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Last modified: January 26, 2018

Zeroth Law of Thermodynamics
----------------------------

Most natural processes are not reversible.

i.e. Thot -> Tcold, Tcold -/-> Thot

First law of Thermodynamics
---------------------------

ΔU = Q - W

U = Internal energy
Q = Heat added/subtracted to/from the system
W = Work done on or by the system

Second law of Thermodynamics
----------------------------

It is impossible to extract an amount of heat QH from a hot reservoir and 
use it all to do work W.  Some amount of heat must be exhausted to a cold 
reservoir. 

Work Done by a Gas
------------------
 
Consider a piston with area A travelling a distance d:

W = Fd
  = (F/A)Ad
  = PΔV
  = the area under the P versus V curve.
  = ∫PdV

Now PV = nRT so P = nRT/V.  Therefore,

W = ∫PdV = nRT∫dV/V

  = nRTln(Vf/Vi)

  = (N/NA)RTln(Vf/Vi)

  = NkBTln(Vf/Vi)

If the gas is expanding, ΔV is positive and constitutes work being
done by the gas.  Since ΔU = Q - W this means that internal energy
of the gas must decrease.  Conversely, if the gas is being compressed, the 
ΔV term is negative.  This constitutes work being done on the gas.
In this case ΔU = Q - (-W) = Q + W so the internal energy of the gas
inreases.

Isothermal, Adiabatic, Isobaric and Isochoric Processes
------------------------------------------------------           



Isothermal:

Isothermal processes take place at constant temperature.   In order for
this to happen the system is assumed to be connected to a heat reservoir
that supplies/absorbs heat to maintain the temperature. Changes are
assumed to take place slowly so that equilibrium with the reservoir is 
maintained at all times.  We can summarize Isothermal processes by 
saying:

     ΔT = 0 and  Q ≠ 0

Since ΔU is proportional to T, this implies that ΔU = 0.  Thus,

    0 = Q - W ∴ Q = W

Thus, all the heat added to the system is used to do work.

Adiabatic:

Adiabatic processes take place at non-constant temperature.   The system 
is insulated and not connected to a heat reservoir so there is no heat flow
to keep the temperature constant.  Changes are assumed to take place
rapidly.  We can summarize Adiabatic processes by saying:

     Q = 0 and ΔT ≠ 0

Correspondingly,

     ΔU = -W

Isobaric:  

Isobaric processes take place at constant pressure.

Isochoric (aka isovolumetric) processes: 

Isochronic processes take place at constant volume.  Since there is no
ΔV involved there is no work done.

Heat Capacities:

U is a funtion of T and V.

dU = (∂U/∂T)V dT + (∂U/∂V)T dV

For constant V, dV = 0. Therefore,

dU = (∂U/∂T)V dT

   = CV dT where CV is the heat capacity at constant V

Define Enthalpy, H = U + PV

H is a funtion of T and P.

dH = (∂H/∂T)P dT + (∂H/∂P)T dP

For constant P, dp = 0.  Therefore,

dH = (∂H/∂T)P dT

   = CP dT where CP is the heat capacity at constant P

Now,

(∂H/∂T)P = (∂U/∂T)P + (∂PV/∂T)P

(∂H/∂T)P = (∂U/∂T)P + (PdV/dT)P

Now PV = nRT so PdV/dT = R if n = 1

CP = (∂U/∂T)P + R

Now, using the Chain Rule, we can write:

(∂U/∂T)P = (∂U/∂T)V + (∂U/∂T)P(∂V/∂T)P 

(∂U/∂T)P = (∂U/∂T)V since there is no change in U when T is constant.

and so,

CP = CV + R

Boyle's Law for Adiabatic process:

dU = dQ - PdV

   = dQ - (nRT/V)dV

For adiabatic process dQ = 0, therefore:

dU = -(nRT/V)dV = CVdT

CVdT/T = -nRdV/V

Integrate both sides and set n= 1 to get:

CVln(Tend/Tbegin) = -Rln(Vbegin/Vend)

∴ CVln(Tend/Tbegin) = Rln(Vend/Vbegin)

∴ ln(Tend/Tbegin)CV = ln(Vend/Vbegin)R

∴ (Tend/Tbegin) = (Vend/Vbegin)R/CV

Now, from before:

CP = CV + R

∴ R/CV = (CP - CV)/CV = γ - 1

which leads to:

(Tend/Tbegin) = (Vend/Vbegin)γ - 1

Using the ideal gas equation PV = nRT we can write:

(PendVend/nR)/(PbeginVbegin/nR) = (Vend/Vbegin)γ - 1

(PendVend)/(PbeginVbegin) = (Vend/Vbegin)γ - 1

Pend/Pbegin = (Vend/Vbegin)γ

which leads to:

PbeginVbeginγ = constant = PendVendγ

This is Boyle's Law where the temperature is not constant.

