• 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