Chap. 2: Thermodynamics

2.2: P-V Work

2.3: Heat

2.4: The 1st Law of Thermodynamics

2.5: Enthalpy

2.6: Heat Capacities

2.7: The Joule and Joule-Thomson

Experiments

2.8: Perfect Gases and the 1st Law

2.9: Calculation of 1st Law Quantities

2.2: P-V Work

1. Work:

2. P-V Work:

xFW

xdFdW

A

Expansion

dx

F = PA

PdVxdAPxdAPxdFdWsys

xdAdV

For a reversible process in a closed system:

A reversible process is one where the system is

always infinitesimally close to equilibrium,

and an infinitesimal change in conditions can

reverse the process to restore both system and

surroundings to their initial states.

2

1 )(

)(

dVPW

dVPdW

Vrev

VrevEqu-2.26*

For an isobaric process, P is a constant, then:

3. Line Integrals:

See Fig-2.3 on p. 44.

The P-V work is path dependent and it is not a

state function. Be careful of this!

VPW

dVPPdVW

rev

rev

2

1

2

1

Exp-2.2 on p. 44: P-V work

Find the work Wrev for processes (a) and (b) of Fig-2.3

if: P1 = 3.00 atm, V1 = 500 cm3,

P2 = 1.00 atm, V2 = 2000 cm3.

Also, find Wrev for the reverse of process (a).

1) Wrev for the process (a):

This process can be splitted into two sub-processes:

[a,1] from the initial point, 1, to the turning point.

[a,2] from the turning point to the ending point, 2.

Note: [a,1] is an isochoric process and [a,2] is an

isobaric process.

Because and dV = 0 for [a,1], we

obtain: Wrev,[a,1] ≡ 0. However, Wrev,[a,2] ≠ 0 in that

[a,2] is an isobaric process. For this change:

Wrev,[a] = Wrev,[a,1] + Wrev,[a,2] = 0 + (-152J) = -152 J

Similarly, for the reverse process (a): Vinitial = V2, Vfinal =

V1, and Pinitial = P2, and Pfinal = P1. Use these relations we

can get Wrev,[-a] = +152 J for the reverse process (a).

2

1 )( dVPW Vrev

J152

10.50010.2000Pa101325

2,,

3636

2,,

122,,

arev

arev

arev

W

mmW

VVPVPW

2) Wrev for the process (b):

Process (b) is very similar to process (a) except

that:

[b,1] is an isobaric expansion process, and

[b,2] is an isochoric process.

Thus,

Wrev,[b] = Wrev,[b,1] + Wrev,[b,2]

Wrev,[b] = -P1(V2 – V1) + 0

Wrev,[b] = -(3 101325 Pa)(2000. – 500.) 10-6 m3

Wrev,[b] = -456J

4. Irreversible P-V work:

The work W in a mechanically irreversible volume change

sometimes cannot be calculated with thermodynamics.

According to the Law of The Conservation of Energy:

It is very hard to figure out the change of the kinetic

energy of the piston, ΔKpiston, during its acceleration.

However, if we wait longer enough, Δkpiston → 0. We

then have:

pistonexternalirrev dKdVPdw

2

1dVPW externalrev

2.3: Heat

Heat is a means of energy transfer due to the

temperature difference between two objects.

Heat Capacity:

Specific Heat Capacity:

Molar Heat Capacity:

Calorimetric Method: dqin = dqout

dq = mcdT

Calorie (cal): The amount of heat needed to

raise one gram of water from 14.5 ºC to 15.5 ºC

at 1 atm pressure.

dT

dqC

mdT

dqc

ndT

dqc~

2

1

T

TP dTTcmq

2.4: The 1st Law of Thermodynamics

E = K + V + U

E: The energy of a body

K: Kinetic energy

V: Potential energy

U: Internal energy

If focus on thermodynamics: let K = 0, V =0,

then, E = U

Molar Internal Energy: Um = U/n

The 1st Law of Thermodynamics: ΔU = q + w

for closed system, at rest, no fields.

For an infinitesimal process:

dU = dq + dw (for closed systems) Equ-2.40*

ΔU ≡ 0 for a cyclic process.

Example 2.3 on p. 50: Calculation of ΔU

Calculate ΔU when 1.00 mol of H2O goes from 25.0 °C

and 1.00 atm to 30.0 °C and 1.00 atm.

