Lec 1 Thermodynamics

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First Law: If A & C systems, B & C systems have the same temperature. A & B systems have the same temperature.

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Second Law: In an adiabatic process (there is no heat transfer), the work chance is equal to the (final E - initial E). If there is heat transfer, ∆Q = (Final E - Initial E) - ∆W

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In differential form

dE(\vec(x))          =        \bar(d) W + \bar(d) Q

Depends on state         Depend on path

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Quasistatic: Slow (slow enough to maintain equilibrium)

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Pull a string sufficient slowly that the string does not start to vibrate (Too rapidly -> oscillating)

Calculate the amount of work

\bar(W) = \Sigma_i J_i dx_i 

\Sigma_i Generalized Force

\dx_i Generalized Path

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                      x.       J

wire(1d)        L.      F

film(2d)        A.      \sigma (surface tension)

gas(3d).        V.      -P

Magnet.       M.      B.

 

x - extensive 

J - intensive

 

\bar(d)W = \Sigma J_i dx_i 

\bar(d)Q = ?

--

dE = \Sigma_i J_i dx_i + ?

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Ideal gas scale:

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Heat capacity C_{path}= \bar(d)Q_{path}/dT  \bar -> dependent on path

C_v C_p

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C_v = \bar(d)Q_v/dT = (dE + PdV_v)/dT = ∂E/∂T|_v No mechanic work is done

C_p = \bar(d)Q_p/dT = (dE + PdV_p)/dT = ∂E/∂T|_p + P(∂V/∂T)|_p

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Expansion ideal gas

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A gas that is adiabatically isolated.

From a chamber to both chambers.

T_f = T_i = T

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∆Q = 0 = ∆W -> ∆E

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Pressure and volume certainly changed when it goes from one chamber to both chambers.

Ideal gas 

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E(P, V) = E(T, V)

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C_p - C_v = PV/T = N k_b

k_B = 1.4 * 10^{-23}

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Lec2 Thermodynamics

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RECAP

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A. Equilibrium (x_i, J_i)

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x_i generalized displacement

J_i generalized force 

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Ideal Gas

(V, -P)

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Where does it lie in the PV-Plane

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B. 0th Laws (Transitive of Equilibrium) If two objects are in equilibrium with the third object they are also in equilibrium with each other

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\Theta(x_i, J_i) = \Theta'(x_i', J_i')

The equilibrium has to be the same

Like a balance

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(V, -P) 

PV = \Theta (constant)

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C. 1st laws 

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Adiabatic 

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The amount of the work is only dependent on the initial and final state. 

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Conversation of Energy 

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\exist E(x_i, J_i)

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dE = \bar(d)W+ \bar(d)Q

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dE only depends on the state

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\bar(d) blah depends on the path

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Doole's free expansion

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Isolate two gases into two chambers

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Initial the gas is completely isolated on one side

Final the gas is on both sides

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T_f = T_i

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E must only be a product of PV

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P changes V changes

But PV is constant

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NEW

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Quasistatic: slow enough:

\bar(d) W = \Sigma_i J_i dx_i (when the spring is pulled slowly aka. without vibrations)

 

mechanical equilibrium J is the same

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\bar(d) Q = 

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temperature for measuring thermal equilibrium

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the conjugate? Entropy

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Student Qs: 

[The equilibrium is ideal

[The adiabatic process is ideal

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D. 2nd law

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Heat <- -> Work

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Engine

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Hot

    | Q_H

   V

Engine -> W

    | Q_C

   V

Cold

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Efficiency

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\ita = W/Q_H = (Q_H - Q_C)/Q_H < 1

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Reverse Refrigerator

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Performance 

\omega = Q_c / W = Q_c / (Q_H - Q_c)

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Kelvin: No process is possible whose sole result is complete conversion of heat to work.

(No idea engine, \ita < 1)

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Clausius: No process is possible whose sole result is transfer of heat from colder to  hotter body. (No ideal refrigerator)

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E. Carnot Engine is any engine that is 

  (1) Reversible: Can go forward/backward by reversing input/output (frictionless)

  (2) Operates in cycle (start and endpoints are the same)

  (3) All heat input and output at 2 temperature 

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