1

Chapter 2 - The First Law of Thermodynamics I. The basic concepts.

A. Definitions.

1. system = the well-defined portion of the world we focus on.

2. surroundings = everything else.

3. open system = system in which matter can transfer in and out.

4. closed system = no matter transfer possible.

5. isolated system = neither matter nor heat can flow in or out.

6. work = the application of an organized force through a distance.

(force x distance) • work is done by system if system applies the force and expends

its energy. • work is done on system if system receives the energy and

surroundings supply the force. • units of work are energy units. • Symbol w = work done on the system, thereby increasing its

energy by amount w.

7. energy = the capacity of a system to do work. • Symbol U = internal energy of a system. • Absolute value of U is difficult to determine, but its value relative

to a reference standard is used, and we usually only care how much U changes in a transformation,

• o ΔU = Uf – Ui.

8. heat flow = energy flow into or out of system resulting from temperature difference between system and surroundings. • Symbol q = energy transferred to the system as heat (or heat

flow into the system). • Adiabatic system = one whose walls do not permit heat transfer.

2

9. Thus systems can change energy U in two ways, work or heat. This is expressed in the following conversion of energy units:

• This equivalence is an expression of the fact that energy is

energy, whether it is in the form of heat or work done. • SI unit of energy is the Joule (J) = 1 kg m2s-2.

10. “state” property = a property of a system which depends on the

current state of the system and not on the process of how we achieved that state.

For example: U, p, V, T are state variables. w and q are not state variables but refer to processes.

B. The First Law of Thermodynamics.

1. Internal energy U of an isolated system is a constant. (Energy can neither be created not destroyed; i.e., conservation of energy.)

2. Additionally, heat and work are two equivalent ways a system

connected to its surroundings can change energy, or:

ΔU = q + w change in internal = heat transfer + work done energy to system on system Note sign convention (not ΔU = q - w)! q > 0 if heat transfers into system, raising its energy. w > 0 if work is done on system, raising its energy.

3. First Law expressed for an infinitesimal change of state:

dU = dq + dw

3

II. Work and Heat.

A. Mechanical work.

1. Work done on system to move an object against an opposing force F by an infinitesimal distance dz:

dw = −Fdz (note, system loses energy by working)

2. Total work in entire process:

w = − Fdz

zi

zf

∫

3. If force is constant throughout process:

€

w = −F dzzi

z f

∫w = −F(zf − zi)

4. Example of mechanical work:

B. Work of compression and expansion.

1. Relation of w to pressure and volume change: Opposing external force F = pexA = (force/area)*Area dw = -Fdz = -pexAdz = -pexdV

dw = -pexdV

Work of lifting a mass m against a gravitational force field from position zi to zf.

€

w = −F(zf − zi)= −mgh

4

Analogous expression for other types of work. See Table 2A.1 Table 2A.1 Varieties of work*

2. Free expansion:

When a system expands against pex= 0, then dw = 0. (No work is done.) Pressure opposing the system is the important pressure.

3. Expansion against constant pressure:

In all cases dw = -pexV and

€

w = − pexdVV1

V2

∫

If pex constant (e.g. a huge surrounding system)

€

w = −pex dVV1

V2

∫= −pexΔV

5

4. Reversible expansion:

Occurs when the surrounding pressure pex is virtually equal to the system pressure p throughout the expansion:

If p is infinitesimally larger than pex at all times, piston will slowly expand in a process called reversible expansion. “Reversible” because a slight change of pressure on either side could change direction of piston motion at any stage along path.

dw = -pexdV always,

but in reversible expansion, system pressure p = external pressure pex.

pex = p so

dw = -pdV

and

€

w = − pdVV1

V2

∫

Not able to integrate as simply as before, since p changes throughout the process.

