Physics 313
Professor Lee Carkner
Lecture 25
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
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)
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:
Q = mc D T = mc(T f
-T i
)
Conduction: dQ/dt = -KA(dT/dx)
Radiation
Q/t = -KA(T
1
-T
2
)/x dQ/dt = A es (T env
4 -T 4 )
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
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
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
General Relations: dx = ( x/ y) z dy + ( x/ z) y dz
( x/ y) z
( x/ y) z
Specific Relations:
= 1/( y/ x)
( y/ z) x
( z/ x) z y
= -1
Volume Expansivity: b = (1/V)(dV/dT)
P
Isothermal Compressibility: k =-(1/V)(dV/dP)
T
Linear Expansivity: a = (1/L)(dL/dT)
Young’s modulus: Y = (L/A)(d t /dL)
T t
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
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
)
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
D U = U f
-U i
= Q+W dU = dQ +dW dU = CdT - PdV
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/dT)
V dQ = C
V
C
P
= (dU/dV)
T
= C
V
= C
+ nR dT+PdV = C
P
V
= 0 dT-VdP
Can an adiabatic process keep constant
P, V, or T?
PV g = const
TV g -1 = const
T/P ( g -1)/ g = const
W = (P f
V f
- P i
V i
)/ g -1
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
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/Q
H
= (Q
H
Q
H
-Q
L
)/Q
H
= W + Q
L
= 1 - Q
L
/Q
H
Otto
Adiabatic, Isochoric h = 1 - (T
1
/T
2
)
Diesel
Adiabatic, isochoric, isobaric h = 1 - (1/ g )(T
4
-T
1
)/(T
3
-T
2
)
Rankine (steam)
Adiabatic, isobaric
Stirling
Isothermal, isochoric
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
)
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
What two processes make up a Carnot cycle? How many temperatures is heat transferred at?
Adiabatic and isothermal h = 1 - T
L
Most efficient cycle
/T
H
Efficiency depends only on the temperature
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 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)
Can integrate dS to find D S dS = dQ/T
D S = dQ/T (integrated from T i
Examples: to T
Heat reservoir (or isothermal process)
D S = Q/T
Isobaric
D S = C
P ln (T f
/T i
) f
)
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 ( d P/ d V) =0 and ( d 2 P/ d V 2 ) = 0
Triple point
where fusion, sublimation and vaporization curves intersect
c c
P
V
= (dQ/dT)
= (dQ/dT)
P
(per mole)
T
(per mole) b = (1/V)(dV/dT) k = -(1/V)(dV/dP)
c
P
, c
V temperature and then level off
P
T and b are 0 at 0 K and rise sharply to the Debye
c
P and c
V 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:
( d x/ d y) z
( d x/ d y)
( d y/ d z) x f
( d y/ d z)
( d z/ d x) y
=-1 f
( d z/ d x) f
=1 dU = -PdV +T dS dH = VdP +TdS dA = - SdT - PdV dG = V dP - S dT
( d T/ d V)
S
( d T/ d P)
( d S/ d V)
S
T
( d S/ d P)
T
= - ( d P/ d S)
V
= ( d V/ d S)
P
= ( d P/ d T)
V
= -( d V/ d T)
P
Entropy
T dS = C
V
T dS = C
P
Internal Energy dT + T ( d P/ d T)
V dT - T( d V/ d T)
P dV dP
( d U/ d V)
( d U/ d P)
T
Heat Capacity
T
= T ( d P/ d T)
V
= -T ( d V/ d T)
P
- P( d
- P
V/ d P)
T
C
P
- C
V
= -T( d V/ d T) c
P
- c
V
P
2 (
= Tv b 2 / k d P/ d V)
T
Can plot on PT diagram
Isenthalpic curves show possible final states for an initial state m = (1/c
P
- v] = slope
Inversion curve separates heating and cooling region
P
)[T(dv/dT) m = 0
Total enthalpy before and after throttling is the same
For liquefaction: h i
= yh
L
+ (1-y)h f
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 D P/ D T for small changes in P and T
For a steady flow open systems mass and energy are conserved:
S in
S m in
= S m out
[Q + W + m q ] = S out
[Q + W + m
Where q is energy per unit mass or: q ] q = h + ke +pe (per unit mass)
Chemical potential = m = ( d U/ d n) m i
= m f
For open systems in equilibrium: