1. Thermal Expansion D L = a Lo D T D L = change in_______
a = coefficient of _______expansion
Lo = original ________
D T = change in oC or___

2. Thermal Expansion D L = a Lo D T D L = change in_______
a = coefficient of _______expansion
Lo = original ________
D T = change in oC or___

3. Thermal Expansion D L = a Lo D T D L = change in length
a = coefficient of linear expansion
Lo = original length
D T = change in oC or K

4. Thermal Expansion D L = a Lo D T D L = change in length
a = coefficient of linear expansion
Lo = original length
D T = change in oC or K

5. Thermal Expansion D L = a Lo D T D L = change in length
a = coefficient of linear expansion
Lo = original length
D T = change in oC or K

6. Thermal Expansion D L = a Lo D T D L = change in length
a = coefficient of linear expansion
Lo = original length
D T = change in oC or K

7. Thermal Expansion D L = a Lo D T D L = change in length
a = coefficient of linear expansion
Lo = original length
D T = change in oC or K

8. Problem Example
A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C.
a = 12x10-6 m-1K-1
D L = a Lo D T= 12x10-6 m-1K-1(12m)(32K)
D L = m

9. Problem Example
A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C.
a = 12x10-6 m-1K-1
D L = a Lo D T= 12x10-6 m-1K-1(12m)(32K)
D L = m

10. Problem Example
A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C.
a = 12x10-6 m-1K-1
D L = a Lo D T= 12x10-6 m-1K-1(12m)(32K)
D L = m

11. Problem Example
A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C.
a = 12x10-6 m-1K-1
D L = a Lo D T= 12x10-6 m-1K-1(12m)(32K)
D L = m

12. Thermo Equations H = kADT l
Rate of heat transfer is equal to the ratio of the product of the coefficient of thermal conductivity, Area, temperature difference and the Length of the path
H = kADT l
-10oC H 25oC
20oC H oC
Lower heat transfer, A is smaller, DT
Is smaller and window thickness is
Greater (l)
Higher heat transfer, A is larger, DT is larger and the window is not
Thick (l)

13. Thermo Equations H = kADT l
Rate of heat transfer is equal to the ratio of the product of the coefficient of thermal conductivity, Area, temperature difference and the Length of the path
H = kADT l
-10oC H 25oC
20oC H oC
Lower heat transfer, A is smaller, DT
Is smaller and window thickness is
Greater (l)
Higher heat transfer, A is larger, DT is larger and the window is not
Thick (l)

14. Thermo Equations H = kADT l
Rate of heat transfer is equal to the ratio of the product of the coefficient of thermal conductivity, Area, temperature difference and the Length of the path
H = kADT l
-10oC H 25oC
20oC H oC
Lower heat transfer, A is smaller, DT
Is smaller and window thickness is
Greater (l)
Higher heat transfer, A is larger, DT is larger and the window is not
Thick (l)

15. Thermo Equations H = kADT l
Rate of heat transfer is equal to the ratio of the product of the coefficient of thermal conductivity, Area, temperature difference and the Length of the path
H = kADT l
-10oC H 25oC
20oC H oC
Lower heat transfer, A is smaller, DT
Is smaller and window thickness is
Greater (l)
Higher heat transfer, A is larger, DT is larger and the window is not
Thick (l)

16. Chapter 13 Temperature 1717 Fahrenheit – Instrument maker
0 Lowest temperature he could achieve with water,ice, and sea _____
96 body temperature
By chance water froze at ___Fand boiled at____F

17. Chapter 13 Temperature 1717 Fahrenheit – Instrument maker
0 Lowest temperature he could achieve with water,ice, and sea salt
96 body temperature
By chance water froze at 32 and boiled at 212.

