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Physical Chemistry (CHE201)

Lesson No. 15-20

Work, Heat, and the First Law

• Work: w = F ⋅ A

applied force [pic]distance

A

Expansion work

F = pext A

pext

pext

|w = − |(pext A)A = −pext ∆V |

|convention: |Having a “-“ sign here implies w > 0 if ∆V < 0 , that |

| |is, positive work means that the surroundings do |

| |work to the system. If the system does work on the |

| |surroundings (∆V > 0) then w < 0 . |

If pext is not constant, then we have to look at infinitesimal changes

|- |- | | |

|dw = −pext dV |d means this is not an exact differential |

|Integral w = −∫12 pextdV |depends on the path!!! |

|• Path dependence of w | | | |

|Example: assume a reversible process so that pext = p |

|Ar (g, p1, V1) |= |Ar (g, p2, V2) |

|Compression |V1 > V2 |and p1 < p2 |

| | |

| | | |

| |đq = CpathdT |or |đq | |

| | | | | |

| | | |Cpath = | | | |

| | | | | | | |

| | | | | |dT |path |

| | | | |heat capacity is path dependent |

| |Constant volume: |CV | | | | |

| |Constant volume: |Cp | | | | |

| |∴ |q = ∫ |CpathdT | | | | |

| | | |path | | |

|Equivalence of work and heat | |[Joule (1840’s)] | |

| | | |

| | | |

|(b) with only work | |T1 → T2 |

| | | |

(weight falls & churns propeller)

Experimentally it was found that

( - - )

∫ dw +dq = 0

⇒ The sum (w + q) is independent of path

⇒ This implies that there is a state function whose differential is

đw + đq

We define it as U, the “internal energy” or just “energy”

∴ dU = đw + đq

For a cyclic process ∫dU = 0

For a change from state 1 to state 2,

∆U = ∫12dU = U2 −U1 = q + w does not depend on path

each depends on path individually, but not the sum

For fixed n, we just need to know 2 properties, e.g. (T, V), to fully describe the system.

So U = U (T ,V ))

U is an extensive function (scales with system size).

| | |= U |is molar energy (intensive function) |

|U | | |

| | |n | |

THE FIRST LAW

| |- |- |

| |dU =dq +dw |

| |or | |

|Mathematical statement: |∆U = q + w |

| |or |- |

| |- | |

| |− ∫ dq = ∫ dw |

Corollary: Conservation of energy

∆Usystem = q +w ∆Usurroundings = −q −w

⇒ ∆Uuniverse = ∆Usystem + ∆Usurroundings = 0

Clausius statement of 1st Law:

The energy of the universe is conserved.

| | | | |

|Enthalpy |H(T,p) |H ≡ U + pV | |

Chemical reactions and biological processes usually take place under constant pressure and with reversible pV work. Enthalpy turns out to be an especially useful function of state under those conditions.

reversible

gas (p, T1, V1) = gas (p, T2, V2)

const .p

|U1 | | | | |

|∆U + p∆V = qp |p | |( |)| |define as H |

|( |pV |)|= q | |

Choose

What are

|H (T , p ) | |∂H |dT |

| |⇒ dH = | | |

| | |∂T p | |

| |∂H |and | |∂H |? |

| | | | | | |

| |∂T p | | |∂p T | |

| |∂H | |dp |

|+ | | | |

| |∂p | | |

| | |T | |

|• | |∂H |⇒ for a reversible process at constant p (dp = 0) |

| | |∂T | | |

| | | |p | |

|dH = đqp |and |dH | |∂H |dT |

| | | |= | | |

| | | | |∂T p | |

⇒ đqp

∴

| |∂H |dT | |but đqp = C p dT also |

|= |∂T | | | | |

| | |p | | | |

| | | | | | | |

| | | |∂H | |= Cp | |

| | | | |∂T | | | |

| | | | | |p | | |

| | | | | | | | |

| | | | | | | |

| | | | | | | |

| | |

|1 |1 |2 |2 | |

| | | | |∴ | |∆H = 0 |

Joule-Thomson is a constant Enthalpy process.

