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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

- 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

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:

Q = mcDT = mc(Tf-Ti)

- Conduction:

dQ/dt = -KA(dT/dx)

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

- Radiation

dQ/dt = Aes(Tenv4-T4)

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

- 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

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

- 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

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

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

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

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

- 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

- 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

- 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

- 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

- 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

- 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

- 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 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 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

- 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

- 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

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

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 RelationsKey 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

- 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

- 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

- 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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