18. Chapter 13 Temperature
1742 Anders Celsius used the freezing and boiling points of water for reference points and then divided them into 100 equal parts.
100 = freezing pt of ______
0 = boiling pt of ______ at standard pressure
Later changed to 0 C = Fpt and 100 C as Bpt
First known as centigrade scale
1954 recognized as _______scale

19. Chapter 13 Temperature
1742 Anders Celsius used the freezing and boiling points of water for reference points and then divided them into 100 equal parts.
100 = freezing pt of water
0 = boiling pt of water at standard pressure
Later changed to 0 C = Fpt and 100 C as Bpt
First known as centigrade scale
1954 recognized as Celsius scale

20. Gas Laws
Vol
OoC oC
Temperature 0C

21. Gas Laws Volume is directly related to temperature Vol V = constant T
OoC oC
Temperature 0C

22. Gas Laws
Vol
-273 oC OoC oC
Temperature 0C

23. Gas Laws
Pressure
OoC oC
Temperature 0C

24. Gas Laws Pressure is directly related to Pressure P = constant T
temperature
Pressure
P = constant T
OoC oC
Temperature 0C

25. Gas Laws
Pressure
-273 oC OoC oC
Temperature 0C

26. Absolute Zero The temperature at which the volume of
of an ideal gas is zero and the pressure is
zero due to the lack of _______of the
particles
1848 – William Thomson – latter to be known
as Lord Kelvin formalized the concept theoretically.
C is considered absolute ______or degrees below the _____point of water (ice, water and water vapor exists at .010C at a pressure of 610 Pa)

27. Absolute Zero The temperature at which the volume of
of an ideal gas is zero and the pressure is
zero due to the lack of motion of the
particles
1848 – William Thomson – latter to be known
as Lord Kelvin formalized the concept theoretically.
C is considered absolute zero or degrees below the triple point of water (ice water and water vapor exists at .010C at a pressure of 610 Pa)

28. 0C to Kelvin = 0C +_____= Kelvin Kelvin to oC = K - _____= oC
Absolute Zero
0C to Kelvin = 0C +_____= Kelvin
Kelvin to oC = K - _____= oC

29. 0C to Kelvins = 0C + 273 = Kelvins Kelvins to oC = K - 273 = oC
Absolute Zero
0C to Kelvins = 0C = Kelvins
Kelvins to oC = K = oC

30. 0C to Kelvins = 0C + 273 = Kelvins Kelvins to oC = K - 273 = oC
Absolute Zero
0C to Kelvins = 0C = Kelvins
Kelvins to oC = K = oC

31. Gas laws continued P
Volume

32. Gas laws continued The pressure of a gas sample is inversely
proportional to its volume
P PV = constant
Volume

33. Gas law problem
A tank having a volume of 1.00 m3 is filled with air at 00C to 20 times atmospheric pressure. How much volume will that gas occupy at 1.00 atm and 200C?

34. Gas law problem
A tank having a volume of 1.00 m3 is filled with air at 00C to 20 times atmospheric pressure. How much volume will that gas occupy at 1.00 atm and 200C?
P1V1 = P2V2
T T2
1.00 m3(20 atm) = V2(1.00 atm)
273 K K
V2 = m3

35. Ideal Gas Equation PV = nRT P = 1.013 x 105 Pa V =.0224 m3
T = 273 Kelvins
R = 8.31 m3 Pa / mol K

36. Ideal Gas Equation PV = nRT P = 1.013 x 105 Pa V =.0224 m3
n = 1.0 moles
T = 273 Kelvins
R = 8.31 m3 Pa / mol K

37. Ideal Gas Equation Problem
PV = nRT
What is the volume of 1.00 mole of any gas at x105 Pa and 273 K?
V = n R T P

38. Ideal Gas Equation Problem
PV = nRT
What is the volume of 1.00 mole of any gas at x105 Pa and 273 K?
V = n R T = 1.00 mol (8.31 J / mol k) (273 K)
P x105 Pa

39. Ideal Gas Equation Problem
PV = nRT
What is the volume of 1.00 mole of any gas at x105 Pa and 273 K?
V = n R T = 1.00 mol (8.31 J / mol k) (273 K)
P x105 Pa
V= m3

