Lecture 6. Entropy of an Ideal Gas (Ch. 3)

Find Ω (U,V,N,...) – the most challenging step

S (U,V,N,...) = kB ln Ω (U,V,N,...)

Solve for U = f (T,V,N,...)

1,...),,(

−

⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂≡

UNVUST

Today we will achieve an important goal: we’ll derive the equation(s) of state for an ideal gas from the principles of statistical mechanics. We will follow the path outlined in the previous lecture:

So far we have treated quantum systems whose states in the configuration (phase) space may be enumerated. When dealing with classical systems with translational degrees of freedom, we need to learn how to calculate the multiplicity.

Multiplicity for a Single particle- is more complicated than that for an Einstein solid, because it depends on threerather than two macro parameters (e.g., N, U, V).

Example: particle in a one-dimensional “box”

-L L

-L L x

px

Δpx-px Δx

The total number of microstates: x

LpxpL

x

x

Δ=

Δ⋅ΔΔ

∝

hx

x

x pLpxpL Δ

=Δ⋅ΔΔ

∝The number of microstates:

Q.M.

h≥Δ⋅Δ xpx

Quantum mechanics (the uncertainty principle) helps us to numerate all different states in the configuration (phase) space:

pspaceΩΩ=ΩThe total number of ways of filling up the cells in phase space is the product of the number of ways the “space” cells can be filled times the number of ways the “momentum” cells can be filled

Multiplicity of a Monatomic Ideal Gas (simplified)For a molecule in a three-dimensional box: the state of the molecule is a point in the 6D space - its position (x,y,z) and its momentum (px,py,pz). The number of “space” microstates is:

There is some momentum distribution of molecules in an ideal gas (Maxwell), with a long “tail” that goes all the way up to p = (2mU)1/2 (U is the total energy of the gas). However, the momentum vector of an “average” molecule is confined within a sphere of radius p ~ (2mU/N)1/2 (U/N is the average energy per molecule). Thus, for a single “average” molecule:

3xV

zyxV

space Δ=

Δ⋅Δ⋅Δ=Ω

zyxp ppp

p

Δ⋅Δ⋅Δ∝Ω

3

34π

The total number of microstates for N molecules:

NN

xpspace

pVpxpV

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×=⎟⎟

⎠

⎞⎜⎜⎝

⎛

Δ⋅Δ×

≈ΩΩ=Ω 33

33

3

h

For N molecules: N

space xV

⎟⎠⎞

⎜⎝⎛Δ

=Ω 3

p

n

p

However, we have over-counted the multiplicity, because we have assumed that the atoms are distinguishable. For indistinguishable quantum particles, the result should be divided by N! (the number of ways of arranging N identical atoms in a given set of “boxes”):

N

ishableindistingupV

N ⎟⎟⎠

⎞⎜⎜⎝

⎛ ×≈Ω 3

3

!1

h

More Accurate Calculation of ΩpMomentum constraints:

mUppp zyx 2222 =++

mUpppppp zyxzyx 222

22

22

21

21

21 =+++++

1 particle -

2 particles -

The accessible momentum volume for N particles = the “area” of a 3N-dimensional hyper-sphere ×Δp

132/3

!12

32area"" −

⎟⎠⎞

⎜⎝⎛ −

= NN

rNπ

( ) pmU

NV

N

NN

N Δ⎟⎠⎞

⎜⎝⎛=Ω 22

!2/3!1 2/3

2h

π

Monatomic ideal gas:

(3N degrees of freedom)

N =1

( ) ( )[ ]22

2/3

42/!2/1!2/1

2 rr πππ ===

f N- the total # of “quadratic” degrees of freedom( ) 2/NfUU ∝Ω

The reason why m matters: for a given U, a molecule with a larger mass has a larger momentum, thus a larger “volume” accessible in the momentum space.

px

py

pz

( ) ( ) pNU

NV

hmeNVU

N

ishableindistingu Δ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=Ω 23,,

2/3

3

2/5 π

Entropy of an Ideal Gas

f ⇒ 3 (monatomic), 5 (diatomic), 6 (polyatomic)

Monatomic ideal gas: ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

25

34ln),,(

2/3

2 NU

hm

NVkNUVNS B

π

),(ln23ln

34ln

35

23ln 2

2/3

mNNUkN

NVkN

hmkN

NU

NVkN BBBB ϕ

π++=⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛++

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

In general, for a gas of polyatomic

molecules:

the Sackur-Tetrodeequation

),(ln2

ln),,( mNTkNfNVkNTVNS BB ϕ++=

( ) pmU

NV

N

NN

N Δ⎟⎠⎞

⎜⎝⎛=Ω 22

!2/3!1 2/3

2h

π

( )pkNNU

hm

NVkNUVNS BB Δ++

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛= 2ln

25

34ln),,(

2/3

2

π

an average volume per molecule

an average energy per molecule

Problem Two cylinders (V = 1 liter each) are connected by a valve. In one of the cylinders – Hydrogen (H2) at P = 105 Pa, T = 200C , in another one –Helium (He) at P = 3·105 Pa, T=1000C. Find the entropy change after mixing and equilibrating.

