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Review for Final. Physics 313 Professor Lee Carkner Lecture 25. Final Exam. Final is Tuesday, May 18, 9am 75 minutes worth of chapters 9-12 45 minutes worth of chapters 1-8 Same format as other tests (multiple choice and short answer) Worth 20% of grade

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review for final
Review for Final

Physics 313

Professor Lee Carkner

Lecture 25

final exam
Final Exam
  • Final is Tuesday, May 18, 9am
  • 75 minutes worth of chapters 9-12
  • 45 minutes worth of chapters 1-8
  • Same format as other tests (multiple choice and short answer)
  • Worth 20% of grade
  • Three formula sheets given on test (one for Ch 9-12 and previous two)
  • Bring pencil and calculator
exercise 24 maxwell
Exercise #24 Maxwell
  • Set escape velocity equal to maximum Maxwell velocity
    • (2GM/R)½ = 10(3kT/m) ½
    • m = (150KTR/GM)
  • Planetary atmospheres
    • Earth: m > 9.5X10-27 kg (NH3, O2)
    • Jupiter: m > 1.4X10-28 kg (He, NH3, O2)
    • Titan: m > 5.6X10-26 kg (None)
    • Moon: m > 2.2X10-25 kg (None)
thermal equilibrium
Thermal Equilibrium

Two identical metal blocks, one at 100 C and one at 120 C, are placed together. Which transfers the most heat?

  • Two objects at different temperatures will exchange heat until they are at the same temperature
  • Zeroth Law: Two systems in thermal equilibrium with a third are in thermal equilibrium with each other
heat transfer
Heat Transfer
  • Heat:

Q = mcDT = mc(Tf-Ti)

  • Conduction:

dQ/dt = -KA(dT/dx)

Q/t = -KA(T1-T2)/x

  • Radiation

dQ/dt = Aes(Tenv4-T4)

temperature
Temperature

How would you make a tube of mercury into a Celsius thermometer? A Kelvin thermometer?

  • Thermometers defined by the triple point of water
  • A system at constant temperature can have a range of values for the other variables
    • Isotherm
measuring temperature
Measuring Temperature
  • Thermometers

T (X) = 273.16 (X/XTP)

  • Temperature scales

T (R) = T (F) + 459.67

T (K) = T (C) + 273.15

T (R) = (9/5) T (K)

T (F) = (9/5) T (C) +32

equations of state
Equations of State

If the temperature of an ideal gas is doubled while the volume stays the same, what happens to the pressure?

  • Equation of state detail how properties change with temperature
  • Increasing T will generally increase the force and displacement terms
mathematical relations
Mathematical Relations
  • General Relations:

dx = ( x/y)zdy + (x/z)ydz

(x/y)z = 1/(y/x)z

(x/y)z(y/z)x(z/x)y = -1

  • Specific Relations:
    • Volume Expansivity: b = (1/V)(dV/dT)P
    • Isothermal Compressibility: k=-(1/V)(dV/dP)T
    • Linear Expansivity: a = (1/L)(dL/dT)t
    • Young’s modulus: Y = (L/A)(dt/dL)T
slide10
Work

How much work is done in an isobaric compression of a gas at 1 Pa from 2 to 1 m3?

  • The work done a system is the product of a force term and a displacement term
    • No displacement, no work
  • Compression is positive, expansion is negative
  • Work is area under PV (or XY) curve
    • Work is path dependant
calculating work
Calculating Work

dW = -PdV

W = - PdV

  • For ideal gas P = nRT/V
  • Examples:
  • Isothermal ideal gas:

W = -nRT  (1/V) dV = -nRT ln (Vf/Vi)

  • Isobaric ideal gas:

W = -P  dV = -P(Vf-Vi)

first law
First Law

Rank the following processes in order of increasing internal energy:

Adiabatic compression

Isothermal expansion

Isochoric cooling

  • Energy is conserved
  • Internal energy is a state function, work and heat are not
first law equations
First Law Equations

DU = Uf-Ui = Q+W

dU = dQ +dW

dU = CdT - PdV

ideal gas
Ideal Gas
  • If the volume of an ideal gas is doubled and the pressure is tripled isothermally, how does the internal energy change?

lim (PV) = nRT

(dU/dP)T = (dU/dV)T = 0

(dU/dT)V = CV

CP = CV + nR

dQ = CVdT+PdV = CPdT-VdP

adiabatic processes
Adiabatic Processes
  • Can an adiabatic process keep constant P, V, or T?