40. Ideal Gas Equation n = N NA=6.02x1023 PV= N kB T kB = 1.38 x 10-23 J/K
PV = nRT
n = N
NA=6.02x1023
PV= N kB T
kB = 1.38 x J/K

41. Kinetic Theory
How does the microscopic movement of particles determine the macroscopic properties of matter?
James Clerk Maxwell and Ludwig Boltzman
Gas molecules are extremely small, hard, and perfectly _______-

42. Kinetic Theory
How does the microscopic movement of particles determine the macroscopic properties of matter?
James Clerk Maxwell and Ludwig Boltzman
Gas molecules are extremely small, hard, and perfectly elastic-

43. Kinetic Theory Assumptions
Ideal gas molecules – have no ______of their own and have ___attraction for other gas molecules.
Theoretical calculations based on the conservation of momentum associated with the _______of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules.
K.E.avg = 3/2 KBT T is in _____Kb=____________

44. Kinetic Theory Assumptions
Ideal gas molecules – have no volume of their own and have NO attraction for other gas molecules.
Theoretical calculations based on the conservation of momentum associated with the collision of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules.
K.E.avg = 3/2 KBT T is in Kelvin Kb1.38x10-23 J/K

45. Kinetic Theory Assumptions
Ideal gas molecules – have no volume of their own and have NO attraction for other gas molecules.
Theoretical calculations based on the conservation of momentum associated with the collision of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules.
K.E.avg = 3/2 KBT T is in Kelvin Kb1.38x10-23 J/K

46. Kinetic Theory Assumptions
Ideal gas molecules – have no volume of their own and have NO attraction for other gas molecules.
Theoretical calculations based on the conservation of momentum associated with the collision of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules.
K.E.avg = 3/2 KBT T is in Kelvin Kb1.38x10-23 J/K

47. Average speed of gas molecules
K.E.avg =
½ mv2avg=
(v2)avg=
Root mean square velocity =

48. Average speed of gas molecules
K.E.avg = 3/2 KBT
½ mv2avg=3/2 KBT
(v2)avg= 3kBT m
Root mean square velocity = 3kBT

49. Example Problem
What is the root mean square velocity of hydrogen molecules and oxygen molecules at STP? H2 = 2.0g / mole amu
O2 = 32.0g / mole amu
Root mean square velocity= 3kBT m
vrms= (1.38x10-23 J / k) (273 K) =1839 m/s
2.0 amu(1.67x10-27kg)

50. Example Problem
What is the root mean square velocity of hydrogen molecules and oxygen molecules at STP?
O2 = 32.0g / mole amu
Root mean square velocity= 3RT M
vrms= (1.38x10-23 J / k) (273 K) =460 m/s
32.0 g(1 kg/100g)

51. Relative Velocities – Graham law of diffusion
vA = mB
VB mA
The velocity of particle A is inversely related
to the square root of the mass

52. 1st Law of Thermodynamics chapter 15
The internal energy of a sample is defined as the sum total of the __________ ___________energy of the molecules in a sample
U=N(1/2mv2)
U=3/2 NkT N=
U=3/2 nRT n=

53. 1st Law of Thermodynamics chapter 15
The internal energy of a sample is defined as the sum total of the _kinetic energy of the molecules in a sample
U=N(1/2mv2)
U=3/2 NkT N=
U=3/2 nRT n=

54. 1st Law of Thermodynamics chapter 15
The internal energy of a sample is defined as the sum total of the _kinetic energy of the molecules in a sample
U=N(1/2mv2)
U=3/2 NkT N=# of molecules
U=3/2 nRT n=# of moles of molecules

55. 1st Law of Thermodynamics chapter 15
The internal energy of a sample is defined as the sum total of the _kinetic energy of the molecules in a sample
U=N(1/2mv2)
U=3/2 NkT N=# of molecules
U=3/2 nRT n=# of moles of molecules
U=3/2 PV

56. 1st Law of Thermodynamics chapter 15
The internal energy of a sample is defined as the sum total of the _kinetic energy of the molecules in a sample
U=N(1/2mv2)
U=3/2 NkT N=# of molecules
U=3/2 nRT n=# of moles of molecules
U=3/2 PV
2/3 U = PV

57. 1st Law of Thermodynamics chapter 15
The internal energy of a system would _________ if heat were added to the system. Q=____
The internal energy of a system would _________ if work was done on the system. A Piston _______ a gas.