i

fB

i

fB T

TkNf

VV

kNTVNS ln2

ln),,( +=Δ

H2 :

( ) ( ) ( )ftotal TUTUTU =+ 2211

2

2

1

1

21

21

2211

35

35

23

25

23

25

TP

TP

PP

kNkN

TkNTkNT

BB

BB

f

+

+=

+

+=

1

ln252ln

2 TT

kNkNS fBBH +=Δ He :2

22 ln232ln

TT

kNkNS fBBHe +=Δ

( ) ⎥⎦

⎤⎢⎣

⎡+++=Δ+Δ=Δ

22

1121 ln3ln52

2ln2 T

TN

TT

NkkNNSSS ffBBHeHtotal J/K 67.0total =ΔS

The temperatureafter mixing: fBBBB TkNkNTkNTkN ⎟⎠

⎞⎜⎝⎛ +=+ 212211 2

325

23

25

For each gas:

Entropy of Mixing Consider two different ideal gases (N1, N2) kept in two separate volumes (V1,V2) at the same temperature. To calculate the increase of entropy in the mixing process, we can treat each gas as a separate system. In the mixing process, U/Nremains the same (T will be the same after mixing). The parameter that changes is V/N:

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

25

34ln),,(

2/3

2 NU

hm

NVkNUVNS B

π

The total entropy of the system is greater after mixing – thus, mixing is irreversible.

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

Δ+Δ=

Δ

22

11 lnln V

VNVVN

kSS

kS

B

BA

B

total

if N1=N2=1/2N , V1=V2=1/2V 2ln2/ln

22/ln

2N

VVN

VVN

kS

B

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

Δ total

Gibbs Paradox

If two mixing gases are of the same kind (indistinguishable molecules):

( )

VN

VN

VN

VN

NVN

VN

NVN

NVN

NVN

NNVVNNSSkS BAB

===⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛++

+=Δ+Δ=Δ

2

2

1

1

2

22

1

11

2

22

1

11

21

2121

0lnln

lnlnln/

if

total

( ) NNNN

N mUNhV

N32/3

3 2!

23

2!

1

⎟⎠⎞

⎜⎝⎛

=Ωπ

BB kNNU

hm

NVkNUVNS

25

34ln),,(

2/3

2 +⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

π

Quantum-mechanical indistinguishability is important! (even though this equation applies only in the low density limit, which is “classical” in the sense that the distinction between fermions and bosons disappear.

ΔStotal = 0 because U/N and V/N available for each molecule remain the same after mixing.

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

Δ+Δ=

Δ

22

11 lnln V

VNVVN

kSS

kS

B

BA

B

total

- applies only if two gases are different !

ProblemTwo identical perfect gases with the same pressure P and the same number of particles N, but with different temperatures T1 and T2, are confined in two vessels, of volume V1 and V2 , which are then connected. find the change in entropy after the system has reached equilibrium.

BBBBB kNTkhm

NVkNkN

NU

hm

NVkNUVNS

25

23

34ln

25

34ln),,(

2/3

2

2/3

2 +⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

ππ

( ) ( ) BfBf kNTNVVkNS 2

25

2ln2 2/321 +⎥⎦

⎤⎢⎣⎡ += α

( ) ( )( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ⎥⎦

⎤⎢⎣

⎡ +=⎥⎦

⎤⎢⎣⎡ +=+=⎥

⎦

⎤⎢⎣

⎡ ++⎥

⎦

⎤⎢⎣

⎡ +

=⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡ +=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧ +=

Δ

21

22121

2121

221

21

221

21

2

21

221

2/32

2/31

3

21

2221

4ln

25

22

4ln

23

4ln

ln23

4lnln

2ln

TTTTTTNkVVP

TTTT

VVVV

TTT

VVVV

TTT

VVN

NVV

kNS

B

ff

B ααα

at T1=T2, ΔS=0, as it should be (Gibbs paradox)

( ) ( ) BBBBi kNTNVkNkNT

NVkNSSS

25ln

25ln 2/322

2/31

121 +⎥⎦

⎤⎢⎣⎡++⎥⎦

⎤⎢⎣⎡=+= αα

221 TTTf

+= - prove it!

An Ideal Gas: from S(N,V,U) - to U(N,V,T)

( ) ( ) 2/,, NfNUVNfNVU =ΩIdeal gas:(fN degrees of freedom)

( ) ),(lnln2

,, mNNVNk

NUNkfNVUS BB ϕ++=

UNkf

US

TB

21

=∂∂

= ⇒ TkNfTVNU B2),,( =

The heat capacity for a monatomic ideal gas: BNV

V kNf

TUC

2,=⎟

⎠⎞

⎜⎝⎛∂∂

=

- in agreement with the equipartition theorem, the total energy should be ½kBT times the number of degrees of freedom.