Entropy:

Consider an isothermal process such that Q = W 

From before we can write that:

Q = W = NkBTln(Vf/Vi)

  = NkBln(Vf/Vi)

  = kBln(Vf/Vi)N

If we assume that the volume of the gas is proportional to the number 
of states in the system, the RHS looks very similar to the definition of 
the change in entropy derived from statistical mechanics.  Thus, we can 
write an alternative thermodynamics equation for the change in entropy 
as:

ΔS = S - S' = Q/T

S is a state function or variable. A state function is a property of a
system that is not dependent on how the system arrived at its present 
condition (is not time dependent).  Any change in a state function is 
only dependent on the initial and final conditions. P, V, and T are other 
examples of state functions whereas heat transfers, work transfers 
and rates of chemical reactions are not.

Carnot Cycle
------------



A to B -  The gas expands and does work on the surroundings.  The temperature 
          of the gas does not change during the process, and so U remains 
          constant.  The gas expansion is due to the absorption of heat 
          energy, Qhot.  The entropy change, ΔS = Qhot/Thot  
 
B to C -  The gas continues to expand, doing work on the surroundings, and 
          losing an equivalent amount of internal energy, U. The loss of U 
          causes it to cool to the "cold" temperature, Tcold. ΔS = 0.

C to D -  The surroundings do work on the gas, causing an amount of heat 
          energy, Qcold, to flow out of the gas to the low temperature 
          reservoir.  Again, the temperature does not change so U remains 
          constant at its new level.  ΔS = -Qcold/Tcold 

D to A -  During this step, the surroundings do work on the gas, increasing 
          its internal energy and compressing it, causing the temperature 
          to rise to Thot. ΔS = 0.

The Carnot cycle is the most efficient heat engine cycle allowed by physical 
laws. When the second law of thermodynamics states that not all the 
supplied heat in a heat engine can be used to do work, the Carnot efficiency 
sets the limiting value on the fraction of the heat which can be so used.

In order to approach the Carnot efficiency, the processes involved in the heat 
engine cycle must be reversible and involve no change in entropy. This means 
that the Carnot cycle is an idealization, since no real engine processes are 
reversible and all real physical processes involve some increase in entropy. 

Consider the adiabatic processes, Q = 0

ΔU = -PΔV

∴ ΔU + PΔV =0

(3/2)nRΔT + nRTΔV/V = 0

(3/2)dt/T + dV/V = 0

(3/2)∫dt/T + ∫dV/V = 0

          Tcold       C
[(3/2)lnT]    +  [lnV] = 0
          Thot        B

ln(Tcold/Thot)3/2 + ln(Vc/VB) = 0

(Tcold/Thot)3/2(VC/VB) = 1

Take the reciprocal to get:

(Thot/Tcold)3/2(VB/VC) = 1

For D to A we get:

(Thot/Tcold)3/2(VA/VD) = 1

Thus, we can say,

VB/VA = VC/VD

What about the entropy?  Consider the isothermals (ΔU = 0 so Q = W)

ΔS = Qhot/Thot + 0 + Qcol/Tcold + 0

   = Whot/Thot + Wcol/Tcold

   = (nRThot/Thot)∫dV/V + (nRTcold/Tcold)∫dV/V

   = nR{∫dV/V +  ∫dV/V}

   = nR{ln(VB/VA) + ln(VD/VC)}

   = nR{ln[(VB/VA)(VD/VC)]}

   = nR{ln[(VB/VA)/(VC/VD)]}

   = nRln(1) from before

   = 0

The efficiency, E is:

W = Work done during a cycle = Qhot - Qcold (ΔU = 0)

E = W/Qhot = (Qhot - Qcold)/Qhot

  = 1 - Qcold/Qhot

The overall change in internal energy around the cycle is 0.  Therefore, 
ΔU = 0 and Q = PdV

We can write from before that:

Qhot = nRThotln(VB/VA) and Qcold = -nRTcoldln(VD/VC) = nRTcoldln(VC/VD)

∴ E = 1 - nRTcoldln(VC/VD)/nRThotln(VB/VA)  

    = 1 -  Tcoldln(VC/VD)/Thotln(VB/VA)

But from before we have shown that VB/VA = VC/VD.  Thus,

    = 1 - Tcold/Thot

If the efficiency is less than 100% we get:

          W/Qhot ≤ (1 - Tcold/Thot)

        ∴ W ≤ Qhot(1 - Tcold/Thot)