58. 1st Law of Thermodynamics chapter 15
The internal energy of a system would increase if heat were added to the system. Q=+
The internal energy of a system would increase if work was done on the system. A Piston compress a gas.

59. 1st Law of Thermodynamics chapter 15
The internal energy of a system would increase if heat were added to the system. Q=+
The internal energy of a system would increase if work was done on the system. A Piston compress a gas.

60. 1st Law of Thermodynamics chapter 15
The internal energy of a system would increase if heat were added to the system. Q=+
The internal energy of a system would increase if work was done on the system. A Piston compress a gas.

61. 1st Law of Thermodynamics chapter 15
The internal energy of a system would increase if heat were added to the system. Q=+
The internal energy of a system would increase if work was done on the system. A Piston compress a gas.

62. 1st Law of Thermodynamics chapter 15
The internal energy of a system would increase if heat were added to the system. Q=+
The internal energy of a system would increase if work was done on the system. A Piston compress a gas.

63. 1st Law of Thermodynamics chapter 15
The internal energy of a system would increase if heat were added to the system. Q=+
The internal energy of a system would increase if work was done on the system. A Piston compresses a gas.

64. 1st Law of Thermodynamics chapter 15
The internal energy of a system would increase if heat were added to the system. Q=+
The internal energy of a system would increase if work was done on the system. A Piston compresses a gas.

65. 1st Law of Thermodynamics chapter 15
The internal energy of a system would increase if heat were added to the system. Q=+
The internal energy of a system would increase if work was done on the system. A Piston compresses a gas.

66. 1st Law of Thermodynamics chapter 15
First Law of Thermodynamics
Change in Internal Energy = Heat transferred plus work done ON the system
Formula
D U = Q + W

67. 1st Law of Thermodynamics chapter 15
First Law of Thermodynamics
Change in Internal Energy = Heat transferred plus work done ON the system
Formula
D U = Q + W

68. 1st Law of Thermodynamics chapter 15
First Law of Thermodynamics
Change in Internal Energy = Heat transferred plus work done ON the system
Formula
D U = Q + W

69. 1st Law of Thermodynamics chapter 15
First Law of Thermodynamics
Change in Internal Energy = Heat transferred plus work done ON the system
Formula
D U = Q + W

70. 1st Law of Thermodynamics chapter 15
If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will ______. System is _________than environment Q=___
If work is done BY the system then the internal energy of the system with respect to the work done will ___________.System is _________W= ____

71. 1st Law of Thermodynamics chapter 15
If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative
If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

72. 1st Law of Thermodynamics chapter 15
If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative
If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

73. 1st Law of Thermodynamics chapter 15
If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative
If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

74. 1st Law of Thermodynamics chapter 15
If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative
If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

75. 1st Law of Thermodynamics chapter 15
If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative
If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

76. 1st Law of Thermodynamics chapter 15
If heat is REMOVED from a system then the internal energy of the system with respect to the heat transferred will decrease. System is warmer than environment Q= negative
If work is done BY the system then the internal energy of the system with respect to the work done will decrease . System is expands W=negative

77. 1st Law of Thermodynamics chapter 15
If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will ______. System is ________ then the environment. Q=___
If work is done ON the system then the internal energy of the system with respect to the work done will ___________. System is ________ W=___

78. 1st Law of Thermodynamics chapter 15
If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +
If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

79. 1st Law of Thermodynamics chapter 15
If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +
If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

80. 1st Law of Thermodynamics chapter 15
If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +
If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

81. 1st Law of Thermodynamics chapter 15
If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +
If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

