Lecture 23
Goals:
• Chapter 16
•
 Use the ideal-gas law.
 Use pV diagrams for ideal-gas processes.
Chapter 17
 Employ energy conservation in terms of 1st law of TD
 Understand the concept of heat.
 Relate heat to temperature change
 Apply heat and energy transfer processes in real situations
 Recognize adiabatic processes.
• Assignment
 HW9, Due Wednesday, Apr. 15th
 HW10, Due Wednesday, Apr. 22nd (9 AM)
Physics 207: Lecture 23, Pg 1
Thermodynamics: A macroscopic description of matter
 Recall “3” Phases of matter: Solid, liquid & gas
 All 3 phases exist at different p,T conditions
 Triple point of water:
p = 0.06 atm
T = 0.01°C
 Triple point of CO2:
p = 5 atm
T = -56°C
Physics 207: Lecture 23, Pg 2
Modern Definition of Kelvin Scale
 Water’s triple point on the Kelvin scale is 273.16 K
 One degrees Kelvin is defined to be 1/273.16 of the
temperature at the triple point of water
Accurate water phase diagram
Triple point
Physics 207: Lecture 23, Pg 3
Transferring energy to a solid (ice)
1. Temperature increase or
2. State Change
If a gas, then V, p and T are interrelated….equation of state
Physics 207: Lecture 23, Pg 4
Temperature scales
 Three main scales
Farenheit
Celcius
Kelvin
212
100
373.15
32
0
273.15
-459.67
-273.15
0
Water boils
Water freezes
Absolute Zero
Physics 207: Lecture 23, Pg 7
Some interesting facts
 In 1724, Gabriel Fahrenheit made thermometers
using mercury. The zero point of his scale is
attained by mixing equal parts of water, ice, and
salt. A second point was obtained when pure water
froze (originally set at 30oF), and a third (set at
96°F) “when placing the thermometer in the mouth
of a healthy man”.
 On that scale, water boiled at 212.
 Later, Fahrenheit moved the freezing point of
water to 32 (so that the scale had 180
increments).
 In 1745, Carolus Linnaeus of Upsula, Sweden,
described a scale in which the freezing point of
water was zero, and the boiling point 100, making it
a centigrade (one hundred steps) scale. Anders
Celsius (1701-1744) used the reverse scale in
which 100 represented the freezing point and zero
the boiling point of water, still, of course, with 100
degrees between the two defining points.
T (K)
108
Hydrogen bomb
107
Sun’s interior
106
Solar corona
105
104
103
100
10
1
Sun’s surface
Copper melts
Water freezes
Liquid nitrogen
Liquid hydrogen
Liquid helium
0.1
Lowest T~ 10-9K
Physics 207: Lecture 23, Pg 8
Ideal gas: Macroscopic description
 Consider a gas in a container of volume V, at pressure P, and at
temperature T
 Equation of state
 Links these quantities
 Generally very complicated: but not for ideal gas
 Equation of state for an ideal gas
 Collection of atoms/molecules moving randomly
 No long-range forces
 Their size (volume) is negligible
 Density is low
 Temperature is well above the condensation point
PV = nRT
R is called the universal gas constant
In SI units, R =8.315 J / mol·K
n = m/M : number of moles
Physics 207: Lecture 23, Pg 9
Boltzmann’s constant
 Number of moles: n = m/M
m=mass
M=mass of one mole
 One mole contains NA=6.022 X 1023 particles :
Avogadro’s number = number of carbon atoms in 12 g of carbon
 In terms of the total number of particles N
PV = nRT = (N/NA ) RT
PV = N kB T
kB = R/NA = 1.38 X 10-23 J/K
kB is called the Boltzmann’s constant
 P, V, and T are the thermodynamics variables
Physics 207: Lecture 23, Pg 10
The Ideal Gas Law
pV  nRT
What is the volume of 1 mol of gas at STP ?
T = 0 °C = 273 K
5
p = 1 atm = 1.01 x 10 Pa
nRT
V 
P
8.31 J / mol  K  273 K