Need to know what type of process to determine how p changes:

5. Isothermal reversible expansion.

Take a perfect gas as an example in thermal contact with a large heat reservoir:

6

€

w = − pdVV1

V2

∫ where p = nRT /V

so:

w = −nRTdV

VV1

V2

∫

w = −nRTdVV

V1

V2

∫

w = −nRT lnV2

V1

isothermal reversible expansion of a perfect gas:

€

w = −nRT lnV2

V1

6. Irreversible expansion.

When system pressure p significantly exceeds the opposing pressure pex expanded against. Pushing power is wasted when p > pex. Maximum work is done when expanding reversibly, i.e. p = pex at all times.

7

Exercise - 4.50g CH4, V = 12.7 L, T = 310 K.

(1) Calculate w in Joules when gas expands isothermally against a constant external pressure of 200 Torr until V has increased by 3.3 L.

w = -pex ΔV (against constant pressure) w = -200 Torr * 3.3 L = -660 Torr-L

now convert Torr-L to J

w=-660Torr-L*(1atm/760Torr)*(8.314JK-1mol-1/0.08206L-atmK-1mol-1) w = -87.985 J

or:

w = -88 J

(2) Calculate w if expansion had been done reversibly:

w = − pdV pex= p( )

V1

V2

∫ p = nRT / V

w = −nRT dV / V∫= −nRTlnV

2V1

= − 4.50g 16.0 gmol( ) × 8.314JK−1mol−1 ×310Kln16.0L /12.7L

w = −167.44J

or:

w = -167 J C. Heat and Enthalpy.

1. When a system is heated under constant volume conditions (so no work of expansion can happen) and no other kind of work is being performed (e.g. electrical), then all heat transferred goes directly into increasing the internal energy U of the system.

dU = dqv

or: ΔU = qv

= ∫dqv

These are conditions of the “bomb” calorimeter.

8

2. Constant-Volume calorimetry (bomb calorimetry).

Typically used to measure combustion reactions. Reactants ignited in central chamber surrounded by a water jacket which absorbs the heat energy produced. ΔT result is ∝ heat generated.

q = CΔT q - can’t monitor it directly C - constant of proportionality

called the “heat capacity” of the calorimeter

T - can monitor it directly This q is qv, and

-ΔUrxn = ΔUcalorimeter = qv Find C of the calorimeter by doing an electrical heating

experiment.

Also could find C by reacting standard substance with known

heat produced in our calorimeter:

qsample/qstandard = ΔTsample/ΔTstandard

9

3. Enthalpy.

When heating a system that can expand against its surroundings, not all the heat goes into raising U, since the system converts some of that heat into work.

Definition:

The heat qp absorbed in a constant P process equals the enthalpy change.

where enthalpy is defined as:

H = U + pV ↑system p Let’s show this is true.

dH = dU + d(pV)

dH = dU + pdV + Vdp ↑ dqp + dw ↑ - pexdV (no other kinds of work) dH = dqp - pexdV + pdV + Vdp

10

Since system p = pex as it expands: dH = dqp - pdV + pdV + Vdp Further, constant p implies dp = 0, so: dH = dqp (at constant p, no other kinds of work) or write:

dH = dqp ↑signifies process done at constant p or in integrated form: ΔH = ∫dqp Question: System = open beaker of H2O in the lab. If 100J of energy is supplied by electrical heater, compute the ΔH of heating.

ΔH = +100J (ΔH =qp)

Such constant p processes are the most common in the laboratory. No rigid-walled containers required, as bomb calorimeter.

Actually, in this example, only a tiny expansion occurred so ΔH≈ΔU.

4. Measurement of ΔH =

Design calorimeter to monitor ΔT accompanying a process occurring at constant pressure.

11

In other words, the solution calorimeter:

time Now do heating experiment electrically to find C of calorimeter,

as before.

Heat flow into calorimeter from reaction or heating qcal = CΔT. ΔHcalorimeter = qcal ΔHcalorimeter =-ΔHrxn ΔHrxn = - qcal so if ΔT > 0 we have exothermic rxn ΔHrxn < 0

ΔHrxn =enthalpy change of the reactive substances converting to products We typically put this on a per mole basis, so the last step of many

problems is to divide by n, so ΔHrxn = - qcal /n

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5. Relation between ΔH and ΔU.