(Qhot - Qcold) ≤ Qhot(1 - Tcold/Thot)
           
         Qcold ≥ QhotTcold/Thot

    Qcold/Tcold ≥ Qhot/Thot

    Scold ≥ Shot ... The 2nd Law of Thermodynamics

This implies that the entropy at the end of the process is greater 
than that at the beginning.  The more efficient the process, the less 
the increase in entropy will be.  In other words, entropy provides a 
measure of the amount of thermal energy that cannot be used to do 
work.  The change in entropy will only be 0 for a reversible process 
(100% efficiency).  The connection between reversibility and efficiency 
can be thought of as follows: 

Consider an upright cylinder and piston in a vacuum.  


   |  W  |
   |=====|   
   |     |
   |     |
    -----

The gas in the cylinder is compressed by a weight, W.  If W is suddenly 
removed, the molecules in the gas will not instantaneously rearrange 
themselves and there will be a period of non-equilibrium in P, V and T.  
If, on the other hand, W is removed slowly in very small increments, 
we can consider the system to be in constant equilibrium. Now, if there 
is friction between the piston and the cylinder, any movement of the 
piston will generate heat.  Removing increments of W and adding 
increments of W to return to the original state both result in the 
generation of heat.  Thus, returning both the system and its surroundings 
to their original state simultaneously cannot be achieved.  In summary, a 
reversible process has to take place slowly and be frictionless.

An implication of the fact that entropy can never decrease is that at the 
time of the 'Big Bang', the universe had low entropy and a high degree 
of order.  Since then, however, the entropy has increased (and continues 
to increase) with time.

It is worthwhile noting that systems that are not isolated may decrease 
in entropy, provided they increase the entropy of their environment by 
at least that same amount.  An example is the freezing of water.  The
entropy of the water decreases as it freezes, but the heat extracted
increases the entropy of the environment.

The Heat Pump
-------------

The heat engine cycle described can be run in the opposite direction to
become the refrigeration cycle. The cycle remains exactly the same 
except that the directions of any heat and work flows are reversed. Heat 
is absorbed from the low temperature reservoir and dumped into the high
temperature reservoir with work input required to accomplish all this. The 
P-V diagram is the same except that the directions of the processes are 
reversed.

In the case of an a/c system, the cold reservoir is the inside of the 
building and the outdoors is the hot reservoir.  In the case of a heat 
pump, the cold reservoir is the outdoors and the inside of the building 
is the cold reservoir.  In both cases the work input is the electrical 
power.  



- The compressor compresses the refrigerant and raises its pressure 
  and temperature.  Latent heat is also released as the gas condenses
  to a liquid (the boiling point is raised).  This corresponds to CB 
  of the Carnot cycle.
- While passing through the condenser, the heated refrigerant yields 
  part of its heat to the hot surroundings with a lower temperature.
  This corresponds to BA of the Carnot cycle.
- The expansion phase reduces the pressure of the fluid, and consequently 
  its temperature. Latent heat is also absorbed as the liquid evaporates 
  to a gas (the boling point is lowered). This corresponds to AD of the 
  Carnot cycle.
- When it passes through the evaporator, as its temperature is lower 
  than the cold surroundings, the fluid picks up heat and the cycle 
  repeats. This corresponds to DC of the Carnot cycle.

Clausius Statement of the 2nd Law
---------------------------------

|   Thot    | 
 -----------
       |
       Q
       |
 -----------
|  Tcold    |


Q flowing from hot to cold:

ΔS = -Q/Thot + Q/Tcold

Since Thot > Tcold  ΔS > 0

Q flowing from cold to hot:

ΔS < 0 ... not allowed.

Spontaneous Action
------------------

Consider a gas in a container with a partition.  After the partition
is removed we get:

 --------------           ------------
| gas  |  vac  |         |     gas    |
|      |       |         |            |
 --------------           ------------

We always get the situation of left proceeding to right but never the
other way around in which the gas collects in the left side of the
container.  Why is this?

For the gas expanding we get:

U = Q - PdV

or

Q = U + PdV

Now, ΔS ≥ Q/T  ∴ Q ≤ TΔS.  Thus,

U + PdV ≤ TΔS

or

U + PdV - TΔS ≤ 0

This formula is satisfied for the left to right process but can never be
satisfied in the reverse direction. 

A Little More About Enthalpy
----------------------------

From before we had:

H = U + PV

∴ ΔH = ΔU + Δ(PV)

     = Q - W + Δ(PV)

     = Q - PdV + Δ(PV)

For constant P (isobaric) Δ(PV) = PΔV

∴ ΔH = Qp ... the heat added at constant P.

Enthalpy is useful in chemistry because most reactions occur at constant P 

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