Download presentation

Presentation is loading. Please wait.

1
Review for Final Physics 313 Professor Lee Carkner Lecture 25

2
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

3
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 (NH 3, O 2 ) Jupiter: m > 1.4X10 -28 kg (He, NH 3, O 2 ) Titan: m > 5.6X10 -26 kg (None) Moon: m > 2.2X10 -25 kg (None)

4
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

5
Heat Transfer Heat: Q = mc T = mc(T f -T i ) Conduction: dQ/dt = -KA(dT/dx) Q/t = -KA(T 1 -T 2 )/x Radiation dQ/dt = A (T env 4 -T 4 )

6
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

7
Measuring Temperature Thermometers T (X) = 273.16 (X/X TP ) 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

8
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

9
Mathematical Relations General Relations: dx = ( x/ y) z dy + ( x/ z) y dz ( x/ y) z = 1/( y/ x) z ( x/ y) z ( y/ z) x ( z/ x) y = -1 Specific Relations: Volume Expansivity: = (1/V)(dV/dT) P Isothermal Compressibility: =-(1/V)(dV/dP) T Linear Expansivity: = (1/L)(dL/dT) Young’s modulus: Y = (L/A)(d /dL) T

10
Work How much work is done in an isobaric compression of a gas at 1 Pa from 2 to 1 m 3 ? 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

11
Calculating Work dW = -PdV W = - PdV For ideal gas P = nRT/V Examples: Isothermal ideal gas: W = -nRT (1/V) dV = -nRT ln (V f /V i ) Isobaric ideal gas: W = -P dV = -P(V f -V i )

12
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

13
First Law Equations U = U f -U i = Q+W dU = dQ +dW dU = CdT - PdV

14
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 = C V C P = C V + nR dQ = C V dT+PdV = C P dT-VdP

15
Adiabatic Processes Can an adiabatic process keep constant P, V, or T? PV = const TV -1 = const T/P ( -1)/ = const W = (P f V f - P i V i )/ -1

16
Kinetic Theory If the rms velocity of gas molecules doubles what happens to the temperature and internal energy (1/2)mv 2 = (3/2)kT U = (3/2)NkT T = mv 2 /3k

17
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 = W/Q H = (Q H -Q L )/Q H = 1 - Q L /Q H Q H = W + Q L

18
Types of Engines Otto Adiabatic, Isochoric = 1 - (T 1 /T 2 ) Diesel Adiabatic, isochoric, isobaric = 1 - (1/ )(T 4 -T 1 )/(T 3 -T 2 ) Rankine (steam) Adiabatic, isobaric Stirling Isothermal, isochoric

19
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 = Q L /W K = Q L /(Q H -Q L )

20
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

21
Carnot What two processes make up a Carnot cycle? How many temperatures is heat transferred at? Adiabatic and isothermal = 1 - T L /T H Most efficient cycle Efficiency depends only on the temperature

22
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

23
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)

24
Determining Entropy Can integrate dS to find S dS = dQ/T S = dQ/T (integrated from T i to T f ) Examples: Heat reservoir (or isothermal process) S = Q/T Isobaric S = C P ln (T f /T i )

25
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 ( P/ V) =0 and ( 2 P/ V 2 ) = 0 Triple point where fusion, sublimation and vaporization curves intersect

26
Properties of Pure Substances c P = (dQ/dT) P (per mole) c V = (dQ/dT) T (per mole) = (1/V)(dV/dT) P = -(1/V)(dV/dP) T c P, c V and are 0 at 0 K and rise sharply to the Debye temperature and then level off c P and c V end up near the Dulong and Petit value of 3R is constant at a finite value at low T and then increases linearly

27
Characteristic Functions and Maxwell’s Relations Legendre Transform: df = udx +vdy g= f-ux dg = -xdu+vdy Useful theorems: ( x/ y) z ( y/ z) x ( z/ x) y =-1 ( x/ y) f ( y/ z) f ( z/ x) f =1 dU = -PdV +T dS dH = VdP +TdS dA = - SdT - PdV dG = V dP - S dT ( T/ V) S = - ( P/ S) V ( T/ P) S = ( V/ S) P ( S/ V) T = ( P/ T) V ( S/ P) T = -( V/ T) P

28
Key Equations Entropy T dS = C V dT + T ( P/ T) V dV T dS = C P dT - T( V/ T) P dP Internal Energy ( U/ V) T = T ( P/ T) V - P ( U/ P) T = -T ( V/ T) P - P( V/ P) T Heat Capacity C P - C V = -T( V/ T) P 2 ( P/ V) T c P - c V = Tv 2 /

29
Joule-Thomson Expansion Can plot on PT diagram Isenthalpic curves show possible final states for an initial state = (1/c P )[T(dv/dT) P - v] = slope Inversion curve separates heating and cooling region = 0 Total enthalpy before and after throttling is the same For liquefaction: h i = yh L + (1-y)h f

30
Clausius-Clapeyron Equation Any first order phase change obeys: (dP/dT) = (s f -s i )/(v f - v i ) = (h f - h i )/T (v f -v i ) dP/dT is slope of phase boundary in PT diagram Can change dP/dT to P/ T for small changes in P and T

31
Open Systems For a steady flow open systems mass and energy are conserved: m in = m out in [ Q + W + m ] = out [ Q + W + m ] Where is energy per unit mass or: = h + ke +pe (per unit mass) Chemical potential = = ( U/ n) i = f For open systems in equilibrium:

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google