1. Ch6.1 The Nature of Energy (hustle!)
Energy – the capacity to do work or to produce heat.
Law of Conservation of Energy – energy can be converted from
one form to another, but can be neither created or destroyed.
1st Law of thermodynamics – the energy of the universe is constant.
Potential energy – stored energy due to position or composition
Kinetic energy – energy of motion KE = ½mv2
Heat – (a form of energy) transfer of energy due to a temp difference
Temperature – property that reflects the randomness of the particles
(NOT a form of energy)

2. Enthalpy Enthalpy, H – the ‘heat content’ of a substance
ΔH = Hproducts – Hreactants
1. Exothermic reaction –
CH4(g) + O2(g) 
Enthalpy
time

3. Enthalpy Enthalpy, H – the ‘heat content’ of a substance
ΔH = Hproducts – Hreactants
1. Exothermic reaction – heat flows out of the system (feel hot)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + energy
In any exothermic reaction, some of the potential energy stored
in chemical bonds is converted to thermal energy.
Activation
Energy
Reactants ∆H
Products
Enthalpy
time

4. Enthalpy Enthalpy, H – the ‘heat content’ of a substance
ΔH = Hproducts – Hreactants
Endothermic reaction –
N2(g) + O2(g) 
Enthalpy
time

5. Enthalpy Enthalpy, H – the ‘heat content’ of a substance
ΔH = Hproducts – Hreactants
Endothermic reaction – absorb heat from their surroundings (feels cold)
- heat flows into the system
N2(g) + O2(g) + energy  2NO(g)
Activation
Energy
Products ∆H
Reactants
Enthalpy
time

6. 1st Law of Thermodynamics – the energy of the universe is constant.
ΔE = q + w
Internal heat work
energy energy
“All energy movements reflect the system’s point of view.”
ΔE = Internal energy of the system increased
ΔE = – Internal energy of the system decreased
q = + Heat flowed into the system
q = – Heat flowed out of the system
w = + The surroundings do work on the system
w = – The system does work on the surroundings
”The system is usually a gas in a container.”

7. Ex1) Calculate ΔE for a system undergoing an endothermic process
in which 15.6kJ of heat flows
and where 1.4kJ of work is done on the system.

8. Work associated with gases: Pressure: Work:
release
energy
(( ))
(( ))
(( ))
(( ))
(( ))

10. Ex2) Calc the work associated with the expansion of a gas
from 46L to 64L at a constant pressure of 15 atm.

11. 1.3x108J of heat are added. Atmospheric pressure is 1.0 atm.
Ex3) A balloon is being inflated to its full extent by heating the air inside it.
The volume changes from 4.00x106L to 4.50x106L when
1.3x108J of heat are added. Atmospheric pressure is 1.0 atm.
Calculate ΔE for the process.
Ch6 HW#1 p283 21,23,25,26

12. Ch6 HW#1 p283 21,23,25,27
21. Calculate ΔE for each of the following cases.
a. q = +51 kJ, w = -15kJ
b. q = +100 kJ, w = -65kJ
c. q = -65kJ, w = -20kJ
d. In which of these does the system do work on the surroundings?

13. Ch6 HW#1 p283 21,23,25,26
21. Calculate ΔE for each of the following cases.
ΔE = q + w
a. q = +51 kJ, w = -15kJ
ΔE = +51 kJ + (-15kJ) = +36kJ (Internal energy increases)
b. q = +100 kJ, w = -65kJ
ΔE = +100 kJ + (-65kJ) = +35kJ (Internal energy increases)
c. q = -65kJ, w = -20kJ
ΔE = -65kJ + (-20kJ) = -85kJ (Internal energy decreases)
d. In which of these does the system do work on the surroundings?
All 3 have w = (–) so all systems did work on the surroundings

14. 23. A gas absorbs 45kJ of heat and does 29kJ of work. Calculate ΔE.
ΔE = q + w

15. 25. The volume of an ideal gas is decreased from 5.0 L to 5.0 mL
at a constant pressure of 2.0 atm.
Calculate the work associated with this process.
ΔV = (0.005L – 5L) = L

16. 25. The volume of an ideal gas is decreased from 5.0 L to 5.0 mL
at a constant pressure of 2.0 atm.
Calculate the work associated with this process.
ΔV = (0.005L – 5L) = L
W = –P. ΔV =

17. 26. Consider the mixture of air and gasoline vapor in a cylinder with
a piston. The original volume is 40. cm3. If the combustion of this
mixture releases 950. J of energy, to what volume will the gases
expand against a constant pressure of 650. torr if all the energy of
combustion is conserved into work to push back the piston.
ΔV = (V2 – 0.040L)
q  W =

18. 26. Consider the mixture of air and gasoline vapor in a cylinder with
a piston. The original volume is 40. cm3. If the combustion of this
mixture releases 950. J of energy, to what volume will the gases
expand against a constant pressure of 650. torr if all the energy of
combustion is conserved into work to push back the piston.
ΔV = (V2 – 0.040L)
q  W = –950J (expands – gas doing work)
W = –P. ΔV

19. Ch6.2A – Enthalpy Enthalpy, H – the ‘heat content’ of a substance
H is related to q.
For practical purposes:
q is the heat measured in an experiment
∆H is the heat related to a chem rxn.

20. Ex1) When 1 mol of methane is burned at constant pressure,
890kJ of energy is released as heat. Calculate ΔH for a process
in which a 5.8g sample is burned.

21. Ex2) a. Write an eqn for the combustion of ethanol, C2H5OH.
1235 kJ of heat is given off.
b. Draw an energy diagram.
c. How much heat is produced when 12.5g of ethanol reacts?
Ch6 HW#2 p283 29,30,31,33,34

22. Ch6 HW#2 p283 29,30,31,33,34
29. The equation for the fermentation of glucose to alcohol
and carbon dioxide is
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)
The enthalpy change for the reaction is -67kJ.
Is the reaction exothermic or endothermic?
Is energy, in the form of heat, absorbed or evolved as the reaction occurs?

23. 29. The equation for the fermentation of glucose to alcohol
and carbon dioxide is
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g) + 67kJ
exothermic
heat is evolved as the reaction occurs?

24. 30. The reaction SO3(g) + H2O(l) → H2SO4(aq)
is the last step in the commercial production of sulfuric acid.
The enthalpy change for this reaction is -227 kJ.
In the designing a sulfuric acid plant, is it necessary to provide
for heating or cooling of the reaction mixture? Explain.

25. 30. The reaction SO3(g) + H2O(l) → H2SO4(aq)
is the last step in the commercial production of sulfuric acid.
The enthalpy change for this reaction is -227 kJ.
In the designing a sulfuric acid plant, is it necessary to provide
for heating or cooling of the reaction mixture? Explain.
SO3(g) + H2O(l) → H2SO4(aq) kJ
Cool the plant!