82. 1st Law of Thermodynamics chapter 15
If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +
If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

83. 1st Law of Thermodynamics chapter 15
If heat is ADDED to a system then the internal energy of the system with respect to the heat transferred will increase. System is cooler then the environment. Q= +
If work is done ON the system then the internal energy of the system with respect to the work done will increase. System is compressing W=Positive

84. 1st Law of Thermodynamics chapter 15
Isothermal Process-A process carried out at constant ___________. The heat reservoir is very _________ which essentially keeps the _________ constant. P V

85. 1st Law of Thermodynamics chapter 15
Isothermal Process-A process carried out at constant temperature. The heat reservoir is very large which essentially keeps the temperature constant. D U = O P V

86. 1st Law of Thermodynamics chapter 15
Isothermal Process-A process carried out at constant temperature. The heat reservoir is very large which essentially keeps the temperature constant. D U = O

87. 1st Law of Thermodynamics chapter 15
Isothermal Process-A process carried out at constant temperature. The heat reservoir is very large which essentially keeps the temperature constant. D U = O

88. 1st Law of Thermodynamics chapter 15
Adiabatic process is one in which no _____ is allowed to flow into or out of the system. Q = ___ These processes usually occur very ________.

89. 1st Law of Thermodynamics chapter 15
Adiabatic process is one in which no heat is allowed to flow into or out of the system. Q = zero. These processes usually occur very rapidly . D U = + if compressed
D U = - if expands

90. 1st Law of Thermodynamics chapter 15
Adiabatic process is one in which no heat is allowed to flow into or out of the system. Q = zero. These processes usually occur very rapidly . D U = + if compressed
D U = - if expands

91. 1st Law of Thermodynamics chapter 15
Adiabatic process is one in which no heat is allowed to flow into or out of the system. Q = zero. These processes usually occur very rapidly . D U = + if compressed
D U = - if expands

92. 1st Law of Thermodynamics chapter 15
Adiabatic process is one in which no heat is allowed to flow into or out of the system. Q = zero. These processes usually occur very rapidly . D U = + if compressed
D U = - if expands

93. 1st Law of Thermodynamics chapter 15
Isobaric Processes-one in which the _______ is keep ________.

94. 1st Law of Thermodynamics chapter 15
Isobaric Processes-one in which the pressure is keep constant. Work = -PD V = + if compressed
Work =-PD V = - if expanded

95. 1st Law of Thermodynamics chapter 15
Isobaric Processes-one in which the pressure is keep constant. Work = -PD V = + if compressed
Work = -PD V = - if expanded

96. 1st Law of Thermodynamics chapter 15
Isobaric Processes-one in which the pressure is keep constant. Work = -PD V = + if compressed
Work = -PD V = - if expanded

97. 1st Law of Thermodynamics chapter 15
Isobaric Processes-one in which the pressure is keep constant. Work = -PD V = + if compressed
Work =-PD V = - if expanded

98. 1st Law of Thermodynamics chapter 15
Isochoric-A process in which there is no change in _________

99. 1st Law of Thermodynamics chapter 15
Isochoric-A process in which there is no change in volume. No work is done
W=O

100. 1st Law of Thermodynamics chapter 15
Isochoric-A process in which there is no change in volume. No work is done
W=O

101. Work and volume changes
If the systems volume increases (it ________) then the work is done ___ the system and Work has a ___ sign. The change in volume(__ V is ___)
If the systems volume decreases (it ________) then the work is done ___ the system and Work has a ___ sign. The change in volume(__V is ____)

102. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

103. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

104. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

105. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

106. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

107. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

108. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

109. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

110. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

111. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

112. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

113. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

114. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

115. Work and volume changes
Work = force *distance
Work = Force *Volume ___
Area
Work= - Pressure x Change in Volume = - PD V
If the systems volume increases (it expands) then the work is done by the system and Work has a - sign. The change in volume(__ V is + )
If the systems volume decreases (it compresses) then the work is done on the system and Work has a + sign. The change in volume(__V is _- )

116. 2nd Law of Thermodynamics
The 1st law of thermodynamics states that ________ is conserved.
Many processes which could obey the 1st law of thermodynamics do not occur ________________.
The second law of thermodynamics is a statement of which processes_______ in nature and which ________.
The second law of thermodynamics - the total ______of a system plus that of the environment _________ as the result of any natural process.
Natural processes tend to move toward a state of ________.