Note: U is a state function. You can always choose your

favorite path to do the calculation.

ΔU = q + w, figure out q and w separately and then

calculate the ΔU.

2.5: Enthalpy

Enthalpy, H ≡ U + PV Equ-2.45*

Why we define enthalpy, H?

Expansion

Fixed V but

P increases

Fixed P but

V increases

q q

U H

Note: ΔH = qP , ΔU = qV (Equ-2.46*/2.49*, p. 52)

At fixed pressure:

ΔH = Δ(U + PV) = ΔU + Δ(PV) = ΔU + PΔV

For solids and liquids, ΔV is very small when

temperature changes (incompressible), thus:

ΔH ≈ ΔU (for solids and liquids)

For gases, ΔV changes a lot when temperature

changes, the PΔV term cannot be overlooked,

thus: ΔH ≠ ΔU (for gases)

2.6: Heat Capacities

For an infinitesimal process:

For a closed system, we can define two types of

capacities:

Heat capacity at constant pressure:

Heat capacity at constant volume:

dT

dqC

pr

pr

P

PP

T

H

dT

Hd

dT

dqC

V

VV

T

U

dT

Ud

dT

dqC

CP(T, P) and CV(T,V) ↔ H(T,P) and U(T,V)

Molar Heat Capacity:

The relations between CP and CV

VV

mVPP

mP Cn

CCC

n

CC

~

~..

VPP

VP

VP

VP

VP

VP

T

U

T

VP

T

UCC

T

U

T

PVUCC

T

U

T

HCC

How do we figure out ?

Thus

VP T

U

T

U

PTVP

P

P

P

TVP

P

P

T

P

V

P

TV

T

V

V

U

T

U

T

U

dT

dV

V

U

T

U

dT

dU

dVV

UdT

T

UdU

dVV

UdT

T

UdU

:P pressureconstant At the

PT

VPT

VP

V

UCC

2.7: The Joule & Joule-Thomson Experiment

1. Joule Experiment (1843): Try to find

See Fig-2.6 on p. 55.

Actually, the Joule Experiment measures the

T change with change in V at constant U.

From the experiment, we can measure the

Joule Coefficient: Equ-2.62

How could we obtain from µJ?

Apply Euler’s Chain Rule to U(T,V)!

TV

U

U

JV

T

TV

U

Because

Thus

U

TV

1TUV U

V

V

T

T

U

JV

UVT

CV

T

T

U

V

U

2. Joule-Thomson Experiment (1853):

See Fig-2.7 on p. 56.

The Joule-Thomson Experiment measures

the T change with change in P at constant H.

From the experiment, we can measure the

Joule-Thomson Coefficient:

Equ-2.64*

How could we obtain from µJ?

Apply Euler’s Chain Rule to H(T,P)!

TP

H

H

JTP

T

TP

H

Because

Thus

H

TP

1THP H

P

P

T

T

H

JTP

HPT

CP

T

T

H

P

H

2.8: Perfect Gases and The 1st Law

1. Perfect Gas: see Eq-2.66*, Eq-2.69*, Eq-2.70*

This implies: U = U(T) → U only depends on

T for perfect gases.

Since , at constant V, this equation

becomes an ordinary derivative,

Thus, . Similarly, .

0 and TV

UnRTPV

V

VT

UC

dT

dUCV

dTCdU vdTCdH P

Combine Eq-2.61* with Eq-2.66*:

Since ,

Therefore, for perfect gases: CP,m – CV,m = R

The 1st Law for perfect gases:

dU = CVdT = dq + dw = dq - pdV

nRTPV

T

VP

V

UCC

PT

VP

0 TV

UnR

T

VPCC

P

VP

0 ,0 0 ,0 0; ,0 JTJP

T

V

T

CP

HC

V

U

Example 2.4, p. 59: Calculation of q, w and ΔU

Suppose 0.100 mol if a perfect gas having CV,m =

1.50R independent of temperature undergoes the

reversible cyclic process 1 → 2 → 3 → 4 → 1

shown in Fig-2.10, where either P or V is held

constant in each step. Calculate q, w, and ΔU for

each step and for the complete cycle.