26. 31. Are the following processes exothermic or endothermic?
a. When solid KBr is dissolved in water, the solution gets colder.
KBr(s) + _____ → K+(aq) + Br-(aq) + _____
b. Natural gas (CH4) is burned in a furnace.
CH4(g) + O2(g) + _____ → CO2(g) + H2O(g) + _____
c. When concentrated H2SO4 is added to water, the solution gets
very hot.
H2SO4(aq) + _____ → 2H+(aq) + SO42-(aq) + _____
d. Water is boiled in a teakettle.
H2O(l) + _____ → H2O(g) + _____

27. 33. For the reaction
S(s) + O2(g) → SO2(g) ΔH = -296 kJ/mol
a. How much heat is evolved when 275g sulfur is burned in excess O2?
b. How much heat is evolved when 25mol sulfur is burned in excess O2?
c. How much heat is evolved when 150. g sulfur dioxide is produced?

28. 33. For the reaction
S(s) + O2(g) → SO2(g) ΔH = -296 kJ/mol
a. How much heat is evolved when 275g sulfur is burned in excess O2?
q =
b. How much heat is evolved when 25mol sulfur is burned in excess O2?
c. How much heat is evolved when 150. g sulfur dioxide is produced?

29. 34. The overall reaction in commercial heat packs can be represented
as 4Fe(s) + 3O2(g) → 2Fe2O3(s) ΔH = -1652kJ
a. About how much heat is released when 4.00 mol iron is reacted
with excess O2?
b. How much heat is released when 1.00 mol Fe2O3 is produced?
c. How much heat is released when 1.00 g iron is reacted with
excess O2?
d. How much heat is released when 10.0 g Fe and 2.00 g O2
are reacted?

30. Ch6.3 – Calorimetry Calorimetry – the measure of heat flow
Heat Capacity (C) – of heat flow to a substance,
compared to its temp change.
(In lab 6.2 you will find C for a foam cup)

31. Calorimetry – the measure of heat flow
Heat Capacity (C) – of heat flow to a substance,
compared to its temp change.
Molar Heat Capacity – heat flow per mole, per temp change
units:

32. Calorimetry – the measure of heat flow
Heat Capacity (C) – of heat flow to a substance,
compared to its temp change.
Specific heat capacity (s or Cp) – measure of heat flow to 1 gram
of a substance, compared to its temp change.
Exs:
(In lab 6.1 you will find the Cp for an unknown…per gram of it)

33. Ex1) A 46.2 g sample of copper is heated to 95.4˚C and then placed in
a calorimeter containing 75.0 g water at 19.6˚C. The final temperature
of the metal and water is 21.8˚C. Calculate the specific heat capacity of
copper, assuming that all the heat lost by the copper is gained by water.

34. Ex2) 50ml of 1.00M HCl at 25.0oC is mixed with 50ml of 1.00M NaOH
at 25.0oC in a calorimeter. The temp of the mix reaches 31.9oC.
If the Cp of the solution is ,
and the density of the final soln is 1.0 g/ml, calc the enthalpy change
per mole of HCl for this neutralization reaction.
Ch6 HW#5 p284 27,42,43, + 2 more problems

35. Ex2) 50ml of 1.00M HCl at 25.0oC is mixed with 50ml of 1.00M NaOH
at 25.0oC in a calorimeter. The temp of the mix reaches 31.9oC.
If the Cp of the solution is ,
and the density of the final soln is 1.0 g/ml, calc the enthalpy change
per mole of HCl for this neutralization reaction.
HCl + NaOH  NaCl + HOH + ______J
0.05mol mol
(100ml)(1.0g/ml) = 100g
q = (100g)(4.18 J/g.oC)(5.9oC) = 2466J
ΔH = 2466J/0.050mol = 49,320J
Ch6 HW#3 p284 27,42, more problems

36. Ch6 HW#3 p284 27,42, more problems
27. A balloon filled with 39.1 mol helium has a volume of 876L a 0.0˚C
and 1.00 atm pressure. The temperature of the balloon is increased
to 38.0˚C as it expands to a volume of 998 L, the remaining pressure
remaining constant. Calculate q, w, ΔE for the helium in the balloon.
(The molar heat capacity for helium gas is 20.8 J/˚C·mol.)
n = 39.1 mol C/mol = 20.8 J/˚C·mol
Vi = 876L Vf = 998 L
Ti = 0.0˚C Tf = 38.0˚C
P = 1.00 atm = 101.3kPa (stays const)
q = ΔH =
W = –P.ΔV =
ΔE = q + W =

37. 42. A 15.0 g sample of nickel metal is heated to 100˚C and dropped
into 55.0 g of water, initially at 23˚C. Assuming that all the heat lost
by the nickel is absorbed by the water, calculate the final temp of
the nickel and water. (Specific heat of nickel = J/˚C · g)
Heat lost by nickel = Heat gained by water
– m.cpNi.ΔT = + m.cpw.ΔT

38. 42. A 15.0 g sample of nickel metal is heated to 100˚C and dropped
into 55.0 g of water, initially at 23˚C. Assuming that all the heat lost
by the nickel is absorbed by the water, calculate the final temp of
the nickel and water. (Specific heat of nickel = J/˚C · g)
Heat lost by nickel = Heat gained by water
– m.cpNi.ΔT = + m.cpw.ΔT

39. 43. A 150. 0-g sample of metal at 75. 0˚C is added to 150
43. A g sample of metal at 75.0˚C is added to g of H2O at 15.0˚C. The temperature of water rises to 18.3˚C. Calculate the specific heat capacity of the metal, assuming that all the heat lost is gained by the water.
Heat lost by nickel = Heat gained by water
– m.cpNi.ΔT = + m.cpw.ΔT

40. 1. Given that we have 1200 cm3 of helium at 15°C and 99 kPa,
what will be its volume at STP?
2. Three moles of a gas are injected into an evacuated container,
with a fixed volume, and maintained at a temp of 60°C. If the pressure
increases to 103,700 Pa what volume is the container?
PV = nRT

41. Ch6.4 Calorimetry and Enthalpy
Ex1) 1.00L of 0.10M Ba(NO3)2 solution at 25.0oC is mixed with
1.00L of 0.10M Na2SO4 soln at 25.0oC in a calorimeter. A ppt forms.
The temp of the mix reaches 28.1oC.
If the Cp of the solution is , and the density of the final soln
is 1.0 g/ml, calc the enthalpy change
per mole of ppt.

42. Ex2) 15.00 g of CaCl2 at 21.1oC is dissolved in 50.0mL of water
also at 21.1oC in a calorimeter. The temp of the mix reaches 42.5oC.
Heat absorbed by the calorimeter is negligible.
If the Cp of the solution is ,
and the density of the final soln is 1.0 g/ml,
calc the enthalpy change per mole of ppt.
Ch6 HW#6 p284 37,45,47,48

43. Ch6 HW#6 p284 37,45,47,48
37. The specific heat capacity of aluminum is 0.9 J/˚C·g.
a. Calculate the energy needed to raise the temperature of
an 8.5x102 g block of aluminum from 22.8˚C to 94.6˚C.
cp = 0.9 J/˚C·g
m = 8.5x102 g
ΔT = 71.8˚C
energy needed = ? q = m.cp.ΔT
b. Calculate the molar heat capacity of aluminum.

