Lecture 2: Heat Capacities/State functions

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Lecture 2 Heat Capacities/State functions

• Reading Zumdahl 9.3
• Outline
• Definition of Heat Capacity (Cv and Cp)
• Calculating DE and DH using Cv and Cp
• Example of Thermodynamic Pathways
• State Functions

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Heat Capacity at Constant V

• Recall from Chapter 5 (section 5.6)
• (KE)ave 3/2RT (ideal monatomic gas)
• Temperature is a measure of molecular speed
• In thermodynamic terms, an increase in system
  temperature corresponds to an increase in system
  kinetic energy ( i.e., q is proportional to ltKEgt)

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Heat Capacity at Constant V

• (KE)ave 3/2RT (ideal monatomic gas)
• How much energy in the form of heat is required
  to change the gas temperature by an amount DT?
• Heat required 3/2R DT
• 3/2R
  (for DT 1K)
• Therefore, Cv 3/2 R is the heat required to
  raise one mole of an ideal gas by 1K. Cv is
  called constant volume molar heat capacity.

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Heat Capacity at Constant P

• What about at constant pressure? In this
  case, PV type work can also occur
• PDV nRDT RDT (for 1 mole)
• R (for
  DT 1 K)
• Cp heat into translation work to expand
  the gas Cv R 5/2R (for ideal monatomic
  gas)

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Cv for Monatomic Gases

• What are the energetic degrees of freedom for a
  monatomic gas?
• Just translations, which contribute 3/2R to Cv.

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Cv for Polyatomics

• What are the energetic degrees of freedom for a
  polyatomic gas?
• Translations, rotations, and vibrations. All of
  which may contribute to Cv (depends on T).

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Cv for Polyatomics

• When heat is provided , molecules absorb energy
  and the translational kinetic energy increases
• In polyatomic gases, rotational and vibrational
  kinetic energies increase as well (depending on
  T).

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Cv for Polyatomics

• T measures the average translational kinetic
  energy
• Increases in rotational and vibrational kinetic
  energies do not increase T directly
• It takes more heat to increase T by the same
  amount (Cv/Cp larger)

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Variation in Cp and Cv

• Monatomics
• Cv 3/2 R
• Cp 5/2 R
• Polyatomics
• Cv gt 3/2 R
• Cp gt 5/2 R
• But.Cp Cv R

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Energy and Cv

• Recall from Chapter 5
• Eave 3/2 nRT (average translational
  energy)
• DE 3/2 nR DT
• DE n Cv DT (since 3/2 R
  Cv)
• Why is CvDE/DT
• When heating our system at constant volume,
  all heat goes towards increasing E (no work).

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Enthalpy and Cp

• What if we heated our gas at constant pressure?
  Then, we have a volume change such that work
  occurs
• q p n Cp DT
• n (Cv R) DT
• DE nRDT DE PDV
• DH or DH nCpDT

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Keeping Track

• All Ideal Gases
• DE nCvDT
• DH nCpDT

• Ideal Monatomic Gas
• Cv 3/2R
• Cp Cv R 5/2 R
• Polyatomic Gas
• Cv gt 3/2R
• Cp gt 5/2 R

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Example

• What is q, w, DE and DH for a process in which
  one mole of an ideal monatomic gas with an
  initial volume of 5 l and pressure of 2.0 atm is
  heated until a volume of 10 l is reached with
  pressure unchanged?

Pinit 2 atm
Pfinal 2 atm
Vinit 5 l
Vfinal 10 l
Tinit ? K
Tfinal ? K
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Example (cont.)

• Since PDV nRDT (ideal gas law) we can determine
  DT

• Then DV (10 l - 5 l) 5 l

• And

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Example (cont.)

• Given this

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To Date.

• Ideal Monatomic Gas
• Cv 3/2R
• Cp Cv R 5/2 R
• Polyatomic Gas
• Cv gt 3/2R
• Cp gt 5/2 R

All Ideal Gases DE q w w
-PextDV (for now) DE nCvDT qV DH
nCpDT qP If DT 0, then DE 0 and
q -w
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State Functions

• If we start in Seattle and end in Chicago, but
  you take different paths to get from one place to
  the other ..
• Will the energy/enthalpy, heat/work we spend be
  the same along both paths?

