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