44. 37. The specific heat capacity of aluminum is 0.9 J/˚C·g.
a. Calculate the energy needed to raise the temperature of
an 8.5x102 g block of aluminum from 22.8˚C to 94.6˚C.
cp = 0.9 J/˚C·g
m = 8.5x102 g
ΔT = 71.8˚C
energy needed = ?
b. Calculate the molar heat capacity of aluminum.

45. 37. The specific heat capacity of aluminum is 0.9 J/˚C·g.
a. Calculate the energy needed to raise the temperature of
an 8.5x102 g block of aluminum from 22.8˚C to 94.6˚C.
cp = 0.9 J/˚C·g
m = 8.5x102 g
ΔT = 71.8˚C
energy needed = ?
b. Calculate the molar heat capacity of aluminum.

46. 45. In a coffee-cup calorimeter, 50.0 mL of 0.100 M AgNO3 and
50.0 mL of M HCl are mixed to yield the following reaction:
NIE: Ag+(aq) + Cl-(aq) →AgCl (s)
The two solutions were initially at 22.60˚C, and the final temperature is 23.40˚C. Calculate the heat that accompanies this reaction in kJ/mol of AgCl formed. Assume that the combined solution has a mass of g and has a specific heat capacity of 4.18 J/˚C·g.
Ag+(aq) + Cl-(aq) water AgCl (s) + heat (ΔH)
V=0.05L V=0.05L
M=0.100M M=0.100M

47. cp = 4.18 J/˚C·g n = 0.005mol of each → n = 0.005mol product
ΔT = 0.8 ˚C Ag+(aq) + Cl-(aq) water AgCl (s) + ____J
cp = 4.18 J/˚C·g n = 0.005mol of each → n = 0.005mol product
ΔH: kJ/mol = ? (q/n)
q = + m.cp.ΔT
0.005 moles yielded 334J of energy to the 100g of solution,
so how much heat would 1mol yield?

48. cp = 4.18 J/˚C·g n = 0.005mol of each → n = 0.005mol product
ΔT = 0.8 ˚C Ag+(aq) + Cl-(aq) water AgCl (s) + heat (q)
cp = 4.18 J/˚C·g n = 0.005mol of each → n = 0.005mol product
kJ/mol = ? (q/n) g total soln
q = + m.cp.ΔT
While the entire mass of solution had its temperature changed
because of the rxn, only the moles of the rxn provided the heat!
moles yielded 334J of energy to the 100g of solution,
- 1 mol of reactants woulda provided 66,900J to the solution!

49. m=11g m=125g 47.Consider the dissolution of CaCl2:
CaCl2(s) → Ca2+(aq) + 2Cl-(aq) ΔH = kJ
An 11.0g sample of CaCl2 is dissolved in 125 g of water, with both substances at 25.0˚C. Calculate the final temperature of the solution assuming no heat lost to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/˚C·g
CaCl2(s) water Ca2+(aq) + 2Cl-(aq) kJ
m=11g m=125g
Ti = 25C Ti = 25C per 1 mol CaCl2

50. 2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l) ΔH = -118 kJ
48. Consider the reaction
2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l) ΔH = -118 kJ
Calculate the heat when mL of M HCl is mixed with
300.0 mL of M Ba(OH)2. Assuming that the temperature of both
solutions was initially 25.0˚C and that the final mixture has a mass
of g and a specific heat capacity of 4.18 J/˚C · g, calculate the
final temperature of the mixture.
2HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2H2O(l) +118 kJ
0.05mol mol
Ti = 25.0˚C
this will run out q = – m.cp.ΔT

51. Ch6.5 – Heat & Changes of State
120
For H2O:
100
Temp
(°C)
-10
Time

52. Specific heat capacity ice: Cp = 2.1 J/g.°C
5) Raise temp of gas:
Δ H = m ∙Cp ∙ ∆T
120
For H2O:
l g
100
4) Vaporize it: ΔHvap = n ∙ Hv
Temp
(°C)
3) Raise temp of liquid:
Δ H = m ∙Cp ∙ ∆T
s l
2) Melt it: ΔHfus = n ∙ Hf
1) Raise temp of solid: ΔH = m ∙Cp ∙ ∆T
-10
Time
6) Add steps
UNITS!
Specific heat capacity ice: Cp = 2.1 J/g.°C
Heat of fusion for H2O: Hfus = 6.1 kJ/mol
Specific heat capacity water: Cp = 4.2 J/g.°C
Heat of vaporization for H2O: Hvap = 40.7 kJ/mol
Specific heat capacity steam: Cp = 2.0 J/g.°C

53. Liquid changing into a solid
Temperature of water versus time as thermal energy is removed
100
Vapor Changing
into a liquid
Temperature of sample (°C) 50
Liquid changing into a solid
Liquid
Time elapsed
Vapor
solid

54. at 120.0°C? (No Calc on MC AP Test!)
Ex1) How much heat is required to turn 90.0g of ice at –10.0°C to vapor
at 120.0°C? (No Calc on MC AP Test!)
Temp
(°C)
Temp
(°C)
Time
1) Heat ice ∆H =
2) Melt ∆H =
3) Heat water ∆H =
4) Boil ∆H =
5) Heat gas ∆H =
6) Add :
Ch6 HW#5

55. at 120.0°C? (No Calc on MC AP Test!)
Ex1) How much heat is required to turn 90.0g of ice at –10.0°C to vapor
at 120.0°C? (No Calc on MC AP Test!) 4 5
Temp
(°C)
Temp
(°C) 3 2 1
Time J
g ∙ °C
1) Heat ice ∆H = m∙Cp∙∆T = (90g)( )(10°C) = 1890J
2) Melt ∆H = m∙Hf J
g ∙ °C
3) Heat water ∆H = m ∙Cp ∙ ∆T = (90g)( )(100°C) = 37,800J
4) Boil ∆H = n ∙ Hv = J
g ∙ °C
5) Heat gas ∆H = m∙Cp∙ ∆T = (90g)( )(20.0°C) = 3636J
6) Add : 277,326J
Ch6 HW#5

56. Ch6 HW#5 (Ch10 p508 87), + 2 more
87. How much energy does it take to convert kg ice at -20˚C
to steam at 250˚C?
Temp
(°C)
Heat Energy (J)
Add:

57. 87. How much energy does it take to convert 0.500 kg ice at -20˚C
to steam at 250˚C?
Temp
(°C)
Heat Energy (J)
Add:

58. + 2 more:
1. How much heat is absorbed by 0.10kg of ice at -20°C
to become water at 0°C?
1) Heat ice ΔH = m∙Cp∙∆T =
2) Melt ΔH = n ∙Hf =
3) Add:

59. 2. A 0.20kg sample of water at 60°C is heated
to steam at 140.0°C. How much heat is absorbed?
1) Heat water ΔH = m∙Cp∙∆T =
2) Vaporize ΔH = n ∙Hv =
3) Heat gas ΔH = m∙Cp∙∆T =
4) Add:

60. Ch6.6 – Phase Diagrams (Ch10)
“At what temperature does water boil?”
For H2O:
218
374
100 1
Pressure
(atm)
Temperature (°C)

61. For H2O:
218
374
100
Solid
Liquid
Gas 1
Pressure
(atm)
Temperature (°C)
Normal boiling point – temp at which the liquid boils.
what that really means is that the liquid
has a Vapor Pressure of 1atm.