117. 2nd Law of Thermodynamics
The 1st law of thermodynamics states that energy is conserved.
Many processes which could obey the 1st law of thermodynamics do not occur _spontaneously _.
The second law of thermodynamics is a statement of which processes occur in nature and which do not.
The second law of thermodynamics - the total entropy of a system plus that of the environment increases_ as the result of any natural process.
Natural processes tend to move toward a state of disorder / choas/entropy________.

118. 2nd Law of Thermodynamics
The 1st law of thermodynamics states that energy is conserved.
Many processes which could obey the 1st law of thermodynamics do not occur _spontaneously _.
The second law of thermodynamics is a statement of which processes occur in nature and which do not.
The second law of thermodynamics - the total entropy of a system plus that of the environment increases_ as the result of any natural process.
Natural processes tend to move toward a state of disorder / choas/entropy________.

119. 2nd Law of Thermodynamics
The 1st law of thermodynamics states that energy is conserved.
Many processes which could obey the 1st law of thermodynamics do not occur _spontaneously _.
The second law of thermodynamics is a statement of which processes occur in nature and which do not.
The second law of thermodynamics - the total entropy of a system plus that of the environment increases_ as the result of any natural process.
Natural processes tend to move toward a state of disorder / choas/entropy________.

120. 2nd Law of Thermodynamics
The 1st law of thermodynamics states that energy is conserved.
Many processes which could obey the 1st law of thermodynamics do not occur _spontaneously _.
The second law of thermodynamics is a statement of which processes occur in nature and which do not.
The second law of thermodynamics - the total entropy of a system plus that of the environment increases_ as the result of any natural process.
Natural processes tend to move toward a state of disorder / choas/entropy________.

121. 2nd Law of Thermodynamics
The 1st law of thermodynamics states that energy is conserved.
Many processes which could obey the 1st law of thermodynamics do not occur _spontaneously _.
The second law of thermodynamics is a statement of which processes occur in nature and which do not.
The second law of thermodynamics - the total entropy of a system plus that of the environment increases_ as the result of any natural process.
Natural processes tend to move toward a state of disorder / choas/entropy________.

122. 2nd Law of Thermodynamics
The 1st law of thermodynamics states that energy is conserved.
Many processes which could obey the 1st law of thermodynamics do not occur _spontaneously _.
The second law of thermodynamics is a statement of which processes occur in nature and which do not.
The second law of thermodynamics - the total entropy of a system plus that of the environment increases_ as the result of any natural process.
Natural processes tend to move toward a state of disorder / choas/entropy________.

123. 2nd Law of Thermodynamics
The 1st law of thermodynamics states that energy is conserved.
Many processes which could obey the 1st law of thermodynamics do not occur _spontaneously _.
The second law of thermodynamics is a statement of which processes occur in nature and which do not.
The second law of thermodynamics - the total entropy of a system plus that of the environment increases_ as the result of any natural process.
Natural processes tend to move toward a state of disorder / choas/entropy________.

124. 2nd Law of Thermodynamics
Heat Engines- Devices designed to obtain ______ energy from __________ enegy
1700’s ________ Engine TH TL QH
Q L W
QH =

125. 2nd Law of Thermodynamics
Heat Engines- Devices designed to obtain mechanical energy from thermal enegy
1700’s Steam Engine
TH temperature of high temperature reservoir
TL temperature of low temperature reservoir
QH heat released by high temperature reservoir
Q Lheat absorbed by high temperature reservoir
W = work output
QH = heat input

