1. Applications of the First Law
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2. The energy equation u u = u(T,v) v T
Internal energy, u
Property of substance
Function only of the state u
u = u(T,v)
Equation of state relates P, v, and T
Any pair defines the state
u can be expressed in terms of any pair of these variables e.g. u = u(T,v) v T
Energy equation,
Defines an energy surface
Energy equation cannot be derived from the equation of state but must be determined independently
Equation of state + Energy equation 
determine all properties of a substance
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3. Choosing T and v as independent variables
Can be calculated from the eq. of state using the 2nd law
Must be determined experimentally
1st law for a reversible process:
In the special case of an isochoric process:
Experimental measurements of cv determine the slope of the isochoric line on a u-T-v surface
For any reversible process:
The derivative
can be replaced by cv in any equation
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4. Dividing by and replacing with
In an isobaric process,
Thus,
Dividing by and replacing with
Does not refer to a process between two equilibrium states.
Simply a general relation that must hold between quantities that are all properties of a system in any one equilibrium state.
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5. Relationships involving enthalpy (choosing T and P as independent variables)
Can be calculated from eqn of state
1st law 
Case when:
Thus, for any reversible process:
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6. Case: or
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7. How does the internal energy of a gas depend on its volume?
Earlier…
Can be calculated from the eq. of state using the 2nd law
Can be calculated from eqn of state
Questions:
How does the internal energy of a gas depend on its volume?
(Gay-Lussac-Joule Experiment)
How does the enthalpy of a gas depend on its pressure?
(Joule-Thomson Experiment)
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8. There are no instruments that can measure internal energy and enthalpy directly.
Therefore, must express the derivatives in terms of measurable properties.
Making use of:
Measure the rate of change of temperature with volume
Similarly,
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9. Gay-Lussac-Joule Experiment
Free expansion W=0
If gas temperature changes,
Heat will flow between gas and water bath,
and reading of thermometer will change
BUT no change in temperature found (or too small to be detected)
Actually, the experiment is not precise due to high heat capacity of water
Direct measurement of temperature change with free expansion is very difficult.
Thermometer
evacuated
Water (known mass)
gas
Heat losses from tank to surrounding assumed negligible
stopcock
Postulate additional property for an ideal gas: no ΔT in free expansion
no heat flow to surrounding i.e Q=0.
Since W=0  ΔU =0
For ideal gas
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10. Since cv is finite  =0 for an ideal gas ≠0 for a real gas
Joule coefficient:
Since cv is finite 
for ideal gas
i.e.
only
Thus,
becomes
and
If cv can be considered constant,
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11. The Joule-Thomson Experiment
(Throttling process)
Process takes place in an insulated cylinder
P1,V1,T1
P2,V2,T2
Initial state
Final state
Porous plug
Specific work done in forcing the gas through plug:
Work done by gas in expansion:
Therefore, total work done is
i.e. the initial and final enthalpies are equal
Note: cannot say that the enthalpy remains constant during the process, since one cannot speak of the enthalpy of the system while it is passing through non-equilibrium states during the irreversible process.
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12. In the Joule-Thomson experiment, P1 and T1 are kept constant while P2 is varied and T2 is measured.
Since h1=h2=h3, etc., the enthalpy is the same at all these points
The locus of points  isenthalpic state of the gas
The numerical value of the slope on the curve at any point is called the Joule-Thomson coefficient μ
inversion point
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13. When the properties of real gases  an ideal gas,
By performing other series of experiments, keeping the initial pressure and temperature the same in each series but varying them from one series to another, a family of curves corresponding to different values of h can be obtained
If the initial temperature is not too great, the curves pass through a maximum point called the inversion point.
The locus of the inversion points is the inversion curve
When the properties of real gases  an ideal gas,
the isenthalpic curves nearly horizontal
i.e. μ≈0 or
Shown earlier
Therefore, for an ideal gas:
and
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14. If cP can be considered constant,
Thus,
becomes
and
If cP can be considered constant,
For an ideal gas
Since for an ideal gas
From eqn of state:
Therefore:
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15. Liquefaction of Gases If you want something really cold:
you don’t use a refrigerator
Instead you put it on dry ice (195 K)
Or immerse it in liquid nitrogen (77 K)
Or even liquid helium (4.2 K)
But how are gases like nitrogen and helium liquefied (or in the case of CO 2 solidified ) in the first place?
The most common methods all involve the throttling process.
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16. Dry ice
You can liquefy Carbon dioxide at room temperature, simply by compressing it isothermally, and then throttle it back to low pressure and it cools and partially evaporates.
However, at pressures < 5.1 atm, liquid CO2 cannot exist; instead the condensed phase is solid, dry ice
So to make dry ice, simply hook up a tank of liquid CO2 to a throttling valve and watch the frost form around the nozzle as the gas rushes out.
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17. Liquid Nitrogen Liquefying nitrogen (air) isn’t so simple
compressing it will only make it denser in a continuous way – no sudden phase transformation to a liquid
Example: Ti = 300K and Pi = 100 atm throttle to Pf = 1 atm
 Tf = 280K only
To get any liquid, must start at Ti < 160K
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18. Liquid Helium
To produce Joule-Thomson cooling of in helium, the helium needs to be cooled below 43K.
Maximum inversion temperature
Once a gas has been pre-cooled to a temperature lower than the maximum inversion temperature, the optimum pressure from which to start throttling corresponds to a point on the inversion curve.
Starting at this pressure and ending at atmospheric pressure, the process produces the largest temperature drop (which may not be large enough to produce liquefaction)
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19.  countercurrent heat exchanger
Consequently, the gas that has been cooled by throttling is used to cool the incoming gas, which after throttling becomes still cooler.
After successive cooling process, the temperature of the gas is lowered to such a temperature that after throttling, it becomes partly liquefied.
 countercurrent heat exchanger
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20. Comparison of properties of u and h for a hydrostatic system
Adiabatic process:
For an ideal gas:
Isobaric process:
Isochoric process:
In general
Throttling process (irreversible)
hi=hf
Free expansion (irreversible)
ui=uf
Enthalpy h
Internal energy u
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21. Reversible adiabatic processes
Since d’q=0
Therefore: or
Cannot be integrated until we know the structure of the gas, which determines γ
Monatomic gases: γ is constant
Diatomic and polyatomic gases : γ varies with T  A very large change in T produces an appreciable change in γ
e.g. T : 300  3000 K , γ: 1.40  1.29 (for carbon monoxide)
For an adiabatic process that involves only a moderate temperature change (as in most cases), γ ≈ constant
Integrating:
Other relations:
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22. Slope of any adiabatic curve:
A family of curves representing quasistatic adiabatic processes may be plotted on a P-v diagram by assigning different values to the constant in the equation Pvγ = constant.
adiabat
isotherm
Slope of any adiabatic curve:
Slope for isothermal curves:
From Pv=RT,
Since γ > 1, P falls off more rapidly with v for an adiabatic process.
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23. The specific work done in a reversible adiabatic process:
But
at both limits 
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