| | | | |∂H | |dp |

|dH = C p dT + | | | |

| | | | |∂p |T | | |

|⇒ | |∂H | | | | |∂T |

| | | | |= −Cp | |

| | |∂p |T | | | |∂p |

|⇒ C |dT |= − | |∂H |dp |

|p | | | | |H |

| | | | |∂p T | |

| |← can measure this | |∆T | |

| | | | | |

|H | | |∆p H |

|Define |lim | |∆T |= | |

|H ≡ U (T ) + pV = U (T ) + nRT | | |

| | |only depends on T, |no p dependence |

|H (T ) |⇒ | |∂H | |= 0 |for an ideal gas |

| | | | |= µJT | | |

| | | |∂p |T | | |

For a van der Waals gas:

|∂H | | | | |a | | |

| | | |≈ b − | | | |

| | | | | | | |

| | |T | | | | |

| | | | | |RT |

| | |then ∆T | |

| | | |∆p | |

| | | | | | | |

| | | | | |RT |

| | |then ∆T | |

| | | |∆p | |

| | | | | | | |

| | | | | | |then|∆T > 0 |

|> b |⇒ | |T < | | | |a |

| | | | |

| | |, |∫dU = 0 for cyclic process ⇒ q = −w |

| |U = q + w | | |

Suggests engine can run in a cycle and convert heat into useful work.

⇒ Second Law

∫ Puts restrictions on useful conversion of q to w

∫ Follows from observation of a directionality to natural or spontaneous processes

∫ Provides a set of principles for

( determining the direction of spontaneous change

( determining equilibrium state of system

|Heat reservoir |Definition: A very large system of uniform |

| |T, which does not change regardless of |

| | |

| |the amount of heat added or withdrawn. |

Also called heat bath. Real systems can come close to this idealization.

Different statements of the Second Law

Kelvin : It is impossible for any system to operate in a cycle that takes heat from a hot reservoir and converts it to work in the surroundings without at the same time transferring some heat to a colder reservoir.

| | | |q>0 | | | |

| | | | | | | | |

| | | | | | | |q1 0, w > 0, T2 < T1 |

| | | |T |−T | | | |

| | | | | | | | |

| | | | | |1 | | |

| | |q1 | |q2 |

| | | | | |

|q1 | |q1′ | | |

|Assume |ε ′ > ε |(left |

engine less efficient than Carnot cycle)

Since the engine is reversible, we can run it backwards. Use the work (-w’) out of the Carnot engine as work input (w) to run the left engine backwards.

| |∴ Total work out = 0 |(-w’ = w > 0) |

|But ε ' > ε ⇒|−w ′ |> |

| |T |2 |1 |∫1 T |

| | | | | |

|Note: |Entropy is a state function, but to calculate S requires a |

| |reversible path. | | | |

• An irreversible Carnot (or any other) cycle is less efficient than a reversible one.

|p | | | |1 → 2 | | |

| | | | | | | | |

|1 | |irreversible |(−w )irrev < (−w )rev |⇒ wirrev > wrev |

| | | |isotherm with pext = p2 | | |

| |adiabat | |2 | |U = qirrev + wirrev |= qrev + wrev |

| | |4 |adiabat | | | | |

| | | | |∴ |qirrev |< qrev | |

| |

| | | | |than a reversible one. |

| | |= 1 + |q rev|< 1 + |q rev |

| | |

Connect two metal blocks thermally in an isolated system ( U = 0)

