Chapter 3: First laws of Thermodynamics

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Chapter 3 First laws of Thermodynamics

• Energy and forms of energy
• - Energy-work-heat
• - System-surrounding-universe
• - work
• - heat
• Laws of conservation of energy The first law
• - Internal energy
• - Statement of the first law
• - Internal energy work heat
• - Partial differential definition of
  internal energy

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? Enthalpy - definition - relation to
internal energy - partial differential
definition of enthalpy ? Thermochemistry -
Hesss law - Bond enthalpies -
Enthalpies of formation - Temperature
dependent
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Energy and forms of energy

• Energy-work-heat
• Put Zn metal inside a container and pour in
  HCl (aq),
• H2 gas will be generated according to reaction.
• Zn (s) 2HCl (aq) ZnCl2 (aq) H2 (g )

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Case A
If we put a movable lid on the container, the lid
will be raised due to pressure generated by gas.
h
If the lid has mass m there will be the work
done w F?l mgh
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Case B
If we put a non-movable lid on and place the
container in an ice bath, we will see ice
melted. The ice bath temperature is raised. The
heat is transferred from the container. Energy
is the capacity to make change. change in
position work change in
temperature heat
Energy from reaction comes from chemical bonds
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? System surrounding -universe
universe
universe everything, physical
universe system - part of universe
which have special interest surrounding
the rest of universe
system
Surrounding
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There can be energy and matters transferring from
and to system. According to energy and matter
exchange, systems can be classified into opened
system both energy and matters are exchanged
(with surrounding) closed system only
energy can be exchanged isolated system neither
energy nor matter are exchanged
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• Work
• motion against an opposing force
• - mechanical work

w -F? l Opposing force has the same amount with
force F but opposite sign (-F)
- gravitational work
w -mgh Opposing force -mg (gravitation)
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• electrical work
• w fdq
• f ? electric potential
• q ? charge

- expansion work (PV work)
external pressure
W -F?l -(PexA) ?l -Pex ?V
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- Process
?T0 constant T
isothermal process
?q0 constant q
adiabatic process
work
?p0 constant p
isopiestic process
reversible process can move back to original
position (slow and infinitesimal
change) Irreversible process cannot move back
to original position (fast and large
change)
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- Irreversible work (wirrev)
Pf,Vf,T
Pin internal pressure Pin gt Pex
Pin Pex
wirrev -Pex ?V -Pf ?V
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• Reversible work (wrev)
• gt To be reversible Pex must almost be equal to
  Pin all the time
• gt One-step expansion from Vi

Vf where Vf gtgt Vi is not
possible

• Reversible expansion is multi-step process and
  for each step
• Pex ? Pin P (Pex varies)

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• To guarantee that Pex ? Pin in each step, the
  piston should move by very small distance
• The charge in gas volume should be as small as
  possible or
• ?V 0 dV

• For each step
• dw -PexdV -PdV

• For overall expansion

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For ideal gas, one can replace P by P nRT/V

Only true for ideal gas and isothermal expansion
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- Reversible VS irreversible work
wrev gt wirrev wrev is the maximum work No
process can produce more work than reversible
process work depends on path, not on initial and
final state work is the path function
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Example 3.1 Calculate the work done when 50 g of
iron reacts with hydrochloric acid to produce
hydrogen gas in (a) a closed vessel of fixed
volume (b) an open beaker at 25 0C
(a) w -PexDV since Vgas
Vbeaker DV 0 w 0
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(b) w -PexDV DV Vgas - Vbeaker ?
Vgas nRT/Pex (Pex Pf) w -Pex nRT/Pex
-nRT -(50 g/55.8 g mol-1)(8.314 J
K-1mol-1)(298 K) -2,220 J -2.22 kJ
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Example 3.2 One mole of ideal gas expands from 5
to 1 bar at 298 K. Calculate w (a) for a
reversible expansion (b) for an expansion against
a constant external press of 1 bar, and (c) for a
free expansion (a) w -nRTln(Vf/Vi) from
PiVi PfVf Vf/Vi Pi/Pf w -nRT ln
(Pi/Pf) -(1 mol)(8.314 JK-1mol-1)(298
K)ln(5/1) -3988 J -3.99 kJ
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• (b) w -PexDV
• -Pex (nRT/Pf nRT/Pi)
• -Pex nRT (1/Pf 1/Pi)
• -(1bar)(1mol)(8.314JK-1mol-1)(298K)(1/1)-(1/5
  )bar-1
• -1.98 kJ
• Free expansion ? expansion without opposing
  force (F0)
• w 0

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• Heat (q)
• gt heat can be measured from the temperature of a
  material
• q ? DT
• gt same amount of heat can not raise temperature
  of different material equally
• gt different material have different heat
  capacity (C)
• q CDT
• For same amount of heat, material with large C
  have smaller ?T. Heat capacity is defined as
• C q/?T

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Heat capacity of common materials
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• Measurement of heat
• gt if C is known, heat can be measured by
  recording ?T.
• gt Heat that released or adsorbed by chemical
  reaction (system) is not measured directly but
  the lose and gain of heat by medium (surrounding)
  due to the reaction is measured.
• gt Water is used as medium and temperature of
  water during the course of reaction is measured.
• gt Water has specific heat (s) of 1 cal/C?g or
  4.184 J/ C?g
• ms C m weight of water

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The loss of energy into surroundings.
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The instrument that measure heat is called
Calorimeter
Simple calorimeter
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Example 3.3 A 466 g sample of water is heated
from 8.50 C? to 74.60 C?. Calculate amount of
heat absorbed by water. q ms ?T (466 g)
(4.184 J/ C?g) (74.60 - 8.50 C?) 1.2 x 10 5
J 129 kJ
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IUPAC sign convention

• Since system can lose and gain heat/work
• Direction of heat/work transferred to/from system
  could be followed from sign of the quantity
• Heat absorbed by system has sign (qgt0)
• Heat lost by system has sign (qlt0)
• Work received by system has sign (wgt0)
• Work performed by system has sign (wlt0)

system
work/ heat
work/ heat
-