PVg = const

TVg-1 = const

T/P(g-1)/g = const

W = (PfVf - PiVi)/g-1

kinetic theory
Kinetic Theory
  • If the rms velocity of gas molecules doubles what happens to the temperature and internal energy

(1/2)mv2 = (3/2)kT

U = (3/2)NkT

T = mv2/3k

engines
Engines
  • If the heat entering an engine is doubled and the work stays the same what happens to the efficiency?
  • Engines are cycles
  • Change in internal energy is zero
  • Composed of 4 processes

h = W/QH = (QH-QL)/QH = 1 - QL/QH

QH = W + QL

types of engines
Types of Engines
  • Otto
    • Adiabatic, Isochoric

h = 1 - (T1/T2)

  • Diesel
    • Adiabatic, isochoric, isobaric

h = 1 - (1/g)(T4-T1)/(T3-T2)

  • Rankine (steam)
    • Adiabatic, isobaric
  • Stirling
    • Isothermal, isochoric
refrigerators
Refrigerators
  • Transfer heat from low to high T with the addition of work
  • Operates in cycle
  • Transfers heat with evaporation and condensation at different pressures

K = QL/W

K = QL/(QH-QL)

second law
Second Law
  • Is an ice cube melting at room temperature a reversible process?
  • Kelvin-Planck
    • Cannot convert heat completely into work
  • Clausius
    • Cannot move heat from low to high temperature without work
carnot
Carnot
  • What two processes make up a Carnot cycle? How many temperatures is heat transferred at?
  • Adiabatic and isothermal

h = 1 - TL/TH

  • Most efficient cycle
  • Efficiency depends only on the temperature
second law1
Second Law
  • The second law of thermodynamics can be stated:
    • Engine cannot turn heat completely into work
    • Heat cannot move from low to high temperatures without work
    • Efficiency cannot exceed Carnot efficiency
  • Entropy always increases
entropy
Entropy
  • Entropy change is zero for all reversible processes
    • All real processes are irreversible
  • Can compute entropy for an irreversible process by replacing it with a reversible process that achieves the same result
  • Entropy change of system + entropy change of surroundings = entropy change of universe (which is > 0)
determining entropy
Determining Entropy
  • Can integrate dS to find DS

dS = dQ/T

DS =  dQ/T (integrated from Ti to Tf)

  • Examples:
  • Heat reservoir (or isothermal process)

DS = Q/T

  • Isobaric

DS = CP ln (Tf/Ti)

pure substances
Pure Substances
  • Can plot phases and phase boundaries on a PV, PT and PTV diagram
  • Saturation
    • condition where substance can change phase
  • Critical point
    • above which substance can only be gas
    • where (dP/dV) =0 and (d2P/dV2) = 0
  • Triple point
    • where fusion, sublimation and vaporization curves intersect
properties of pure substances
Properties of Pure Substances

cP = (dQ/dT)P (per mole)

cV = (dQ/dT)T (per mole)

b = (1/V)(dV/dT)P

k = -(1/V)(dV/dP)T

  • cP, cV and b are 0 at 0 K and rise sharply to the Debye temperature and then level off
    • cP and cV end up near the Dulong and Petit value of 3R
  • k is constant at a finite value at low T and then increases linearly
characteristic functions and maxwell s relations
Legendre Transform:

df = udx +vdy

g= f-ux

dg = -xdu+vdy

Useful theorems:

(dx/dy)z(dy/dz)x(dz/dx)y=-1

(dx/dy)f(dy/dz)f(dz/dx)f=1

dU = -PdV +T dS

dH = VdP +TdS

dA = - SdT - PdV

dG = V dP - S dT

(dT/dV)S = - (dP/dS)V

(dT/dP)S = (dV/dS)P

(dS/dV)T = (dP/dT)V

(dS/dP)T = -(dV/dT)P

Characteristic Functions and Maxwell’s Relations
key equations
Key Equations
  • Entropy

T dS = CV dT + T (dP/dT)V dV

T dS = CP dT - T(dV/dT)P dP

  • Internal Energy

(dU/dV)T = T (dP/dT)V - P

(dU/dP)T = -T (dV/dT)P - P(dV/dP)T

  • Heat Capacity

CP - CV = -T(dV/dT)P2 (dP/dV)T

cP - cV = Tvb2/k

joule thomson expansion
Joule-Thomson Expansion
  • Can plot on PT diagram
  • Isenthalpic curves show possible final states for an initial state

m = (1/cP)[T(dv/dT)P - v] = slope

  • Inversion curve separates heating and cooling region

m = 0

  • Total enthalpy before and after throttling is the same
  • For liquefaction:

hi = yhL + (1-y)hf

clausius clapeyron equation
Clausius-Clapeyron Equation
  • Any first order phase change obeys:

(dP/dT) = (sf -si)/(vf - vi)

= (hf - hi)/T (vf -vi)

  • dP/dT is slope of phase boundary in PT diagram
  • Can change dP/dT to DP/DT for small changes in P and T
open systems
Open Systems
  • For a steady flow open systems mass and energy are conserved:

Smin = Smout

Sin[Q + W + mq] = Sout [Q + W + mq]

  • Where q is energy per unit mass or:

q = h + ke +pe (per unit mass)

  • Chemical potential = m = (dU/dn)

mi = mf

    • For open systems in equilibrium:
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