1. Review Universe = system + surroundings into system = + -
out of system =
E =
internal energy
E =
 kinetic energy
+  potential energy
K.E. - energy of motion
P.E. - energy of position

2. 1st Law of Thermodynamics
Energy can not be created or destroyed
Euniverse = 0
Euniverse =
Esystem
+ Esurroundings
= 0
Esystem =
- Esurroundings
Energy is exchanged as :
heat
= q
work
= w
E = q + w

3. heat flow of energy along a T gradient If T1 > T2 a) > b) <
System at T1
Surroundings at T2
a) >
b) <
c) =
q system
q surroundings
Exothermic reaction
> 0

4. heat flow of energy along a T gradient If T1 < T2 q system > 0
Surroundings at T2
q system
> 0
q surroundings
Endothermic reaction
< 0

5. work - Pext V Wsystem = Electrical work Mechanical work =
force x distance
force =
pressure x area =
P x m2
distance = m
work = P
x m2
x m
= P x V -
Pext V
Wsystem =

6. State Functions Property that depends only
on the initial and final states
Temperature, T
: raise T from 298  300 K
path 1
T =
Tfinal - Tinitial =
= 2 K
path 2
: raise T 298  500 K
lower T from 500  300 K
T =
Tfinal - Tinitial =
= 2 K

7. Extensive v.s. Intensive
Proportional to the mass of the system
Intensive:
Independent of the mass of the system
Temperature:
Intensive or Extensive
Volume :
Intensive or Extensive
Pressure:
Intensive or Extensive
Internal Energy:
Intensive or Extensive

8. System 1 How much work will be done as the gas expands
against the piston?
Pext = 1.5 atm,
T = 298 K
P1 = 6.0 atm
V1 = 0.4 L
V2 =
P2 =
1.5 atm
1.6 L
P1 V1 = P2 V2
PV = nRT
(6.0 atm)
(0.4 L)
= (1.5 atm)
(V2)
w = -Pext V
w =
-(1.5 atm)
(1.6 -
0.4)L =
-1.8 L atm
(-1.8 L atm)
(101.3 J/L atm) =
-182 J

9. System 2 How much work is done when the stopcock is opened?
ideal gas
vacuum
0.4 L
1.2 L
6.0 atm
P1 = 6 atm
P2 =
1.5 atm
V1 = 0.4 L
V2 =
1.2
+ 0.4 L
= 1.6 L
T1 = 298 K
= T2
P1 V1 = P2 V2
w = -Pext V
= -(0 atm)
(1.6 L L)
w = 0

10. Same initial and final conditions 
System 1
System 2
P1 = P2 =
V1 = V2 =
T1 = T2 =
6.0 atm
1.5 atm
P1 = P2 =
V1 = V2 =
T1 = T2 =
6.0 atm
1.5 atm
0.4 L
1.6 L
0.4 L
1.6 L
298 K
298 K
w = -182 J
q = +182 J
w = 0
q = 0
Same initial and final conditions 
all State functions are the same
E will be the same
Ideal gases :
if T = 0,
E =
E = q + w = 0

11. Thermochemistry State 1 = State 2 = reactants products
E = Eproducts - Ereactants
Endothermic reaction
Ba(OH)2•8H2O (s)
+ 2NH4SCN (s) 
2NH3(g)
+ Ba(SCN)2(l)
+ 10H2O(l)
a) <
b) >
c) =
a) <
b) >
c) = q w
reaction lowered T of system

12. Endothermic Reaction Tprod < Treact K.E.prod < K.E.react
P.E.prod > P.E.react
products less stable than reactants

13. Exothermic Reaction C12H24O12 (s) + KClO3(s)  mix of gases
Tprod > Treact
K.E.prod > K.E.react
P.E.prod < P.E.react
C12H24O12 (s)
+ KClO3(s)
 mix of gases q
< w
<
products more stable than reactants