126. 2nd Law of Thermodynamics
Heat Engines- Devices designed to obtain mechanical energy from thermal enegy
1700’s Steam Engine
TH temperature of high temperature reservoir
TL temperature of low temperature reservoir
QH heat released by high temperature reservoir
Q Lheat absorbed by high temperature reservoir
W = work output
QH = heat input

127. 2nd Law of Thermodynamics
Heat Engines- Devices designed to obtain mechanical energy from thermal enegy
1700’s Steam Engine
TH temperature of high temperature reservoir
TL temperature of low temperature reservoir
QH heat released by high temperature reservoir
Q Lheat absorbed by high temperature reservoir
W = work output
QH = heat input

128. 2nd Law of Thermodynamics
Heat Engines- Devices designed to obtain mechanical energy from thermal enegy
1700’s Steam Engine
TH temperature of high temperature reservoir
TL temperature of low temperature reservoir
QH heat released by high temperature reservoir
Q Lheat absorbed by high temperature reservoir
W = work output
QH = heat input

129. 2nd Law of Thermodynamics
Heat Engines- Devices designed to obtain mechanical energy from thermal enegy
1700’s Steam Engine
TH temperature of high temperature reservoir
TL temperature of low temperature reservoir
QH heat released by high temperature reservoir
Q Lheat absorbed by high temperature reservoir
W = work output
QH = heat input

130. 2nd Law of Thermodynamics
Heat Engines- Devices designed to obtain mechanical energy from thermal enegy
1700’s Steam Engine
TH temperature of high temperature reservoir
TL temperature of low temperature reservoir
QH heat released by high temperature reservoir
Q Lheat absorbed by high temperature reservoir
W = work output
QH = heat input

131. 2nd Law of Thermodynamics
Heat Engines- Devices designed to obtain mechanical energy from thermal enegy
1700’s Steam Engine
TH temperature of high temperature reservoir
TL temperature of low temperature reservoir
QH heat released by high temperature reservoir
Q Lheat absorbed by high temperature reservoir
W = work output
QH = heat input

132. 2nd Law of Thermodynamics
Heat Engines- Devices designed to obtain mechanical energy from thermal enegy
1700’s Steam Engine
TH temperature of high temperature reservoir
TL temperature of low temperature reservoir
QH heat released by high temperature reservoir
Q Lheat absorbed by high temperature reservoir
W = work output
QH = heat input

133. 2nd Law of Thermodynamics
Efficiency e=
Ideal carnot efficiency
e ideal =
No device is possible whose sole effect is to transform a given amount of heat completely into _______

134. 2nd Law of Thermodynamics
Efficiency
e= W/QH
e=QH-QL/QH
e= 1-Ql/QH
Ideal carnot efficiency
e ideal = TH - TL / TH
No device is possible whose sole effect is to transform a given amount of heat completely into work

135. 2nd Law of Thermodynamics
Efficiency
e= W/QH
e=QH-QL/QH
e= 1-Ql/QH
Ideal carnot efficiency
e ideal = TH - TL / TH
No device is possible whose sole effect is to transform a given amount of heat completely into work

136. 2nd Law of Thermodynamics
Efficiency
e= W/QH
e=QH-QL/QH
e= 1-Ql/QH
Ideal carnot efficiency
e ideal = TH - TL / TH
No device is possible whose sole effect is to transform a given amount of heat completely into work

137. 2nd Law of Thermodynamics
Efficiency
e= W/QH
e=QH-QL/QH
e= 1-Ql/QH
Ideal carnot efficiency
e ideal = TH - TL / TH
No device is possible whose sole effect is to transform a given amount of heat completely into work

138. 2nd Law of Thermodynamics
Efficiency
e= W/QH
e=QH-QL/QH
e= 1-Ql/QH
Ideal carnot efficiency
e ideal = TH - TL / TH
No device is possible whose sole effect is to transform a given amount of heat completely into work

139. 2nd Law of Thermodynamics
Efficiency
e= W/QH
e=QH-QL/QH
e= 1-Ql/QH
Ideal carnot efficiency
e ideal = TH - TL / TH
No device is possible whose sole effect is to transform a given amount of heat completely into work