E.g. If ΔU of reaction is determined in bomb calorimeter, but we want ΔH instead:

H = U + pV

ΔH = ΔU + Δ(pV) ↑crucial term

Δ(pV) small for solids and liquids so ΔH ≈ ΔU for reactions in which no gases are involved If gases are involved, let’s assume they are behaving ideally. Δ(pV) =(Δng)RT ↑ ↑constant crucial term Δng =change in moles of gas going from reactant to product

ΔH = ΔU + Δng RT CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) n=3 n=3 Δng = 0 ΔH ≈ ΔU If under different conditions, H2O(l) produced Δng = 1-3 = -2 ΔH = ΔU - 2RT

Exercise: Calculate ΔU combustion 1 mole propene at 298.15K if ΔH=-2058 kJ:

C3H6(g) + 9/2O2(g) → 3CO2(g) + 3H2O(l) Δn = 3 - 5.5 = -2.5 ↑Note liquid product ΔU = ΔH - Δng RT = -2058kJ + (2.5 mol * 8.314 x 10-3kJ/mol K * 298.15 K) ↑+6.2kJ

Answer: ΔU = -2052kJ

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D. Work of Adiabatic Expansion.

1. Already dealt with isothermal expansions. 2. Adiabatic: dq = 0, no heat flow at any stage of expansion.

3. In adiabatic process dU = dw. 4. A useful rule in thermo ! when calculating value of a path-

dependent variable, (w or q), find the state function related to it and calculate change in that function by any convenient pathway.

5. Here, since dw = dU (since dq = 0):

€

w = dUi

f

∫

i.e. we can’t integrate dw easily but we can integrate dU, a state

function. Now dU = CVdT (for constant V process) and

€

dU = πTdV + CVdT in general ↑= 0 for ideal gas So, for adiabatic expansion of an ideal gas:

w = dUT1

T2∫ = CVdT

T1

T2∫

and if over a small T-range such that CV is T-independent over range

of T)

w = CVdT

T1

T2∫ ≈ CVΔT

6. From here on we will assume ideal gas. 7. Therefore, if we can find ΔT, we’ve got w.

14

E. Special cases which simplify calculation.

1. Irreversible adiabatic expansion.

Irreversible is when system pressure and surrounding pressure are not equal. Adiabatic means q = 0. Trivial irreversible case: expansion against pex = 0. Then w = 0, so ΔT = 0. (this case is both adiabatic (q=0) & isothermal (ΔT=0)) Expansion adiabatically against pex constant: w = -pexΔV as before. Now setting -pexΔV = CVΔT

ΔT = -pexΔV/CV = T-change of system expansion when q=0. This is saying that when no heat is allowed to flow into system and it expands, doing work, temperature will go down. V↑ T↓

2. Reversible adiabatic expansion.

Reversible means p = pex always. Since dw always = -pexdV. Here, then dw = -pdV also. Therefore: CVdT = -pdV CVdT = -(nRT/V) dV CVdT/T = - nRdV/V Integrate both sides, assuming CV ~ constant over T range again.

€

CV dT /TT1

T2

∫ = −nR dV /VV1

V2

∫

15

CVln

T2

T1

= −nRlnV

2

V1

CV

/ nR( ) ln T2

T1

= − lnV

2

V1

lnT

2

T1

"

#$$

%

&''

CV /nR( )= + ln

V1

V2

T2

T1

"

#$$

%

&''

CV /nR( )=

V1

V2

Let c = CV

/ nR

V2T

2c = V

1T

1c

T2=

V1

V2

"

#$$

%

&''

1/c

T1

So now that we have final temperature, we can get ΔT, and then

€

w = CVΔT

w = CVT1V1

V2

#

$ %

&

' (

1/c

−1*

+ , ,

-

. / /

Assumptions: reversible expansion p = pex adiabatic q = 0 perfect gas pV = nRT CV is T-independent small ΔT range

16

3. Isotherms versus Adiabats for ideal gas.

As gas expands doing work, and if adiabatic (q = 0), it will lose

energy and T will go down. It will move between isotherms T1 and T2. Assume reversible adiabatic expansion (& ideal gas):