|Initially |T1 ≠T2 | | | | | |

| | | | | |T −T | |

|dS = dS1 |+ dS2 = |đq1 |− |đq2 = đq1 |( 2 |1 ) |(đq1 = −đq2 ) |

| | | | | |TT | |

| | |T | |T | | |

| | |1 | |2 |1 |2 | |

|dS > 0 for spontaneous process | |

|⇒ if |T2 >T1 |⇒ | |đq1 > 0 |in both cases heat flows |

| |T2 0 | | | | |

| | | | |

| | | | |

| | |T |

Entropy and Disorder

• Mixing of ideal gases at constant T and p

|nA A (g, VA, T) |+ |nB A (g, VB, T) |= n (A + B) (g, V, T) |

| | | | | | | |

| |VA |VB | |mixing |V = VA + VB | |

| | | | |

|A + B | | |

|piston |piston |back to initial state |

|permeable |permeable | | |

|to A only |to B only | | |

Sdemix = − Smix function of state

For demixing process

|⇒U = 0 ⇒ q |rev|= −w |rev|= p dV + p dV |

| | | | |AA |BB |

work of compression of each gas

| | | | |d| |

| | | | |q| |

| | | |

| |Smix > 0 | |

The mixed state is more “disordered” or “random” than the demixed state.

Smixed > Sdemixed

This is a general result ⇒

Entropy is a measure of the disorder of a system

∴ For an isolated system (or the universe)

S > 0 Spontaneous, increased randomness S = 0 Reversible, no change in disorder

S < 0 Impossible, order cannot “happen” in isolation

There is an inexorable drive for the universe to go to a maximally disordered state.

Examples of S calculations

In all cases, we must find a reversible path to calculate ∫ đTqrev

⇒ Mixing of ideal gases at constant T and p

nA A (g, VA, T) + nB A (g, VB, T) = n (A + B) (g, V = VA + VB, T)

Smix = −nR[ XA ln XA +XB ln XB ]

(b) Heating (or cooling) at constant V

A (T1, V) = A (T2, V)

| |đq |rev|T2 |C dT |if CV is |T |

|S = ∫ | | |= ∫T1 |V |= |CV ln | |

| |T | |T -independent | | |

[Note S > 0 if T2 >T1 ]

• Reversible phase change at constant T and p

|e.g. H2O (l, 100°C, 1 bar) = |H2O (g, 100°C, 1 bar) |

|qp = |Hvap | | |

|Svap ( 100°C ) |= |qpvap|= |H vap |(Tb = boiling Temp at 1 bar) |

| | |T | |T | |

| | | | | | |

| | |b | |b | |

(d) Irreversible phase change at constant T and p e.g. H2O (l, -10°C, 1 bar) = H2O (s, -10°C, 1 bar)

This is spontaneous and irreversible.

• We need to find a reversible path between the two states to calculate S.

H2O (l, -10°C, 1 bar)

đqrev = Cp ( A )dT

H2O (l, 0°C, 1 bar)

|irreversible |H2O (s, -10°C, 1 bar) |

|= | | |

| | | | |đqrev = Cp (s)dT |

|reversible | | | | |

| |H2O (s, 0°C, 1 bar) |

|= | |

qprev = − Hfus

| | |S =|Sheat|Sfus |+|Scoo| |

| | | |ing | | |ling| |

| | | |+ | | | | |

| | | | | | | | |

| |T | | | | | | |

∴ Absolute Entropies

Absolute entropy of an ideal gas

|From dU = TdS – pdV |⇒ |dS = dU + pdV | | | |

| | | | | |

| | |

| | | | |

| |∫|p |

| | | | | | | |

| | | | | |o (T) − Rln p (p in bar) |

| | |S(p,T) = |S | |

|But to finish, we still need | | |o (T) ! |

| | |S | |

Suppose we had S o (0K) (standard molar entropy at 0 Kelvin)

| |∂S | |Cp | | |o| |

| | | | |we should be able to get S | |(T) |

|Then using | |= | | | | |

| | | |T | | | |

| |∂T p | | | | | | |

Consider the following sequence of processes for the substance A:

