it’s graph look and is there any work?
• Isothermal process – Expansion or Compression
• Since T is constant the internal energy is constant so
• Q = W
• Any work done by the gas results in heat flowing out to the surroundings and vice versa.
ln f
o
VW nRT
V
• Adiabatic Processes – Expansion/Compression
• Since no heat is transferred the internal energy is related only to the work
• ∆U = -W
• When the gas does work the T decreases and the internal energy of the gas has decreased
3
2 o fW nR T T
• If 2 moles of an ideal gas expands from 0.020 to 0.050 m3 at a pressure of 101300 Pa, how much work is done?
• W = P∆V• W = 101300Pa(0.050 m3 -0.020 m3)• W = 3039Pa m3 = J
• 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
• How much heat was transferred?
• The same as the work. Q = W
• Q = 3039 J
• What is the temperature of the gas?
• 3039J = (2n)(8.31J/nK)T ln(0.050/0.020)•
T = 199.6 K
ln f
o
VW nRT
V
• Specific Heat Capacities
• Gases use a molar heat capacity at constant pressure and another for constant volume
• Cp and Cv
Ideal Gases
• At constant pressure the heat is related to both the change in internal energy and work thus Cp = 5/2R
• At constant volume its only the internal energy and Cv = 3/2R
• So Cp – Cv = R
• Isobaric (P const) W = P∆V
• Isochoric (V const) W = 0
• Isothermal (T const) W = nRT ln(Vf/Vo)
• Adiabatic (no Q) W = 3/2nR(To – Tf)
• 2nd Law of Thermodynamics
• Heat flows spontaneously from a higher temperature to a lower temperature
• 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/QH
• Efficiency can be multiplied by 100 to make it a percentage.
• Since QH = W + QC W = QH – QC
• e = 1 – QC/QH
• 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 QC/QH = TC/TH
• ecarnot = 1 – TC/TH
• If absolute zero could be maintained while depositing heat in then a 100% efficiency would be possible but it’s not.
• If my truck operates at a running temperature of 94 oC and the outside air is only -5 oC, what is the maximum efficiency for the engine?
• TH = 273 +94 = 367 K
• TC = 273 + -5 = 268 K
• e = 1 – TC/TH
• e = 1 – 268K / 367 K = 0.27 or 27%
• 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.
• It’s the reverse of the heat engine.
• Why can’t you cool your house by running an air conditioner without having it exhaust outside?
• Coefficient of performance = QC/W
• Heat pumps warm up a space by moving heat from the cold outside to the warm inside.
• 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 oC while it is -5oC outside, how much work is required?
• W = QH – QC and QC/QH = TC/TH
• So QC = QHTC/TH and
• W = QH – QHTC/TH
• W = QH(1-TC/TH) • W = 2500J (1- 268 K/293K) = 210 J
• Entropy
• Randomness or disorder gas>>>liquids>solids
• The entropy of the universe increases for irreversible process but stays constant for reversible
QS
T
Ssys Ssurr Suniv
c H
c H
Q QSuniv
T T
• Since carnot engines are reversible
• QC/TC = QH/TH Thus
0c c
c c
Q QSuniv
T T
• 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?
c H
c H
Q QSuniv
T T
4700 4700
295 372
J JSuniv
K K
3.3 /Suniv J K
• 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
• Wunavailable = Tc∆Suniv
• So how much energy was “lost” to do work in the earlier example?
• Wunav = (295K)(3.3J/K) = 970 J