62. For H2O:
218
374
100
Solid
Liquid
Gas 1
Pressure
(atm)
Temperature (°C)
The point where the solid and the liquid have identical
vapor pressures is where the melting point is.
Normal melting point – temp at which the s and l
have identical vp’s and the total pressure is 1atm.

63. 1
218
Critical Point
374
100
Solid
Liquid
Gas
Triple Point
Pressure
(atm)
Temperature (°C)
Critical point for water – At 374°C and 22,100kPa (218 atm)
is the last point where water can still be condensed to a liquid.
Triple point – all 3 phases are in equilibrium – at 0.01oC and 4.58 torr

64. Supercooling – cool a liquid below its normal melting point.
Molecules haven’t organized.
Superheating – heat a liquid above its normal BP.
Hi KE molecules haven’t congregated enuf.

65. Critical Point
374
100
Solid
Liquid
Gas
Triple Point
218 1
Pressure
(atm)
Temperature (°C)
Ex: What is the physical state of H2O at 50°C and .70 atm?_____
Name a temp and pressure that H2O boils at: _____________

66. Vapor Pressure
100
80 chloroform water
VP ethanol ethanoic acid
(kPa)40 20
Temp (°C)
If a solid can go directly to the gaseous state, called sublimation.
Ch6 HW#6

67. Lab6.1 Calorimetry
- unknown end of period.
- lab write up due 2 days
- Ch6 HW#6 due tomorrow

68. Ch6 HW#6 (Ch10 p508) 91,104, + 2 more problems
91. Consider the phase diagrams given below. What phases are present from point A through H? Identify the triple point, normal boiling point, normal freezing point, and critical point. Which phase is denser solid or liquid?
1.0atm
n.m.p n.b.p. H A B G F C E D

69. - as pressure increases …
91. Consider the phase diagrams given below. What phases are present from point A through H? Identify the triple point, normal boiling point, normal freezing point, and critical point. Which phase is denser solid or liquid?
- as pressure increases …
H (g)
A (solid)
B (gas)
G (l,g)
F (l,g)
C (liquid)
E Triple pt
D (s,g)

70. 104. Use the accompanying phase diagram to answer the following
104. Use the accompanying phase diagram to answer the following questions.
a. How many triple points are
in the following diagram?
b. What phases can coexist at
each triple point?
c. What happens if graphite is
subjected to a very high
pressure at room temp?
d. If we assume that the density
increases with an increase
in pressure, which is more
dense, graphite or diamond?

71. a) What is the physical state of H2O at 100°C and 2 atm?_____
Problem 1:
218
374
100
Solid
Liquid
Gas 1
Pressure
(atm)
Temperature (°C)
a) What is the physical state of H2O at 100°C and 2 atm?_____
b) If H2O is raised to a temperature of 375°C and the pressure is raised
to an absurd 1000atm, what physical state will it exist in? _____

72. Problem 2:
100
80 chloroform water
VP ethanol ethanoic acid
(kPa)40 20
Temp (°C)
a) What is the normal boiling point of chloroform? ______
b) Which liquid will boil at 60°C when the pressure is lowered
to 40kPa? ______

73. Ch6.7 – Hess’s Law
In going from a particular set of reactants to a particular set of products,
the change in enthalpy is the same whether the reaction takes place
in one step or a series of steps.
1 step: N2(g) + 2O2(g)  2NO2(g) ΔH1 = 68 kJ
2steps:

74. In going from a particular set of reactants to a particular set of products,
the change in enthalpy is the same whether the reaction takes place
in one step or a series of steps.
1 step: N2(g) + 2O2(g)  2NO2(g) ΔH1 = 68 kJ
2steps: N2(g) + O2(g)  2NO(g) ΔH2 = 180 kJ
2NO(g) + O2(g)  2NO2(g) ΔH3 = –112 kJ
Net rxn: N2(g) + 2O2(g)  2NO2(g) ΔH2 + ΔH3 = 68 kJ

75. C(s)diamond + O2(g)  CO2(g) ΔH = -396kJ
Ex1) Calculate ΔH for converting graphite to diamond,
C(s)graphite  C(s)diamond
using the enthalpies for combustion given:
C(s)graphite + O2(g)  CO2(g) ΔH = -394kJ
C(s)diamond + O2(g)  CO2(g) ΔH = -396kJ

76. 2B(s) + 3/2O2(g)  B2O3(s) ΔH = –1273 kJ
Ex2) Calculate ΔH for the synthesis of diborane (B2H6),
given the following reactions:
2B(s) + 3/2O2(g)  B2O3(s) ΔH = –1273 kJ
B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g) ΔH = –2035 kJ
H2(g) + ½O2(g)  H2O(l) ΔH = –286 kJ
H2O(l)  H2O(g) ΔH = kJ

77. NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -199 kJ
HW#55) Given the following data:
2O3(g) → 3O2(g) ΔH = -427 kJ
O2(g) → 2O(g) ΔH = +495 kJ
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -199 kJ
calculate ΔH for the reaction
NO(g) + O(g) → NO2(g)
Ch6 HW#7 p284 51,53,55,57

78. NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -199 kJ
HW#55) Given the following data:
2O3(g) → 3O2(g) ΔH = -427 kJ
O2(g) → 2O(g) ΔH = +495 kJ
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -199 kJ
calculate ΔH for the reaction
2NO(g) + 2O3(g) → 2NO2(g) +2O2(g) ΔH = -398 kJ
3O2(g) → 2O3(g) ΔH = +427 kJ
2O(g) → O2(g) ΔH = -495 kJ
2NO(g) + 2O(g) → 2NO2(g) ΔH = -466 kJ
NO(g) + O(g) → NO2(g) ΔH = -233 kJ
Ch6 HW#7 p284 51,53,55,57

79. Ch6 HW#7 p284 51,53,55,57
51. The enthalpy of combustion of solid carbon to form carbon dioxide is kJ/mol carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is kJ/mol CO. Use these data to calculate ΔH for the reaction

80. Ch6 HW#7 p284 51,53,55,57
51. The enthalpy of combustion of solid carbon to form carbon dioxide is kJ/mol carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is kJ/mol CO. Use these data to calculate ΔH for the reaction
2C(s) + O2(g) → 2CO(g)