€

p1V1 = nRT1

p2V2 = nRT2

p1V1

p2V2

=T1

T2

"

# $

%

& ' =

V2

V1

"

# $

%

& '

1/c

p1V1V11/c = p2V2V2

1/c

p1V11+1/c = p2V2

1+1/c

But :

1+1/c =c +1

c=

CV /nR +1CV /nR

=CV + nR

CV

=CP

CV

= γ

so :p1V1

γ = p2V2γ

or :pVγ = constant for an adiabat

whereas:

pV=constant for an isotherm.

Since γ > 1, pVγ (p goes down faster as V↑ along an adiabat).

In isothermal reversible expansion from V1 to V2 more work is done than in adiabatic reversible expansion, since in isothermal, heat can flow into system to maintain T constant. This replenishes internal energy lost by system as it does work. Remember w ∝ area under p-V curve.

17

F. Summary.

1. dw = -pexdV always! So:

€

w = − pex

1

2

∫ dV

2. Reversible means system pressure p = pex throughout expansion or

contraction process. Substitute p for pex in work integration. 3. Irreversible pex ≠ p. For expansion: p > pex.

Can’t substitute p for pex. w is less for irreversible than reversible expansion.

4. Isothermal means T of system is maintained constant because system is connected to surrounding bath.

5. U of perfect gas is dependent only on T, therefore in isothermal

expansion (T constant) dU = 0 and dq = -dw. 6. Adiabatic means q = 0 for process, therefore dU = dw.

Summarizing table: (*) indicates may not be exact unless gas is ideal gas. (**) indicates not exact unless over small T range Type of work w q ΔU ΔT Exp. against pex=0 isothermal 0 *0 *0 0 adiabatic 0 0 0 *0 Exp. against constant pex isothermal -pexΔV *+pexΔV *0 0 adiabatic -pexΔV 0 -pexΔV -pexΔV/CV Rev. expansion (p=pex) isothermal *-nRT ln V2/V1 *nRT ln V2/V1 0 0 adiabatic **CVΔT 0 **CVΔT *T1{(V1/V2)1/c-1} etc.

18

III. Thermochemistry - thermodynamics of chemical reactions.

A. Standard enthalpy changes.

1. Standard enthalpy change = ΔHo = change in enthalpy for a process in which the initial and final substances are in their standard states.

2. Standard state (recent definition):

= substance in its pure form at 1 bar pressure (no temperature is part of this definition.)

T must be supplied as separate info, though usually it is at 25°C, 298.15K, called Tc (conventional temp).

Example: Standard enthalpy of vaporization ΔHo

vap: = ΔH per mole when pure liquid at 1 bar → pure gas at 1 bar

ΔHovap(H2O, 373K) = +40.66 kJmol-1

H2O(l) → H2O(g)

Consider rxn previously at Tc (25°C):

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ↑pure H2O at 25°C at standard p = 1 bar is liquid!

Standard reaction enthalpy ΔHo = enthalpy change when 1 mole pure CH4(g) combines completely with 2 mole pure O2(g) to produce 1 mole pure CO2(g) and 2 mole pure liquid H2O, all at 1 bar.

All assumed unmixed (no heat of mixing involved).

(unmixed reactants) → (unmixed products)

3. Enthalpies of physical changes:

fusion (s → l) ΔHfuso

sublimation (s → g) ΔHsubo

solution (g, l, s → aq) ΔHsolo

when substance dissolves in specified amount of solvent, since heat of dilution may be involved.

Forming infinitely dilute solution is the limiting enthalpy of solution. vaporization (l → g) ΔHvap

o

19

4. Enthalpies of ionization:

ΔHio = for removal of an electron from gas phase species.

