Transcript Physics 207: Lecture 2 Notes
• Goals: Chapter 16 Use the ideal-gas law.
Lecture 23 • • Use
pV
diagrams for ideal-gas processes.
Chapter 17 Employ energy conservation in terms of 1 st 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. 14 th HW10, Due Wednesday, Apr. 21 st (9 AM) Exam 3 on Wednesday, Apr. 21 at 7:15 PM (Statics, Ang. Mom., Ch.14 through 17) 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
Phase diagram
Triple point of water: p = 0.06 atm T = 0.01
°C Triple point of CO 2 : 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
Changes due to Heat or Thermal Energy Transfer
Change the temperature (it gets hotter) Change the state of matter (solid liquid, liquid gas) Accurate water phase diagram Path on heating Physics 207: Lecture 23, Pg 4
Energy Transfer 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 5
Defining a temperature scale
Three main scales Farenheit Celsius 212 100 Kelvin 373.15
Water boils 32 -459.67
0 -273.15
273.15
0 Water freezes Absolute Zero Physics 207: Lecture 23, Pg 8
Some interesting facts
T (K) 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 30 o F), 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.
10 8 10 7 10 6 10 5 10 4 10 3 100 10 1 0.1
Hydrogen bomb Sun’s interior Solar corona Sun’s surface Copper melts Water freezes Liquid nitrogen Liquid hydrogen Liquid helium Lowest T~ 10 -9 K Physics 207: Lecture 23, Pg 9
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
Physics 207: Lecture 23, Pg 10
Ideal gas: Macroscopic description
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 n ≡ number of moles In SI units,
R
=8.315 J / mol·K Physics 207: Lecture 23, Pg 11
Boltzmann’s constant
Number of moles: n = m/M m = mass (kg) M= mass of one mole (kg/mol) One mole contains N A =6.022 X 10 23 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/N A
)
RT pV = N k B T k B = R/N A = 1.38
X 10 -23 J/K
k B
is called the Boltzmann’s constant
p, V,
and
T
are thermodynamic variables
Physics 207: Lecture 23, Pg 12
The Ideal Gas Law
pV
nRT
What is the volume of 1 mol of gas at STP ?
T = 0 °C = 273 K p = 1 atm = 1.01 x 10 5 Pa
V
nRT p
8 .
31 J / mol 1 .
01 10 5 K Pa 273 K 0 .
0224 m 3 22 .
4 Physics 207: Lecture 23, Pg 13
The Ideal Gas Law
pV
nRT
There are four things that can vary: p, V, n & T Typically, two of these are held constant and the relationship between the remaining two is studied
Physics 207: Lecture 23, Pg 14
Example
A spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having a volume of 125.00 cm 3 is at 27 o C. It is then tossed into an open fire. When the temperature of the gas in the can reaches 327 o C, what is the pressure inside the can? Assume any change in the volume of the can is negligible.
Steps 1.
2.
3.
Convert to Kelvin (From 300 K to 600 K) Use P/T = nR/V = constant P 1 /T 1 = P 2 /T 2 Solve for final pressure P 2 = P 1 T 2 /T 1
WD40 foolishness
Physics 207: Lecture 23, Pg 15
Example problem: Air bubble rising
A diver produces an air bubble underwater, where the absolute pressure is p 1 = 3.5 atm the pressure is p 2 . The bubble rises to the surface, where = 1.0 atm . The water temperatures at the bottom and the surface are, respectively, T 1 = 4 °C, T 2 = 23 °C What is the ratio of the volume,V 2 , of the bubble just as it reaches the surface to its volume at the bottom, V 1 ? 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 16
Example problem: Air bubble rising
A diver produces an air bubble underwater, where the absolute pressure is p 1 = 3.5 atm the pressure is p 2 . The bubble rises to the surface, where = 1 atm . The water temperatures at the bottom and the surface are, respectively, T 1 = 4 °C, T 2 = 23 °C What is the ratio of the volume of the bubble as it reaches the surface,V pV=nRT 2 , to its volume at the bottom, V pV/T = const so p 1 V 1 /T 1 = p 1 2 ? (Ans.V
V 2 /T 2 2 /V 1 = 3.74) V 2 /V 1 = p 1 T 2 / (T 1 p 2 ) = 3.5 296 / (277 1) If thermal transfer is efficient. [More than likely the expansion will be “adiabatic” and, for a diatomic gas, PV g = const. where g = 7/5, see Ch. 17 & 18] Physics 207: Lecture 23, Pg 18
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/m 3 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.
p, V and R do not change!
Balloon weight = Buoyant force – Weight of hot air Ideal gas law: pV = nRT or r T = const. = 1.2 x 273 kg K/m 3 nT= pV/R = const.
Physics 207: Lecture 23, Pg 19
Buoyancy and the Ideal Gas Law
m balloon g = r air at 273 K V g – r air at T V g m balloon = r air at 273 K V – r air at T V m balloon = (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 20
pV diagrams: Important processes
Isochoric process:
V = const
(aka isovolumetric) Isobaric process:
p = const
Isothermal process:
T = const pV T
constant
Isochoric 2 p
1
T
1
p T
2 2
1 Volume 1 Isothermal p
1
V
1
p
2
V
2
Volume 1 Isobaric V
1
T
1
V T
2 2
2 2 Volume
Physics 207: Lecture 23, Pg 21
pV vs. Fx diagrams
Work (on the system) remains the area under the curve
dW = F dx
or
dW = F/A (A dx) = P dV world = -p dV system 1 Isothermal p
1
V
1
p
2
V
2
1 pV
constant
T Volume 2 2 Position
Physics 207: Lecture 23, Pg 22
Work and Energy Transfer (Ch. 17)
K reflects the kinetic energy of the system
Δ
K
=
W conservative
+
W dissipative + W external
W conservative
= ΔU (e.g., gravity)
W dissipative = -
Δ
E Thermal
W external
Typically, work done by contact forces Δ
K +
ΔU
+
Δ
E Th = W external =
Δ
E sys
Physics 207: Lecture 23, Pg 23
Work and Energy Transfer
Δ
K +
ΔU
+
Δ
E
Th
= W
external
=
Δ
E
sys
But we can transfer energy without doing work
Q ≡ thermal energy transfer
Δ
K +
ΔU
+
Δ
E
Th
= W + Q =
Δ
E
sys
If Δ
K +
ΔU
=
Δ
E
Mech
= 0
Δ
E
Th
= W + Q
Physics 207: Lecture 23, Pg 24
1 st Law of Thermodynamics
Δ
E th
=
W
+
Q W & Q with respect to the system
Thermal energy E th : Microscopic energy of moving molecules and stretching molecular bonds. ΔE th 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 25
Lecture 23 • Assignment HW9, Due Wednesday, Apr. 14 th HW10, Due Wednesday, Apr. 21 st (9 AM) Exam 3 on Wednesday, Apr. 21 at 7:15 PM (Statics, Ang. Mom., Ch.14 through 17) Physics 207: Lecture 23, Pg 26