81. 2CO2(g) → 2CO(g) + O2(g) ΔH = +566.6 kJ/mol
Ch6 HW#7 p284 51,53,55,57
51. The enthalpy of combustion of solid carbon to form carbon dioxide is kJ/mol carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is kJ/mol CO. Use these data to calculate ΔH for the reaction
2C(s) + O2(g) → 2CO(g)
2C(s) + 2O2(g) → 2CO2(g) ΔH = kJ/mol
2CO2(g) → 2CO(g) + O2(g) ΔH = kJ/mol
2C(s) + O2(g) → 2CO(g) ΔH =

82. 2so2(g) + O2(g) → 2SO3(g) ΔH = -198.2 kJ
53. Given the following data:
S(s) + 3/2O2(g) → SO3(g) ΔH = kJ
2so2(g) + O2(g) → 2SO3(g) ΔH = kJ
calculate ΔH for the reaction
S(s) + O2(g) → SO2(g)

83. 2so2(g) + O2(g) → 2SO3(g) ΔH = -198.2 kJ
53. Given the following data:
S(s) + 3/2O2(g) → SO3(g) ΔH = kJ
2so2(g) + O2(g) → 2SO3(g) ΔH = kJ
calculate ΔH for the reaction
S(s) + O2(g) → SO2(g)
2S(s) + 2(3/2O2(g)) → 2SO3(g) ΔH = kJ
2SO3(g) → 2so2(g) + O2(g) ΔH = kJ
2S(s) + 2O2(g) → 2SO2(g) ΔH = kJ
S(s) + O2(g) → SO2(g) ΔH =

84. H2(g) + 1/2O2(g) → H2O(l) ΔH = -285.8 kJ
55. In class
57. Given the following data:
H2(g) + 1/2O2(g) → H2O(l) ΔH = kJ
N2O5(g) + H2O(l) → 2HNO3(l) ΔH = kJ
1/2N2(g) + 3/2O2(g) + 1/2H2(g) → HNO3(l) ΔH = kJ
calculate the ΔH for the reaction
2N2(g) + 5O2(g) → 2N2O5(g)

85. 4(1/2N2(g))+4(3/2O2(g))+4(1/2H2(g))→4(HNO3(l)) ΔH=4(-174.1)kJ
57. Given the following data:
4(1/2N2(g))+4(3/2O2(g))+4(1/2H2(g))→4(HNO3(l)) ΔH=4(-174.1)kJ
10(H2(g) + 1/2O2(g) → H2O(l)) ΔH=10(-285.8)kJ
2(2HNO3(l) → N2O5(g) + H2O(l)) ΔH =2(+76.6)kJ

86. 4(1/2N2(g))+4(3/2O2(g))+4(1/2H2(g))→4(HNO3(l)) ΔH=4(-174.1)kJ
57. Given the following data:
4(1/2N2(g))+4(3/2O2(g))+4(1/2H2(g))→4(HNO3(l)) ΔH=4(-174.1)kJ
10(H2(g) + 1/2O2(g) → H2O(l)) ΔH=10(-285.8)kJ
2(2HNO3(l) → N2O5(g) + H2O(l)) ΔH =2(+76.6)kJ
2N2(g) + 6O2(g) + 2H2(g)→ 4HNO3(l) ΔH=-696.4kJ
10H2(g) + 5O2(g) → 10H2O(l) ΔH =-2858kJ
4HNO3(l) → 2N2O5(g) + 2H2O(l) ΔH =+153.2kJ
12H2(g) + 6O2(g) → 12H2O(l) +
2N2(g) + 5O2(g) → 2N2O5(g) ΔH = kJ total

87. 4(1/2N2(g))+4(3/2O2(g))+4(1/2H2(g))→4(HNO3(l)) ΔH=4(-174.1)kJ
57. Given the following data:
4(1/2N2(g))+4(3/2O2(g))+4(1/2H2(g))→4(HNO3(l)) ΔH=4(-174.1)kJ
10(H2(g) + 1/2O2(g) → H2O(l)) ΔH=10(-285.8)kJ
2(2HNO3(l) → N2O5(g) + H2O(l)) ΔH =2(+76.6)kJ
2N2(g) + 6O2(g) + 2H2(g)→ 4HNO3(l) ΔH=-696.4kJ
10H2(g) + 5O2(g) → 10H2O(l) ΔH =-2858kJ
4HNO3(l) → 2N2O5(g) + 2H2O(l) ΔH =+153.2kJ
12H2(g) + 6O2(g) → 12H2O(l) +
2N2(g) + 5O2(g) → 2N2O5(g) ΔH = kJ total
12(H2O(l) → H2(g) + ½O2(g)) ΔH=12(+285.8)kJ
2N2(g) + 5O2(g) → 2N2O5(g) ΔH =-28.4kJ

88. Ch6.8 – Standard Enthalpies of Formation
C(s)graphite  C(s)diamond
Enthalpy
time

89. Standard Enthalpy of Formation (ΔHfo)
- change in enthalpy that accompanies the formation
of 1 mole of a substance from its elements.
(All substances in their standard states)

90. Standard Enthalpy of Formation (ΔHfo)
- change in enthalpy that accompanies the formation
of 1 mole of a substance from its elements.
(All substances in their standard states)
- use the state of the substance as it exists at 1 atm and 25oC.
Exs) oxygen: _____ sodium: ______ mercury: ______
- for gases use a pressure of 1 atm.
- for solutions use a concentration of 1M.

91. Standard Enthalpy of Formation (ΔHfo)
- change in enthalpy that accompanies the formation
of 1 mole of a substance from its elements.
- use the state of the substance as it exists at 1 atm and 25oC.
- for gases use a pressure of 1 atm.
- for solutions use a concentration of 1M.
- elements in their standard states = 0.
- multiply # of moles to ΔH
ΔHoreaction = ΣnpΔHfo(products) – Σnr Δhfo(reactants)

92. Ex1) Use the standard enthalpies of formation listed in Table6
Ex1) Use the standard enthalpies of formation listed in Table6.2 to calc
the standard enthalpy change for the given reaction:
4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(l)

93. Ex2) Use the standard enthalpies of formation listed in Table6
Ex2) Use the standard enthalpies of formation listed in Table6.2 to calc
the standard enthalpy change for the given reaction:
2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s)
Ch6 HW#8 p285 61,65,66
Ch6 Rev p283 28,35
(Go over B4 lab pt 1)

94. a. 2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g)
Ch6 HW#8 p285 61,65,66 Ch6 Rev p283 28,35 (Go over B4 lab pt 1)
61. Use the values of ΔHfo in Appendix 4 to the calculate ΔHo for the following reactions.
a. 2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g)
2mol(-80kJ/mol)+ 3(0) + 2mol(-75kJ/mol) 2mol(135.1kJ/mol) + 6mol(-242kJ/mol)
ΔHoreaction = (products) – (reactants)