A(g) → A+(g) + e-(g) Relationship to “ionization energy” Ei: Ei = internal energy change for same process at T = 0 K (quantum mechanical or spectroscopic quantity)

Ei ≈ ΔUio (room temperature)

ΔHio = ΔUi

o+ RT ↑small at room temperature compared to ioniz

energies

so: ΔHio ≈ ΔUi

o ≈ Ei

5. Bond formation and dissociation:

ΔHo(A-B) = standard bond dissoc. enthalpy Process: A-B(g) → A(g) + B(g) where A,B may be atoms or groups of atoms. Example: Interested in energy of O-H bonds. Problem: HOH → HO + H ΔHo = 499kJmol-1 HO → H + O ΔHo = 428kJmol-1 CH3OH → CH3O + H ΔHo = 368kJmol-1 Must take some sort of average for a given bond type, the mean bond enthalpy. Tables of these mean bond enthalpies are developed to help estimate ΔH of rxns.

6. Atomization enthalpy ΔHao.

= standard enthalpy change accompanying the separation of all

atoms of a substance. Example: H2O ΔHa

o = 499 + 428 = +927kJ/mol 1st H 2nd H

20

7. Enthalpies of chemical reactions.

Tables exist for enthalpies of combustion, hydrogenation and other common reaction types. ΔHc

o, standard enthalpy of combustion is for combination of one mole of organic compound with O2 to produce CO2 and H2O. ΔHo of hydrogenation is for reaction of one mole of unsaturated organic compounds with H2(g).

8. Hess’ Law = if a process can be envisioned as sum of steps, overall ΔH is the sum of ΔH of the steps.

• Consequence of H being state fct, and the First Law. • Individual steps may be hypothetical ones. • Allows us to use tables of data to determine ΔH for rxn never

before measured. (Enthalpies of formation are the most useful; will discuss next.)

• Series of steps constructed is called “thermodynamic cycle.”

Problem: Calculate the standard enthalpy of the partial oxidation of methane to methanol given the ΔHc

o of CH4(g) is -890 kJmol-1 and CH3OH(l) is -726 kJmol-1 at Tc.

Rxn of interest: (1) CH4 + ½ O2 → CH3OH ΔH1

o = ? (2) CH4 + 2O2 → CO2 + 2H2O ΔH2

o = -890kJ/mol (3) CH3OH + 3/2 O2 → CO2 + 2H2O ΔH3

o = -726kJ/mol Rx(1) = Rx(2) - Rx(3) so: ΔH1

o = ΔH2o - ΔH3

o = -890 - (-726)kJ/mol = -164kJ/mol

B. Enthalpies of formation.

1. Symbol ΔHfo = standard reaction enthalpy to form a substance from

its elements in their reference states. 2. Reference state of an element = its most stable state at the specified

T and 1 bar pressure. Example: at T = 298K; H2(g), O2(g), N2(g), C(s,gr)=carbon graphite, not diamond.

Elements must be in their most stable allotrope.

21

3. Tables of these formation reaction enthalpies become extremely useful when using Hess’ Law.

Problem: How much more energy can be liberated if 1 mol of n-octane were completely combusted to CO2 instead of CO at Tc. Use Table 2C.3 and 2C.4 in back of Atkins.

(1) C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(l) (2) C8H18(l) + 17/2 O2(g) → 8CO(g) + 9H2O(l) Let x = ΔH1

o - ΔH2o ‚not in table, use formation rxns to find

↑ΔHco of octane = -5471kJ/mol

(2) C8H18(l) + 17/2 O2(g) → 8CO(g) + 9H2O(l) Can show that: ΔH2

o = {8ΔHfo (CO(g)) + 9ΔHf

o(H2O(l))} - {ΔHf

o (C8H18) + 17/2 ΔHfo(O2(g))}

Now look these up in tables of ΔHf

o.