Solution: see p. 59

2. Reversible Isothermal Process in a Perfect Gas

Isothermal → dU = 0, thus dq = -dw

Because dw = -pdV, dq = pdV

Ideal gas: PV = nRT, thus, P = nRT/V

2

1

1

2

1

2

2

1

12

lnln

lnln

lnlnln2

1

2

1

2

1

2

1

P

PnRT

V

VnRTWq

P

PnRT

V

VnRTW

VVnRTVnRTW

V

dVnRTdV

V

nRTPdVW

V

V

V

V

V

V

V

V

Example 2.5 on p. 60.

A cylinder fitted with a frictionless piston

contains 3.00 mol of He gas at P = 1.00 atm and

is in a large constant-T bath at 400 K. The

pressure is reversibly increased to 5.00 atm.

Find q, w, and ΔU for this process.

For an isothermal process, ΔU = 0, q = -w

3. Reversible Constant-P (or Constant-V)

Process in a Perfect Gas: See Example 2.4!

1

2

1

2 lnln2

1

2

1 P

PnRT

V

VnRTdV

V

nRTPdVw

T

T

V

V

4. Reversible Adiabatic Process in a Perfect Gas

For an adiabatic process: dq ≡ 0

For a reversible process with only P-V work:

dw = -PdV

For a perfect gas: dU = CVdT

Thus, the 1st law becomes: CVdT = -PdV

For an ideal gas: P = nRT/V = RT/(V/n) = RT/Vm

Then we get:

V

dVR

T

dTC

dVV

RTdTC

mv

mV

,

,

2211

,

,

,

,

,

,,

1

11

1

22

2

1

11

22

11

22

1

2

2

22

1

11

2

1

1

2

2

1

1

2

1

2,

,

:get we, Define

1 thus,, :Note

Hence,

gas,perfect aFor

lnlnln

11

,,

,

,

,

2

1

2

1

VPVPC

C

C

C

C

RRCC

VPVP

V

V

VP

VP

VP

VP

T

T

T

VP

T

VP

V

V

T

T

V

VR

V

VR

T

TC

dVV

RdTT

C

mV

mP

mV

mP

mV

mVmP

CR

CR

CR

mV

T

T

V

VmV

mVCR

mVCR

mV

mV

mV

For an adiabatic process, ΔU = q + w = w. Since

ΔU = CVΔT and CV is a constant for a perfect gas

ΔU = CV(T2 – T1) = w

This formula is applied to a perfect gas,

reversible adiabatic process, and a constant CV.

See Fig-2.12 on p. 61.

2.9: Calculation of 1st Law Quantities

Thermodynamic Processes

Cyclic: ΔT = ΔP = ΔV = ΔU = ΔH = 0

Reversible: dwrev = -PdV infinity small process

Isothermal: T is constant throughout the

process

Adiabatic: dq ≡ 0, q ≡ 0

Constant-V: dV ≡ 0, ΔV ≡ 0, ΔU = qV

(Isochoric Change)

Constant-P: dP ≡ 0, ΔP ≡ 0, ΔH = qP

(Isobaric Change)

The q, w, ΔU, ΔH for various process:

1. Reversible phase change at const. T and P

A Phase change or phase transition

2. Const.-P heating with no phase change

3. Const.-V heating with no phase change

2

1

const.at

2

1

T

TPP

rev

PdTTCqH

VPPdVww

PVUPVUPVUH

VdTTCqUT

TVV

2

1

const.at

4. Perfect gas change of state

5. Reversible isothermal process in a perfect gas

6. Reversible adiabatic process in a perfect gas

7. Adiabatic Expansion of a perfect gas into vacuum

q = 0 (adiabatic process)

w = 0 (w = -PdV, P = 0 for vacuum surroundings)

ΔU = q + w = 0, ΔH = ΔU + Δ(PV) = ΔU + nRΔT = 0

2

1

2

1

T

TP

T

TV dTTCHdTTCU

1

22

11

2 ln ln

0,0 thus,0or 0 implies 0

V

VnRTwq

V

VnRTPdVw

HUqqq PV

2211 implies 0 VPVPwUq

Example 2.6: Calculation of ΔH

CP, m of a certain substance in the temperature

range 250 K to 500 K at 1 bar pressure is given

by CP, m = b + kT, where b and k are certain

known constants. If n moles of this substance is

heated from T1 to T2 at 1 bar (where T1 and T2

are in the range 250 K to 500 K), find the

expression for ΔH.

2

1

2

212

2

21

,

2

1

2

1

2

1

2

1

TTkTTbnH

kTbTndTkTbndTnCHT

T

T

T

T

TmP