95. a. 2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g)
61. Use the values of ΔHfo in Appendix 4 to the calculate ΔHo for the following reactions.
a. 2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g)
2mol(-46kJ/mol)+ 3(0) + 2mol(-75kJ/mol) 2mol(135.1kJ/mol) + 6mol(-242kJ/mol)
(-92kJ) + (-150kJ) (270kJ) + (-1452kJ)
(-242kJ) (-1182kJ)
ΔHoreaction =

96. ΔHoreaction = (products) – (reactants)
61 - b. Ca(PO4)2(s) + 3H2SO4(l) → 3CaSO4(s) + 2H3PO4(l)
1mol(-4126kJ/mol) + 3mol(-814kJ/mol) 3mol(1433kJ/mol)+2mol(-1267kJ/mol)
ΔHoreaction = (products) – (reactants)

97. (-4126kJ) + (-2442kJ) (-4299kJ) + (-2534kJ)
61 - b. Ca(PO4)2(s) + 3H2SO4(l) → 3CaSO4(s) + 2H3PO4(l)
1mol(-4126kJ/mol) + 3mol(-814kJ/mol) 3mol(-1433kJ/mol)+2mol(-1267kJ/mol)
(-4126kJ) + (-2442kJ) (-4299kJ) + (-2534kJ)
(-6568kJ) (-6833kJ)
ΔHoreaction = (products) – (reactants)
ΔHoreaction =

98. 61 - c. NH3(g) + HCl(g) → NH4Cl(s)
1mol(-46kJ/mol) + 1mol(-92kJ/mol) 1mol(-314kJ/mol)
ΔHoreaction = (products) – (reactants)

99. ΔHoreaction = (products) – (reactants) ΔHoreaction =
61 - c NH3(g) + HCl(g) → NH4Cl(s)
1mol(-46kJ/mol) + 1mol(-92kJ/mol) 1mol(-314kJ/mol)
(-46kJ) + (-92kJ) (-314kJ)
(-138kJ) (-314kJ)
ΔHoreaction = (products) – (reactants)
ΔHoreaction =

100. 3Al(s) + 3NH4ClO4(s)→Al2O3(s) + AlCl3(s) + NO(g) +6H2O(g)
65. The reusable booster rockets of the space shuttle use a mixture of aluminum and ammonium perchlorate as fuel. A possible reaction is
Calculate the ΔHo for this reaction.
3Al(s) + 3NH4ClO4(s)→Al2O3(s) + AlCl3(s) + NO(g) +6H2O(g)
3(0) mol(-295kJ/mol) 1mol(-1676)+1mol(-704)+1mol(90)+6(-242)
ΔHoreaction = (products) – (reactants)
ΔHoreaction =

101. 3Al(s) + 3NH4ClO4(s)→Al2O3(s) + AlCl3(s) + NO(g) +6H2O(g)
65. The reusable booster rockets of the space shuttle use a mixture of aluminum and ammonium perchlorate as fuel. A possible reaction is
Calculate the ΔHo for this reaction.
3Al(s) + 3NH4ClO4(s)→Al2O3(s) + AlCl3(s) + NO(g) +6H2O(g)
3(0) mol(-295kJ/mol) 1mol(-1676)+1mol(-704)+1mol(90)+6(-242)
(-885kJ) (-1676kJ) + (-704kJ) + (+90) + (-1452)
(-885kJ) (-3742kJ)
ΔHoreaction = (products) – (reactants)
ΔHoreaction

102. 4N2H3CH3(l) + 5N2O4(l) → 12H2O(g) + 9N2(g) + 4CO2(g)
66. The space shuttle orbiter utilizes the oxidation of methyl hydrazine
by dinitrogen tetroxide for propulsion:
Calculate the ΔHo for this reaction.
4N2H3CH3(l) + 5N2O4(l) → 12H2O(g) + 9N2(g) + 4CO2(g)
4(54) (-20) (-242) (0) (-393.5)
ΔHoreaction = (products) – (reactants)

103. 4N2H3CH3(l) + 5N2O4(l) → 12H2O(g) + 9N2(g) + 4CO2(g)
66. The space shuttle orbiter utilizes the oxidation of methyl hydrazine
by dinitrogen tetroxide for propulsion:
Calculate the ΔHo for this reaction.
4N2H3CH3(l) + 5N2O4(l) → 12H2O(g) + 9N2(g) + 4CO2(g)
4(54) (-20) (-242) (0) (-393.5)
(+116kJ) (-4478kJ)
ΔHoreaction = (products) – (reactants)
ΔHoreaction = (-4478kJ)– (+116kJ) = -4594kJ very exothermic!
Do u think that might be
a very powerful and
very spontaneous reaction?!?
Where does all the energy come from?

104. Ch6 Rev p283 28,35,46,54,64
28. One mole of H2O(g) at 1.00atm and 100°C occupies a volume of
30.6L. When one mole of H2O(g) is condensed to one mole of H2O(l)
at 1.00atm and 100°C, 40.66kJ of heat is released.
If the density of H2O(l) at the this temp and pressure is g/cm3,
calc ∆E for the condensation of one mole of of H2O.
H2O(g) H2O(l)
1mol d= 0.996g/cm3
V=30.6L
P=1atm
T=100°C
∆E =?

105. 3. A 0.05 m3 volume of gas absorbs 10 kJ of heat and expands to a volume
of 0.15 m3, while remaining at a constant pressure. During the process,
the internal energy increases by 4500 J. What was the pressure?
∆V = .10m3 w = ∆E – q = (+4.5kJ) – (+10kJ) = – 5.5kJ = – 5500J
Q = +10kJ
∆U = +4.5kJ
P = ?
4. A piston with a cross-sectional area of 0.10 m2 seals a cylinder that contains a gas at a pressure of 100 kPa J of heat are removed
from the cylinder, and the piston falls 0.05 m isobarically.
a. How much work is done on the gas?
b. What is the change in internal energy?
c. If the gas is then heated, so that its absolute
temperature doubles, what is the new volume of the cylinder?

106. 4. A piston with a cross-sectional area of 0
4. A piston with a cross-sectional area of 0.10 m2 seals a cylinder that contains a gas at a pressure of 100 kPa J of heat are removed
from the cylinder, and the piston falls 0.05 m isobarically.
a. How much work is done on the gas? ∆V=A. ∆h = (.1m2)(–.05m)= –0.005m3
w = – P. ∆V =
b. What is the change in internal energy?
∆E = q + w =
c. If the gas is then heated, so that its absolute
temperature doubles, what is the new volume of the cylinder?