140. 2nd Law of Thermodynamics
Efficiency
e= W/QH
e=QH-QL/QH
e= 1-Ql/QH
Ideal carnot efficiency
e ideal = TH - TL / TH
No device is possible whose sole effect is to transform a given amount of heat completely into work

141. 2nd Law of Thermodynamics
Efficiency
e= W/QH
e=QH-QL/QH
e= 1-Ql/QH
Ideal carnot efficiency
e ideal = TH - TL / TH
No device is possible whose sole effect is to transform a given amount of heat completely into work
The greater the difference in temperature the greater the ideal efficiency

142. Carnot Cycle Isothermal expansion D U = O W = neg Q = pos(absorbing heat)

143. Carnot Cycle Adiabiatic expansion D U = - W = neg Q = 0

144. Carnot Cycle Isothermal compression D U = O W = + Q = neg (releases heat)

145. Carnot Cycle Adiabatic compression D U = + W = positive Q = 0

146. Heat
Heat flows from a sample with a high temperature to one with a lower
temperature

147. Heat Measurements
calorie = the amount of _______ necessary to raise the ________ of ___gram of a water ____ degree celsius
____ cal = 1 ____ = 1 kcal
James Prescott Joule-Mechanical Equivalent of heat-Experiment
1 cal = ________ joules

148. Heat Measurements
calorie = the amount of energy necessary to raise the temperature of 1 gram of a water
1 degree celsius
1000cal = 1 kcal
James Prescott Joule-Mechanical Equivalent of heat-Experiment
1 cal =4.18 joules

149. Heat Definition/Formula
Heat is the energy that is ____________ from one body to another because of a difference in ______________
Q =

150. Heat Definition/Formula
Heat is the energy that is transferred from one body to another because of a difference in temperature
Q = mcD t

151. Temperature Definition/Formula
A measure of the average ______energy of matter
Formula: K.E. =______

152. Temperature Definition/Formula
A measure of the average kinetic energy of matter
Formula: K.E. = 3/2 kBT

153. Rate of heat transfer Formula D Q / t = k A ( T1-T2) / l Q= t = k= A=
Can be used to determine the rate of heat transfer across a
__________

154. Rate of heat transfer Formula Q/t = K A (DT) l
Q= heat transfer in joules
t = time heat transferred
k= coefficient of heat transfer
A= Area of material
T1 - T2= temperature difference between sides
l = thickness of material
Can be used to determine the rate of heat transfer across a window

155. Internal Energy-Thermal Energy
The sum of the _________ energy of ___the molecules in a sample
Formula:
U=____________
U = Internal energy______
N = __________of particles
T =___________

156. Internal Energy-Thermal Energy
The sum of the kinetic energy of all the molecules in a sample
Formula:
U=N 3/2 KBT
U = Internal energy change
N = number of particles
T = temperature

157. 1st Law of Thermodynamics chapter 15
Thermodynamics is the study of the ____________ in which energy is transferred as ______ and as _______
Heat is the energy transferred due to a difference in _____________
Work is the energy transferred that is not due to a _____________ difference.

158. 1st Law of Thermodynamics chapter 15
Thermodynamics is the study of the
Processes in which energy is transferred as heat and as work.
Heat is the energy transferred due to a difference in temperature
Work is the energy transferred that is not due to a temperature difference.

159. 1st Law of Thermodynamics chapter 15
Closed system- an object or set of objects for which no __________ enters or leaves. Energy can be __________ with the environment.
Isolated Closed system - an object or set of objects which no __________ enters or leaves and no _____________ is _______
Open system - _______ and ________ may enter or leave.

160. 1st Law of Thermodynamics chapter 15
Closed system- an object or set of objects for which no matter enters or leaves. Energy can be exchanged with the environment.
Isolated Closed system - an object or set of objects which no matter enters or leaves and no energy is
transferred
Open system – matter and energy may enter or leave.