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The First Law Part 3

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Joule-Thompson Effect. Therefore, there is no enthalpy change in a J-T expansion ... This is important because Joule was unsuccessful in evaluating the partial of U ... – PowerPoint PPT presentation

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Title: The First Law Part 3


1
Chapter 3
  • The First Law (Part 3)
  • The Machinery

2
The Joule-Thompson Effect
3
Joule-Thompson Effect
  • Under certain conditions, this leads to cooling
    (T1gtT2)
  • The pipe is insulated adiabatic system
  • q0
  • Work to push 1 mole of gas through the pipe
  • wleftP1V1 (work done on the system by piston)
  • wrightP2V2 (work done by the system on piston)

4
Joule-Thompson Effect
  • The total work may be expressed as
  • Since q0
  • Therefore,

5
Joule-Thompson Effect
  • Therefore, there is no enthalpy change in a J-T
    expansion
  • The definition of m is
  • Since dH0, then

6
Joule-Thompson Effect
  • Recall, CP?

7
Joule-Thompson Effect
  • So, we see that it is possible to measure the
    partial derivative of H with respect to P at
    constant T.
  • This is important because Joule was unsuccessful
    in evaluating the partial of U with respect to P
    at constant T by allowing a gas to expand into a
    vacuum.

8
Problem
  • Show that mK for an ideal gas is zero if

9
Inversion Curve
  • If we choose some P1,T1, vary P2, and plot T2,
    then we get a graph like the one shown on p. 84.
  • The curves plotted are called isenthalps b/c H is
    constant.
  • At some point along each line, the sing will
    change.
  • The temperature at which this occurs is called
    the inversion temperature.
  • At the inversion temp, the sign of m changes.
  • A gas may be used as a refrigerant at temps BELOW
    the maximum inversion temp as shown on p. 85.

10
Relationship of CP and CV
  • Lets go way back in the math and get this
    equation

11
Relationship of CP and CV
This is the work differential
This is an expression of the energy differential
required to separate molecules due to
intermolecular attractions
For an Ideal Gas
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