107. ∆E = q + W work done on system ∆E = q + (-P.∆V)
28. One mole of H2O(g) at 1.00atm and 100°C occupies a volume of
30.6L. When one mole of H2O(g) is condensed to one mole of H2O(l)
at 1.00atm and 100°C, 40.66kJ of heat is released.
If the density of H2O(l) at the this temp and pressure is g/cm3,
calc ∆E for the condensation of one mole of of H2O.
H2O(g) H2O(l) kJ
1mol 1mol
V=30.6L d= 0.996g/cm3 ΔH  q
P=1atm
T=100°C
∆E =?
∆E = q + W work done on system
∆E = q (-P.∆V)
∆E = (-40.66kJ) + (101.3kPa) .∆V

108. ∆E = q + W work done on system ∆E = q + (-P.∆V)
28. One mole of H2O(g) at 1.00atm and 100°C occupies a volume of
30.6L. When one mole of H2O(g) is condensed to one mole of H2O(l)
at 1.00atm and 100°C, 40.66kJ of heat is released.
If the density of H2O(l) at the this temp and pressure is g/cm3,
calc ∆E for the condensation of one mole of of H2O.
H2O(g) H2O(l) kJ
1mol 1mol
V=30.6L d= 0.996g/cm3 q
P=1atm
T=100°C
∆E =?
∆V = 0.018L – 30.6L = L
∆E = q + W work done on system
∆E = q (-P.∆V)
∆E = (-40.66kJ) + (101.3kPa) .∆V

109. ∆E = q + W work done on system ∆E = q + (-P.∆V)
28. One mole of H2O(g) at 1.00atm and 100°C occupies a volume of
30.6L. When one mole of H2O(g) is condensed to one mole of H2O(l)
at 1.00atm and 100°C, 40.66kJ of heat is released.
If the density of H2O(l) at the this temp and pressure is g/cm3,
calc ∆E for the condensation of one mole of of H2O.
H2O(g) H2O(l) kJ
1mol 1mol
V=30.6L d= 0.996g/cm3 q
P=1atm
T=100°C
∆E =?
∆V = 0.018L – 30.6L = L
∆E = q + W work done on system
∆E = q (-P.∆V)
∆E = (-40.66kJ) + (-(101.3kPa) .∆V)
∆E = (-40.66kJ) + (-(101.3kPa) .(-30.58L))
∆E =

110. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) + 1.3x108J
35. Consider the combustion of propane:
C3H8(g) + 5O2(g) CO2(g) + 4H2O(l) + 1.3x108J
Assume that all the heat in sample Exercise 6.3 comes from the combustion of propane. What mass of propane must be burned to furnish this amount of energy assuming the heat transfer process is 60% efficient? 1.3x108J (from 6.3)
60% efficient means you gotta burn a whole lot more
 1.3x108J / 0.60 = 2.17x108J will get you there.

111. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) + 1.3x108J
35. Consider the combustion of propane:
C3H8(g) + 5O2(g) CO2(g) + 4H2O(l) + 1.3x108J ?g

112. Heat absorbed by ammonium nitrate, comes from the soln,
Rev – Day 2 (Lab6.2 Day 2)
46. In a coffee-cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water
at an initial temperature of 25.00oC. After dissolution of the salt, the final
temperature of the calorimeter contents is 23.34o C. Assuming the solution
has a heat capacity of 4.18 J/oC.g and assuming that no heat loss to the
calorimeter. Calculate the enthalpy charge for the dissolution of NH4NO3
in units of kJ/mol .
NH4NO water NH NO3-
m=1.60g m=75.0g ∆H = ? (kJ/mol)
Ti = 25.00C Ti = 25.00C
Tf = 23.34C Tf = 23.34C
Heat absorbed by ammonium nitrate, comes from the soln,
and is measured in the temp change of the total soln:
q = – m.cp.ΔT

113. Heat absorbed by ammonium nitrate, comes from the soln,
46. NH4NO water NH NO3-
m=1.60g m=75.0g
Ti = 25.00C Ti = 25.00C
Tf = 23.34C Tf = 23.34C
Heat absorbed by ammonium nitrate, comes from the soln,
and is measured in the temp change of the total soln :
q = – m.cp.ΔT

114. When 1.60g NH4NO3 dissolved, ___ J (q) was released.
46. NH4NO3 +____ water NH NO3-
m=1.60g m=75.0g
Ti = 25.00C Ti = 25.00C
Tf = 23.34C Tf = 23.34C
Heat absorbed by ammonium nitrate, comes from the soln,
and is measured in the temp change of the total soln :
q = – m.cp.ΔT
When 1.60g NH4NO3 dissolved, ___ J (q) was released.
How much heat (ΔH) would 1 mol produce?

115. C2H2(g) + 5/2O2(g) 2CO2(g) + H20(l) ∆H= -1300kJ
54. Given the following data:
C2H2(g) + 5/2O2(g) CO2(g) + H20(l) ∆H= -1300kJ
C(s) + ½O2(g) CO2(g) ∆H= -394kJ
H2 (g) + ½O2(g) H2O(l) ∆H= -286kJ
Calculate ∆H for the reaction:
2C (s) + H2 (g) C2H2 (g)

116. 64.Calculate ∆H˚ for each of the following reactions using the data provided:
2Na(s)+2H2O(l) → 2NaOH(aq)+H2(g)
2Na(s)+CO2(g) → Na2O(s)+CO(g)

117. 3) Heat water 4) Vaporize 6) Add:
Extra Problem: How much heat is needed to change 0.30kg of ice at -30°C
to steam at 130°C?
1) Heat ice
2) Melt
3) Heat water
4) Vaporize
5) Heat gas
6) Add:

118. Ch6.1B – PV Diagrams
‘A closed system at equilibrium has no tendency to undergo spontaneous
macroscopic change.”
4 basic models where a closed system is changed:
Constant Can change Equations Implications Thermo Eqn
1. Isothermal
2. Isobaric
3. Isovolumetric
4. Adiabatic
(no heat transferred
to or from a system.)
Work equation:
(Units: ______, or ______, or ______)

119. Ch6.1B – PV Diagrams
‘A closed system at equilibrium has no tendency to undergo spontaneous
macroscopic change.”
4 basic models where a closed system is changed:
Constant Can change Equations Implications Thermo Eqn
1. Isothermal T V, P V1P1 = V2P2 If ∆T = 0, q = W
then ∆E=0
2. Isobaric P V, T ∆E = q + W
3. Isovolumetric V P, T No work ∆E = q
done.
4. Adiabatic nothing V, P, T Q = 0 ∆E = W
(no heat transferred
to or from a system.)
Work equation: W = –P . ∆V
(Units: Joules, or Pa.m3, or kPa.L)

120. 50 Work = (area under the PV curve.)
Decide if work is +/-
P based on direction.
(Pa) [gas expands, w = ( – ),
gas compressed, w = (+)]
V (m3)
101.3 P
(kPa)
V (L)
150
P 100
(MPa) 50
V(cm3)

121. 50 Work = –(area under the PV curve.)
Decide +/- based on direction
P W = ½(3m3)(40N/m2)
(Pa) = –60J 10
V (m3)
W = (101.3kPa)(3L)
P (1kPa  1000Pa)
(kPa) = –303.9J (1L  0.001m3)
V (L)
150
No work done
P (Isovolumetric)
(Mpa) (1 Mpa 1,000,000Pa)
(1cm3  m3)
V(cm3)

