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Final Exam_1
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Final Exam
Chapter 14
- Thermodynamics
- 0th law
- Ta=Tb=Tc
- focuses on energy
- conservation of energy
- transfer of energy
- 0th law
- Temperature & Heat
- thermal equilibrium
- Ta=Tb=Tc
- heat flow
- heat energy
- flows between systems
- Joule
- J
- Calorie
- =1000 calories
- 1 cal=4.186J
- =1000 calories
- 3 scales
- fahrenheit
- water freezes at 32
- water boils at 212
- Tf=9/5tc+32
- Kelvin
- water freezes at 273.15
- water boils at 373.14
- Tk=Tc+273.15
- absolute 0 at 0
- Celsius
- water freezes at 0
- water boils at 100
- fahrenheit
- thermal equilibrium
- Phases of Matter
- Solid
- atoms held in place by forces from next atom
- constant volume and shape
- Liquid
- atoms able to move about
- constant volume but not constant shape
- Gas
- density lower than liquid
- not constant in either volume or shape
- Internal Energy U
- phase change
- phase diagram
- melting
- freezing
- evaporation
- sublimation
- heat capacity = q/delta(t)
- specific heat = c=Q/mdelta(T)
- latent heat of fusion
- latent heat of vaporization
- phase change
- Solid
- Thermodynamics
Chapter 15
- Gasses and Kinetic Theory
- Gas Laws
- Ideal Gas Law
- dilute gass
- Pv=nRT
- R=8.31J/mole*K
- n=number of moles present
- Kb=Boltzmanns Constant
- 1.38e-23J/K
- combination of 3 other laws
- Avogadros Law
- V proportional to N
- Boyles Law
- P proportional to 1/v
- for constant temp
- P proportional to 1/v
- Charles law
- V proportional to T
- for constant pressure
- V proportional to T
- Gay-Lussacs Law
- P proportional to T
- for constant volume
- P proportional to T
- Ideal Gas Law
- Kinetic Theory
- 3/2KbT
- Molecular Speeds
- v=sqrt((3KbT/M))
- Gas Laws
- Avogadros Number
- Na=6.023e23
- Pure number
- Number of particles in a mole
- Na=6.023e23
- atomic mass
- M=M/Na
- Gasses and Kinetic Theory
Chapter 16
- Thermodynamics
- Quantities Used to describe systems
- Pressure
- Volume
- Temperature
- Four Laws
- Zeroth Law
- Based on idea of thermal equilibrium
- If ta=tb and tb=tc then ta=tc
- Based on idea of thermal equilibrium
- First Law
- U=3/2KbT or 3/2nRT
- Thermal Reservoir
- Work
- W=Padeltax=PdeltaV
- Thermodynamic Processes
- Isobaric
- Constant Pressure
- W=PdeltaV=P(V1-V2)
- IsoThermal
- Constant Temperature
- W=nRTln(Vf/Vi)
- Expansion
- Compression
- Adibiatic
- Heat Q=0
- P proportional to 1/Vf
- Isochoric
- Constant volume
- W=PdeltaV=0
- Isobaric
- 2nd law
- Heat cannot flow from cold to hot
- Heat Engine
- Qh=W
- Carnot Engine
- Reversible Heat Engine
- Qc must be low
- Qc/Qh=Tc/Tn
- e=1-Tc/th
- Heat Engine
- Heat cannot flow from cold to hot
- 3rd law
- it is impossible to achieve a temperature of absolute 0 (0K)
- Zeroth Law
- Quantities Used to describe systems
- Thermodynamics
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- Mind Map on Final Exam_1, created by Alex Stewart on 09/12/2013.
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- Alex Stewart
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- Created by jvogan · Dec 9, 2013
- Copied by Alex Stewart · Dec 9, 2013
- Copied by Alex Stewart · Dec 10, 2013
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