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Thermodynamic Pathways an Example

• Example 9.2. We take 2.00 mol of an ideal
  monatomic gas undergo the following
• Initial (State A) PA 2.00 atm, VA 10.0 L
• Final (State B) PB 1.00 atm, VB 30.0 L

Well do this two ways Path 1 Expansion
then Cooling Path 2 Cooling then Expansion
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Thermodynamic Jargon

• When doing pathways, we usually keep one variable
  constant. The language used to indicate what is
  held constant is
• Isobaric Constant Pressure
• Isothermal Constant Temperature
• Isochoric Constant Volume

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Thermodynamic Path A series of manipulations of
a system that takes the system from an initial
state to a final state
isochoric
isobaric
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Pathway 1

• Step 1. Constant pressure expansion (P 2 atm)
  from 10.0 l to 30.0 l.
• PDV (2.00 atm)(30.0 l - 10.0 l) 40.0 l.atm
• (40.0 l.atm)(101.3 J/l.atm) 4.0
  x 103 J
• -w (the system does work)
• And DT PDV/nR 4.05 x 103 J/(2 mol)(8.314
  J/mol.K)
• 243.6 K (from the ideal gas law)

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Pathway 1 (cont.)

• Step 1 is isobaric (constant P) therefore,
• q1 qP nCpDT (2mol)(5/2R)(243.6 K)
• 1.0 x 104 J
• DH1
• And DE1 nCvDT (2mol)(3/2R)(243.6 K)
• 6.0 x 103 J
• (check DE1 q1 w1 (1.0 x 104 J) -(4.0 x
  103 J)
• 6.0 x 103 J )

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Pathway 1 (cont.)

• Step 2 Isochoric (const. V) cooling until
  pressure is reduced from 2.00 atm to 1.00 atm.
• First, calculate DT
• Now, DT DPV/nR (note P changes, not V)
• (-1.00 atm)(30.0 l)/ (2
  mol)(.0821 l.atm/mol K)
• -182.7 K

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Pathway 1 (cont.)

• q2 qv nCvDT (2 mol)(3/2R)(-182.7 K)
• - 4.6 x 103 J
• and DE2 nCvDT -4.6 x 103 J
• and DH2 nCpDT -7.6 x 103 J
• Finally w2 0 (isochoricno V change, no PV-type
  work)

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Pathway 1 (end)

• Thermodynamic totals for this pathway are the sum
  of values for step 1 and step 2
• q q1 q2 5.5 x 103 J
• w w1 w2 -4.0 x 103 J
• DE DE1 DE2 1.5 x 103 J
• DH DH1 DH2 2.5 x 103 J

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Next Pathway
Now we will do the same calculations for the
green path.
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Pathway 2

• Step 1 Isochoric cooling from P 2.00 atm to P
  1.00 atm.
• First, calculate DT
• DT DPV/nR (-1.00 atm)(10.0 l)/ (2 mol)R
• -60.9 K

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Pathway 2 (cont.)

• Then, calculate the rest for Step 1
• q1 qv nCvDT (2 mol)(3/2 R)(-60.9 K)
• -1.5 x 103 J DE1

DH1 nCPDT (2 mol)(5/2 R)(-60.9 K) -2.5 x
103 J w1 0 (constant volume)
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Pathway 2 (cont.)

• Step 2 Isobaric (constant P) expansion at 1.0
  atm from 10.0 l to 30.0 l.
• DT PDV/nR (1 atm)(20.0 l)/(2 mol)R
• 121.8 K

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Pathway 2 (cont.)

• Then, calculate the rest
• q2 qp nCPDT (2 mol)(5/2 R)(121.8 K)
• 5.1 x 103 J DH2

DE2 nCvDT (2 mol)(3/2 R)(121.8 K) 3.1 x
103 J w1 -PDV -20 l.atm -2.0 x 103 J
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Pathway 2 (end)

• Thermodynamic totals for this pathway are again
  the sum of values for step 1 and step 2
• q q1 q2 3.6 x 103 J
• w w1 w2 -2.0 x 103 J
• DE DE1 DE2 1.5 x 103 J
• DH DH1 DH2 2.5 x 103 J

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Comparison of Path 1 and 2

• Pathway 1
• q 5.5 x 103 J
• w -4.1 x 103 J
• DE 1.5 x 103 J
• DH 2.5 x 103 J

• Pathway 2
• q 3.6 x 103 J
• w -2.0 x 103 J
• DE 1.5 x 103 J
• DH 2.5 x 103 J

Note Energy and Enthalpy are the same, but
heat and work are not the same!
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State Functions

• A State Function is a function in which the value
  only depends on the initial and final state.NOT
  on the pathway taken

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Thermodynamic State Functions

• Thermodynamic State Functions Thermodynamic
  properties that are dependent on the state of the
  system only. (Example DE and DH)
• Other variables will be dependent on pathway
  (Example q and w). These are NOT state
  functions. The pathway from one state to the
  other must be defined.