122. Ex3) A cylinder contains 264
Ex3) A cylinder contains moles of a monatomic gas that is initially
at state A at standard temperature, and a pressure of 6x105Pa.
a. What volume does the gas occupy? 6
b. Please graph state A
c. A certain amount of heat is added, P 4
bringing the gas isobarically to state B, (x105 Pa) 3
at a volume of 3m3, while its internal energy 2
is increased by 400kJ. How much heat was added? 1
Please graph state B V(m3)
d. The gas is brought isothermally to state C, where the pressure was reduced
to 3x105Pa. What is the new volume? Please graph state C.
Ch6 HW#2 1 – 4

123. Ex3) A cylinder contains 264
Ex3) A cylinder contains moles of a monatomic gas that is initially
at state A at standard temperature, and a pressure of 6x105Pa.
a. What volume does the gas occupy? 6
b. Please graph state A
c. A certain amount of heat is added, P 4
bringing the gas isobarically to state B, (x105 Pa) 3
at a volume of 3m3, while its internal energy 2
is increased by 400kJ. How much heat was added? 1
Please graph state B V(m3)
d. The gas is brought isothermally to state C, where the pressure was reduced
to 3x105Pa. What is the new volume? Please graph state C.
a. PV=nRT
c. ∆E = q + W W = –(area) = (+2m3)(6x105Pa) = –1,200,000J
(+400,000J) = q + (–1,200,000J) +
q = +1,600,000J
d. VB = 3m VC = ? VB. PB = VC. PC
PB = 6x105Pa PC = 3x105Pa (3)(6) = (VC)(3)
TB = ? same TC = ? VC =6m3

124. Ch6 HW#2 1 – 4
1. A quantity of gas in a cylinder sealed with a piston absorbs 4186J of heat
as it expands by 15L at constant pressure. If the internal energy increases
by 286J, what was the constant pressure?
2. A cylinder sealed with a piston contains 20.0L of pure oxygen
at standard pressure. The gas is then forced into a tiny high pressure container
of negligible volume. What is the amt of work done on the gas?
What do you think happens to the temp of the gas as it is compressed?

125. Ch6 HW#2 1 – 4
1. A quantity of gas in a cylinder sealed with a piston absorbs 4186J of heat
as it expands by 15L at constant pressure. If the internal energy increases
by 286J, what was the constant pressure?
q = J ∆E = q + w
∆V = + 15L  0.015m3 w = (+286J) – (+4186J)
∆U = +286J w = – 3890J
2. A cylinder sealed with a piston contains 20.0L of pure oxygen
at standard pressure. The gas is then forced into a tiny high pressure container
of negligible volume. What is the amt of work done on the gas?
What do you think happens to the temp of the gas as it is compressed?

126. 3. A gas experiences a change from state I to state F.
Determine the work done.
4. A piston with a cross-sectional area of 0.50m2 seals a gas within a cylinder
at a pressure of 300,000Pa. 10kJ of heat is gradually added to the cylinder, and the
piston rises 6.5cm as the gas expands isobarically. What is the work done
by the gas? What is the change in internal energy?
A = 0.50m2
h = 6.5cm 6 F P
(Pa) 3 2
1 I
V (m3)

127. the gas did the work… Work = –12J
3. A gas experiences a change from state I to state F. Determine the work done.
Work = Area
= l.w + ½b.h
= 4m3.1Pa + ½(4m3).(4Pa)
= 4J + 8J
= 12J
Since the gas expanded to a bigger volume,
the gas did the work… Work = –12J
4. A piston with a cross-sectional area of 0.50m2 seals a gas within a cylinder
at a pressure of 300,000Pa. 10kJ of heat is gradually added to the cylinder,
and the piston rises 0.065m as the gas expands isobarically.
What is the work done by the gas? What is the change in internal energy? 6 F P
(Pa) 3 2
1 I
V (m3)
A = 0.50m2
h = 0.065m

128. the gas did the work… Work = –12J
3. A gas experiences a change from state I to state F. Determine the work done.
Work = Area
= l.w + ½b.h
= 4m3.1Pa + ½(4m3).(4Pa)
= 4J + 8J
= 12J
Since the gas expanded to a bigger volume,
the gas did the work… Work = –12J
4. A piston with a cross-sectional area of 0.50m2 seals a gas within a cylinder
at a pressure of 300,000Pa. 10kJ of heat is gradually added to the cylinder,
and the piston rises 0.065m as the gas expands isobarically.
What is the work done by the gas? What is the change in internal energy? 6 F P
(Pa) 3 2
1 I
V (m3)
A = 0.50m2 ∆V=A. ∆h=(.50m2)(.065m) = m3
h = 0.065m
W = –P.∆V =
∆E = q + W =

129. Ch6.1C – PV Diagrams
Ex1) A piston with a radius of 0.05m seals a cylinder that contains a gas
at standard pressure. 100J of heat is added to the cylinder, and the piston
rises 0.04m isobarically.
a. How much work is done by the gas? h
b. What is its change in internal energy?
The total pressure exerted on the gas increases
to 107.7kPa.
c. If the piston seals 1m3, when the pressure increased,
and the piston lowers isothermally, what is the new volume?
Ch6 HW#3 5 – 8

130. b. ∆E = q + w = (+100J) + (– 31.8J) = +68J
Ex1) A piston with a radius of 0.05m seals a cylinder that contains a gas
at standard pressure. 100J of heat is added to the cylinder, and the piston
rises 0.04m isobarically.
a. How much work is done by the gas? h
b. What is its change in internal energy?
The total pressure exerted on the gas increases
to 107.7kPa.
c. If the piston seals 1m3, when the pressure increased,
and the piston lowers isothermally, what is the new volume?
a. ∆V = A. ∆h = π(.05m)2(+.04m) = m3
w = – P. ∆V = – (101,300Pa)( m3) = – 31.8J
b. ∆E = q + w = (+100J) + (– 31.8J) = +68J
c. V1 = 1 m V2 = ___m V1. P1 = V2. P2
P1 = 101,300Pa P2 = 107,700Pa (1m3)(101,300Pa) = V2(107,700Pa)
T1 = ? same T2 = ? V2 = .9m3
Ch6 HW#3 5 – 8

131. Ch6.1C – PV Diagrams (removed this example)
Ex1) The graph shows P vs V for 2 moles of gas cycling from state A to B.
The internal energy increases by 400J.
a. Calc work done
b. Is heat added or removed? How much? P
c. The pressure is reduced to 200Pa (Pa) A B
isovolumetrically as gas goes from state B
to state C. Please graph state C C
d. The graph is compressed isothermally to state A
Draw the lines to represent the cycle. V (m3)
e. Is heat added or removed from C  A? How much?
a. W = -P. ∆V = -(600Pa)(+6m3) = -3600J
b. Added: Why else would the gas expand?
∆E = q + w  q = ∆E – w = (+400J) – (-3600J) = +4000J
c. isovolumetric
d. isothermal – P and V can change, while ∆T = 0, so ∆E = 0.
e. Heat is removed.
∆E = q + w
0 = q + w
Q = -WorkCA
= -(Area) = -(2400J)