4. ΔHrxo in terms of ΔHf

o, a general expression:

€

ΔHrxo = νi

i

products

∑ ΔHfo − νi

i

reactants

∑ ΔHfo

ν = stoichiometric coeff. in balanced “rx” eqn. Proof in above problem. (3) 8C(s,gr) + 9H2(g) → C8H18(l) ΔHf

o(octane) (4) 17/2 O2(g) → 17/2 O2(g) (17/2)ΔHf

o(O2(g)) = 0 (5) 8C(s,gr) + 4O2(g) → 8CO(g) 8*ΔHf

o(CO(g)) (6) 9H2(g) + 9/2O2(g) → 9H20(l) 9*ΔHf

o(H2O(l)) Rx(2) = Rx(5) + Rx(6) - Rx(3) - Rx(4) All elements in their reference stable state cancel out!! ΔH2

o ={8ΔHfo (CO) + 9ΔHf

o (H2O(l))} - {ΔHfo (oct) + 0}

↑O2(g)

!"#$%#&%'(( )*+,-$%'(

./"0"&%'(

!12*3(4(5(

6"#!127(8"#!127(

22

IV. Variation of Enthalpy with Temperature.

A. The Heat Capacity of Substances.

1. General - heat capacity is relationship between heat flow (energy) and temperature change.

q ∝ ΔT

Proportionality constant is Heat Capacity. Units are (energy/degrees). Depends on conditions. Substances have higher heat capacities at high T than at low T. Depends also on whether at constant V or p.

2. Constant volume heat capacity.

dqV = CvdT

heat flow = (constant-vol. heat cap) (temperature change) into system at const V

Since dqv = dU dU = Cv dT

(constant V)

Integration:

ΔU = CvT1

T2

∫ dT

Cv in general is T-dependent, so stays inside integral

Over short ranges of T, Cv may not vary much, so ΔU ≈ Cv ΔT

Can also write:

Cv = (∂U/∂T)v

= differential change in U as T changes with V fixed.

What is Cv of monatomic gas? It’s internal energy is given by

U = (3/2)nRT

Cv=

∂U∂T

"

#$$

%

&''V

=∂(3 / 2)nRT

∂T

"

#$$

%

&''V

= (3 / 2)nR ∂T∂T

"

#$$

%

&''V

Cv= (3 / 2)nR

Constant vol. molar heat capacity then = (3/2)R

23

3. Constant-pressure heat capacity.

dqp = Cp dT heat flow = (constant-pressure h.c.) (temperature change) into system at const p dH = dqp, so

dH = CpdT (const p) Cp = (∂H/∂T)p

4. Relation between Cp and Cv.

When a system is able to expand against surroundings as it is heated (as in const p), the temp will rise more slowly as heat is added, since some heat is being converted into work on the surroundings. Implies in general Cp > Cv. Relation is simple for ideal gases: H = U + pV dH = dU + d(pV) Cp dT = Cv dT + d(nRT) heating at constant n, d(nRT) = nRdT Cp = Cv + nR

Cp - Cv = nR (ideal gas) Remember heat cap ratio expt? γ = Cp / Cv = (Cv + R)/ Cv (n =1)

for He: γ =32R +R32R

=52R32R=53= 1.667

24

B. T-dependence of ΔHrx.

1. Heat capacities may be used to correct ΔHrx(T1) to ΔHrx(T2).

2. Example:

3. Kirchoff’s Law:

ΔHrx(T2) = ΔH

rx(T1)+ ΔC

pT1

T2

∫ dT

Where:

ΔCp= ν

ii

products

∑ Cp(i) − ν

ii

reactants

∑ Cp(i)

ν = stoich coeff Cp are molar heat capacities

4. If T1 and T2 are not too far apart, Cp’s may be fairly constant over T range. Then,

ΔHrx(T2) = ΔH

rx(T1)+ ΔC

pΔT

Problem: If ΔHfo of H2O(g) at 25°C = -241.8kJmol-1, what is ΔHf

o at 50°C. Use equipartition theorem to estimate heat capacities (w/o vibrational contribution).

ΔT = 25° (work with 1 mole)

H2(g) + (1/2)O2(g) ! H2O(g)

Cv(H2O) = 3/2R + 3/2R = 3R (1 mole) trans. rot. Cp(H2O) = 3R + R = 4R (1 mole)

Similarly for H2: Cp(H2(g)) = 3/2R + 2/2R + R = 3.5R trans. rot.