A(s,0K,1bar) = A(s,Tm ,1bar) = A(ℓ,Tm,1bar) = A(ℓ,Tb,1bar) = A(g,Tb,1bar) = A(g,T,1bar)

| | | | | | | | | | | | | | | | | | | | | | | | | | | | | |vap | |T | | | | | | | | | | |Tm |Cp (s)dT | | | | |H | | | |Tb | |Cp (A)dT | | | | |H | | | |Cp (g)dT | | | | |= S o (OK) + ∫ | | | | | | | | | | | | | | | | | | | | | | | |S(T,1bar) | | | | | | | | | |+ | | | |fus |+ ∫ | | | | | | |+ | | | | |+ ∫ | | | | | | | | | | | | | |T | | | | | | | | | |T | | | | |Tb | | | | |T | | | | | | | |0 | | | | | | | | | |Tm |T | | | | | | | | |T | | | | | | | | | | | | | | | | | | | | | | | | | | |m | | | | | | | | | | | |b | | | | | |So(T) |S |= ∫ CpdT | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |T | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |Liquid boils, | | |S = |Hvap | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |T | | | | | | | | | | | | | | | | | | | | |Solid melts, | |S = |Hfus | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |T | | | | | | | | | | | | | | | | | | | | | | |0 | | | | | | | | | | | | | | | | | | | |T | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |

Since S0 is positive for each of these processes, the entropy must have its smallest possible value at 0 K. If we take S o (0K) =

zero for every pure substance in its crystalline solid state, then we could calculate the entropy at any other temperature.

This leads us to the Third Law !!!

THIRD LAW:

First expressed as Nernst's Heat Theorem:

• Nernst (1905): As T → 0 K , S → 0 for all isothermal processes in condensed phases

More general and useful formulation by M. Planck:

• Planck (1911): As T → 0 K , S → 0 for every chemically homogeneous substance in a perfect crystalline state

Justification:

⇒ It works!

• Statistical mechanics (5.62) allows us to calculate the entropy and indeed predicts S o (0K) = 0.

This leads to the following interesting corollary:

It is impossible to decrease the temperature of any system to T = 0 K in a finite number of steps

How can we rationalize this statement?

Recall the fundamental equation, dU = T dS – p dV

dU = Cv dT For 1 mole of ideal gas, P = RT/V

so Cv dT = T dS – (RT/V) dV dS = Cv d (ln T) + R d (ln V)

For a spontaneous adiabatic process which takes the system from T1 to a lower temperature T2,

S = Cv ln (T2/T1) + R ln (V2/V1) ≥ 0

but if T2 = 0, Cv ln (T2/T1) equals minus infinity !

Therefore R ln (V2/V1) must be greater than plus infinity, which is impossible. Therefore no actual process can get you to T2 = 0 K.

But you can get very very close!

In Prof. W. Ketterle's experiments on "Bose Einstein Condensates" (MIT Nobel Prize), atoms are cooled to nanoKelvin temperatures (T = 10-9 K) … but not to 0 K !

Another consequence of the Third Law is that

It is impossible to have T=0K.

How can we rationalize the alternate statement?

Consider the calculation of S starting at T=0K

S(s,T,1bar) = ∫TCp (s)dT

0 T

to prevent a singularity at T=0, Cp → 0 as T → 0 K

in fact, experimentally Cp = γT + AT3 + ...

That is, the heat capacity of a pure substance goes to zero as T goes to zero Kelvin and this is experimentally observed.

Combining the above with dT = đqp/Cp , at T=0 any infinitesimally small amount of heat would result in a finite temperature rise.

In other words, because Cp → 0 as T → 0 K, the heat đ qp needed to achieve a temperature rise dT, (đq p=CpdT) also goes to zero at 0 K. If you somehow manage to make it to 0 K, you will not be able to maintain that temperature because any stray heat from a warmer object nearby will raise the temperature above zero, unless you have perfect thermal insulation, which is impossible.

• Some apparent violations of the third law (but which are not !) Any disorder at T = 0 K gives rise to S > 0

. For example in mixed crystals

Smix = −nR[XA ln XA + XB ln XB ] > 0 Always !!! Even at T=0K

But a mixed crystal is not a pure substance, so the third law is not violated.

a) Any impurity or defect in a crystal also causes S > 0 at 0 K

b) Any orientational or conformational degeneracies such is in a molecular crystal causes S > 0 at 0 K, for example in a carbon monoxide crystal, two orientations are possible:

C O C O C O C O C O C O C O

C O C O C O C O C O O C C O

C O C O C O O C C O C O C O

C O C O C O C O C O C O C O

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