ReviewReview
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
1st Law of Thermodynamics
Euniverse = = 0
Esystem =
Energy is exchanged as : heat
work
E = q + w
Energy can not be created or destroyed
Euniverse = 0Esystem + Esurroundings
- Esurroundings
= q
= w
heat
If T1 > T2
SystematT1
SurroundingsatT2
flow of energy along a T gradient
q system 0
Exothermic reaction
a) >b) <c) => 0q surroundings
heat
If T1 < T2
SystematT1
SurroundingsatT2
flow of energy along a T gradient
q system
Endothermic reaction
> 0
q surroundings < 0
workwork
Electrical work
Mechanical work = force x distance
force = P x m2
distance = m
work = = P x V
WWsystemsystem == -- VV
pressure x area =
P x m2 x m
PPextext
State FunctionsState Functions
Property that depends only on the initial and final states
Temperature, T
: raise T from 298 300 KT = Tfinal - Tinitial = 300 - 298
: raise T 298 500 K
T = Tfinal - Tinitial = 300 - 298 = 2 K
path 1
path 2lower T from 500 300 K
= 2 K
Extensive v.s. Intensive
Extensive: Proportional to the mass of the system
Intensive: Independent of the mass of the system
Temperature: Intensive or Extensive
Volume :Pressure:Internal Energy:
Intensive or Extensive
Intensive or Extensive
Intensive or Extensive
System 1
P1 = 6.0 atm V1 = 0.4 L
Pext = 1.5 atm,
How much work will be done as the gas expands against the piston?
P2 = V2 =
P1 V1 = P2 V2
(6.0 atm)
1.6 L
w = -Pext V
-(1.5 atm) -1.8 L atm(-1.8 L atm)
PV = nRT
1.5 atm
(101.3 J/L atm) = -182 Jw =
(0.4 L) = (1.5 atm) (V2)
(1.6 - 0.4)L=
T = 298 K
System 2
How much work is done when the stopcock is opened?
P1 = 6 atm
V1 = 0.4 L
T1 = 298 K
P2 =
V2 = 1.2
= T2
P1 V1 = P2 V2
1.5 atm
w = -Pext V = -(0 atm)
w = 0
1.2 Lvacuum
0.4 Lideal gas
6.0 atm
(1.6 L - 0.4 L)
+ 0.4 L= 1.6 L
System 1 System 2
P1 = P2 =V1 = V2 =T1 = T2 =
Same initial and final conditions
E will be the same
Ideal gases :
E = q + w = 0
w = -182 J q = +182 J w = 0 q = 0
all State functions are the same
6.0 atm 6.0 atm0.4 L 0.4 L
298 K 298 K
0
1.5 atm 1.5 atm1.6 L 1.6 L
if T = 0, E =
P1 = P2 =V1 = V2 =T1 = T2 =
Thermochemistry
State 1 = State 2 =reactants products
E = Eproducts - Ereactants
Endothermic reaction
q a) <b) >c) =
reaction lowered T of system
w0 0
Ba(OH)2•8H2O (s) 2NH3(g)+ 2NH4SCN (s) + Ba(SCN)2(l)+ 10H2O(l)
a) <b) >c) =
Endothermic Reaction
Tprod < Treact
K.E.prod < K.E.react
P.E.prod > P.E.react
products less stable than reactants
Exothermic Reaction
Tprod > Treact
K.E.prod > K.E.react
P.E.prod < P.E.react
products more stable than reactants
q 0
w 0
<
<
C12H24O12 (s) + KClO3(s) mix of gases