Cp(O2(g)) = 3/2R + 2/2R + R = 3.5R

ΔCp = 4R - {3.5R + 3.5/2R} H2O H2 (1/2)O2

= -1.25R = -1.25 x 8.314Jmol-1K-1 x 10-3kJ/J

= -0.0104kJmol-1K-1

ΔHrx(50°C) = ΔHrx (25°C) + (-0.0104kJ/mol)(50° - 25°) = -241.8kJmol-1 - 0.26kJmol-1 = -242.1kJmol-1 Very slight change.

25

5. If T1 and T2 are widely different, must evaluate integral:

ΔCp

T1

T2

∫ dT

since ΔCp will vary with T from T1 to T2.

6. Standard empirical expression for Cp (T):

Cp (T) = a + bT + c/T2 a, b, c are constants which are substance-dependent. Problem: Calculate ΔU in kJ when 0.400 mol of a certain liquid is

vaporized at 240K and 740 Torr. ΔHvap = 34.0 kJ/mol at that T.

26

V. State functions and differentials.

A. State functions:

1. State functions = Properties that are independent of how system state was achieved: H, U, T, V, p.

2. Path or process functions = properties depending upon the path: q,

w. 3. Exact differential - can be integrated as follows:

€

dU = Uf

i

f

∫ −Ui = ΔU

4. Inexact - cannot:

€

dq≠ qf

i

f

∫ −qi (meaningless)

B. Important differential expressions involving state functions.

1. Alternate differential expressions of U.

U can be viewed as function of V, T. U = U (V,T) Thus responds to both changes in V and T.

€

dU =∂U∂V#

$ %

&

' (

T

dV +∂U∂T#

$ %

&

' (

V

dT

Total change in U

= how U changes when V changes + how U changes when T changes

€

dU = πTdV + CVdT

€

πT ≡∂U∂V%

& '

(

) *

T = “internal pressure”

= coefficient of response of U to volume change.

27

2. Behavior of πT.

Observe for ideal gas. U = function of T only, independent of V

€

πT ≡∂U∂V%

& '

(

) *

T

= 0 (this is one definition of ideal gas)

Helium: U = (3/2) RT (πT = 0) VdW gas

πT = a/

€

Vm2

Apparatus of the Joule Experiment:

To measure change in U in expansion of a gas.

3. The thermal expansion coefficient α.

α ≡1V

∂V∂T

$

%&&

'

())p

Expresses response of V of system to T change at constant p, relative to absolute volume of system.

i.e. α = rate of change of volume with T, per unit volume. Example: perfect gas. V = nRT/p

α ≡1V

∂(nRT / p)∂T

$

%&&

'

())p

=nRpV

=1T

28

4. Expressions of H: View H as function of p and T. H = H(p,T)

dH = ∂H∂p

"

#$$

%

&''T

dp + ∂H∂T

"

#$$

%

&''p

dT

↑ ↑ µT Cp

5. The isothermal compressibility κT.

κT≡ −1V

∂V∂p

%

&''

(

)**T

Meaning: how volume of system responds to pressure change at constant T, per unit volume. (-) makes κT. positive, since V↓ as p↑. For example: perfect gas V = nRT/p

κT≡ −1V

∂(nRT / p)∂p

%

&''

(

)**T

= −nRTV

∂(1 / p)∂p

%

&''

(

)**T

= −p / p2

= 1 / p

6. The Joule-Thomson coefficient µ.

µ ≡∂T∂p

#

$%%

&

'((H

Can be measured in J-T experiment, which holds H constant (isenthalpic). Based on observation of cooling of a real gas when it expands adiabatically. (q=0)

29

7. Now, we can show how H depends on T at constant V.

∂H∂T

"

#$$

%

&''V

= 1− αµ / κT( )Cp

Note: (∂H/∂T)p = Cp, but how H varies with T at constant volume is more complicated.

C. General relation between Cp and CV.

1. Remember for ideal gas:

Cp - CV = nR

2. General case, can be shown that, for any substance:

Cp - CV = α(p + πT)V or Cp - CV = α2TV/κT