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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


Review for final

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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