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

32 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

QST

Ssys Ssurr Suniv

c H

c H

Q QSunivT T

• Since carnot engines are reversible

• QC/TC = QH/TH Thus

0c c

c c

Q QSunivT 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 QSunivT T

4700 4700295 372

J JSunivK 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