1.0110 5 Pa
 0.0224 m 3  22.4 
Physics 207: Lecture 23, Pg 11
Example
 A spray can containing a propellant gas at twice atmospheric
pressure (202 kPa) and having a volume of 125.00 cm3 is at
27oC. It is then tossed into an open fire. When the
temperature of the gas in the can reaches 327oC, what is the
pressure inside the can?
Assume any change in the volume of the can is negligible.
Steps
1. Convert to Kelvin (From 300 K to 600 K)
2. Use P/T = nR/V = constant  P1/T1 = P2/T2
3. Solve for final pressure  P2 = P1 T2/T1
http://www.thehumorarchives.com/joke/WD40_Stupidity
Physics 207: Lecture 23, Pg 12
Example problem: Air bubble rising
 A diver produces an air bubble underwater, where the absolute
pressure is p1 = 3.5 atm. The bubble rises to the surface, where
the pressure is p2 = 1.0 atm. The water temperatures at the
bottom and the surface are, respectively, T1 = 4°C, T2 = 23°C
 What is the ratio of the volume of the bubble as it reaches the
surface,V2, to its volume at the bottom, V1? (Ans.V2/V1 = 3.74)
 Is it safe for the diver to ascend while holding his breath?
No! Air in the lungs would expand, and the lung could rupture.
Physics 207: Lecture 23, Pg 13
Example problem: Air bubble rising
 A diver produces an air bubble underwater, where the absolute
pressure is p1 = 3.5 atm. The bubble rises to the surface, where
the pressure is p2 = 1 atm. The water temperatures at the
bottom and the surface are, respectively, T1 = 4°C, T2 = 23°C
 What is the ratio of the volume of the bubble as it reaches the
surface,V2, to its volume at the bottom, V1? (Ans.V2/V1 = 3.74)
 pV=nRT  pV/T = const so p1V1/T1 = p2V2/T2
V2/V1 = p1T2/ (T1 p2)
= 3.5 296 / (277 1)
If thermal transfer is efficient.
[More than likely the expansion will be “adiabatic” and, for a
diatomic gas, PVg = const. where g = 7/5, see Ch. 17 & 18]
Physics 207: Lecture 23, Pg 15
Buoyancy and the Ideal Gas Law
 A typical 5 passenger hot air balloon has
approximately 700 kg of total mass and the
balloon itself can be thought as spherical
with a radius of 10.0 m. If the balloon is
launched on a day with conditions of 1.0 atm
and 273 K, how hot would you have to heat
the air inside (assuming the density of the
surrounding air is 1.2 kg/m3 and the air
behaves and as an ideal gas) in order to
keep the balloon at a constant altitude?
 Hint: Remember the weight of the air inside
the balloon.
Balloon weight = Buoyant force – Weight of hot air
Ideal gas law:
pV = nRT

nT= pV/R = const.
or r T = const. = 1.2 x 273 kg K/m3
Physics 207: Lecture 23, Pg 16
Buoyancy and the
Ideal Gas Law
 mballoon g = rair at 273 K V g – rair at T V g
 mballoon = rair at 273 K V – rair at T V
 mballoon = (1.2 – 330 / T) V
 700 / 4200 = 1.2 – 330 / T
 330 / T = (1.2 - 0.2)
 T = 330 K  57 C
Physics 207: Lecture 23, Pg 17
PV diagrams: Important processes
 Isochoric process:
V = const (aka isovolumetric)
 Isobaric process:
p = const
pV
 Isothermal process: T = const
 constant
T
1
p1V1  p2V2
1
Volume
Pressure
p1 p2

T1 T2
Isobaric
Isothermal
Pressure
Pressure
Isochoric
2
V1 V2

T1 T2
1
2
2
Volume
Volume
Physics 207: Lecture 23, Pg 18
Work and Energy Transfer (Ch. 16)
 K reflects the kinetic energy of the system
 ΔK =Wconservative + Wdissipative + Wexternal
Wconservative = - ΔU (e.g., gravity)
 Wdissipative = - ΔEThermal
 Wexternal  Typically, work done by contact forces
ΔK + ΔU + ΔETh = Wexternal= ΔEsys
Physics 207: Lecture 23, Pg 19
Work and Energy Transfer (Ch. 17)
ΔK + ΔU + ΔETh = Wexternal= ΔEsys
But we can transfer energy without doing work
Q ≡ thermal energy transfer
ΔK + ΔU + ΔETh = W + Q = ΔEsys
If ΔK + ΔU = ΔEMech = 0  ΔETh = W + Q
Physics 207: Lecture 23, Pg 20
1st Law of Thermodynamics
ΔEth =W + Q
W & Q with respect to the system
 Thermal energy Eth : Microscopic energy of moving molecules
and stretching molecular bonds. ΔEth depends on the initial
and final states but is independent of the process.
 Work W : Energy transferred to the system by forces in a
mechanical interaction.
 Heat Q : Energy transferred to the system via atomic-level
collisions when there is a temperature difference.
Physics 207: Lecture 23, Pg 21
1st Law of Thermodynamics
ΔEth =W + Q
W & Q with respect to the system
 Work W and heat Q depend on the step process by
which the system is changed (path dependent).
 The change of energy in the system, ΔEth depends
only on the total energy exchanged W+Q, not on the
process.
Physics 207: Lecture 23, Pg 22
1st Law: Work & Heat
 Work done on system (an ideal
gas)
Won  
final
 p dV  (area under curve )
initial
 Won system < 0 Moving left to right
[where (Vf > Vi)]
 If ideal gas, PV = nRT, and given Pi &
Vi fixes Ti
 Wby system > 0 Moving left to right
Physics 207: Lecture 23, Pg 23
 Work:
1st Law: Work & Heat
 Depends on the path taken in the PV-diagram
(It is not just the destination but the path…)
 Won system > 0 Moving right to left
Physics 207: Lecture 23, Pg 24
1st Law: Work (“Area” under the curve)
 Work depends on the path taken in the PV-diagram :
3
3
2
1
2
1
(a) Wa = W1 to 2 + W2 to 3 (here either P or V constant)
 Wa (on) = - Pi (Vf - Vi) + 0 > 0
(b) Wb = W1 to 2 + W2 to 3 (here either P or V constant)
 Wb (on) = 0 - Pf (Vf - Vi) > Wa > 0
(c) Need explicit form of P versus V but Wc (on) > 0
Physics 207: Lecture 23, Pg 25
Recap
• Assignment
 HW9, Due Wednesday, Apr. 15th
 HW10, Due Wednesday, Apr. 22nd (9 AM)
Physics 207: Lecture 23, Pg 26