- the “energy”equation of state

Partial Derivatives of the Entropy

We have been considering the entropy changes in the processes where two interacting systems exchanged the thermal energy but the volume and the number of particles in these systems were fixed. In general, however, we need more than just one parameter to specify a macrostate, e.g for an ideal gas

Today we will explore what happens if we let the other two parameters (V and N) vary, and analyze the physical meaning of the other two partial derivatives of the entropy:

NUVS

,⎟⎠⎞

⎜⎝⎛∂∂

VUNS

,⎟⎠⎞

⎜⎝⎛∂∂

We are familiar with the physical meaning only one partial derivative of entropy: TU

S

NV

1

,

=⎟⎠⎞

⎜⎝⎛∂∂

( ) ( )NVUkNVUSS B ,ln, ,, Ω==

When all macroscopic quantities S,V,N,U are allowed to vary:

NdNSVd

VSUd

USdS

VUUNVN ,,,⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=

Thermodynamic Identity for dU(S,V,N)

( )NVUSS ,,= ⇒ if monotonic as a function of U(“quadratic” degrees of freedom!), may be inverted to give

( )NVSUU ,,=

dNNUdV

VUdS

SUdU

VSNSNV ,,,⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=

compare withTU

S

VN

1

,

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

μ≡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−≡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

≡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

SVSNVN NUP

VUT

SU

,,,

pressure chemical potential

This holds for quasi-static processes (T, P, μ are well-define throughout the system).

NddVPSdTUd μ+−=

μ shows how much the system’s energychanges when one particle is added to the system at fixed S and V. The chemical potential units – J.

- the so-called thermodynamic identity for U

The Exact Differential of S(U,V,N)

NdT

dVTPUd

TdS μ−+= 1

The coefficients may be identified as:

( ) NdNSVd

VSUd

USNVUdS

VUUNVN ,,,

,, ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=

TNS

TP

VS

TUS

UVUNVN

μ−=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

,,,

1

Again, this holds for quasi-static processes (T and P are well defined).

NddVPSdTUd μ+−=

Type of interaction

Exchanged quantity

Governing variable

Formula

thermal energy temperature

mechanical volume pressure

diffusive particles chemical potential

NVUS

T ,

1⎟⎠⎞

⎜⎝⎛∂∂

=

NUVS

TP

,⎟⎠⎞

⎜⎝⎛∂∂

=

VUNS

T ,⎟⎠⎞

⎜⎝⎛∂∂

−=μ

connection between

thermodynamics and statistical

mechanics

Mechanical Equilibrium and Pressure

Let’s fix UA,NA and UB,NB , but allow V to vary (the membrane is insulating, impermeable for gas molecules, but its position is not fixed). Following the same logic, spontaneous “exchange of volume” between sub-systems will drive the system towards mechanical equilibrium (the membrane at rest). The equilibrium macropartition should have the largest (by far) multiplicity Ω (U, V) and entropy S (U, V).

UA, VA, NA UB, VB, NB

0=∂∂

−∂∂

=∂∂

+∂∂

=∂∂

B

B

A

A

A

B

A

A

A

AB

VS

VS

VS

VS

VS

( )BA VV ∂−=∂B

B

A

A

VS

VS

∂∂

=∂∂

VNk

TP

VS B

NU

==⎟⎠⎞

⎜⎝⎛∂∂

,

- the sub-system with a smaller volume-per-molecule (larger P at the same T) will have a larger ∂S/∂V, it will expand at the expense of the other sub-system.

In mechanical equilibrium:

- the volume-per-molecule should be the same for both sub-systems, or, if T is the same, P must be the same on both sides of the membrane.

S AB

VAVAeq

S AS B

NVNUNU US

VS

VSTP

,,,

/ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

≡The stat. phys. definition of pressure:

The “Pressure” Equation of State for an Ideal Gas

Ideal gas:(fN degrees of freedom)

VkNT

VSTP B

NU

=⎟⎠⎞

⎜⎝⎛∂∂

=,

TkNPV B=

UkNf

US

T BNV

12

1

,

=⎟⎠⎞

⎜⎝⎛∂∂

=

The “energy” equation of state (U ↔ T):

TkNfU B2=

The “pressure” equation of state (P ↔ T):

- we have finally derived the equation of state of an ideal gas from first principles!

),(ln2

ln),,( mNTkNfNVkNTVNS BB ϕ++=

Quasi-Static Processes

constVTS f =⇒=Δ 2/0 constVT =−1γ

The quasi-static adiabatic process with an ideal gas :

constVT =−1γconstPV =γ - we’ve derived these equations from the 1stLaw and PV=RT

On the other hand, from the Sackur-Tetrode equation for an isentropic process :

Quasistatic adiabatic (δQ = 0) processes: 0=Sd ⇒ isentropic processes

(all processes)

dVPSdTUd −= WQUd δδ +=

(quasi-static processes with fixed N)

SdTQ =δThus, for quasi-static processes :TQSd δ=

TQSd δ=

- is an exact differential (S is a state function). Thus, the factor 1/T converts δQ into an exact differential for quasi-static processes.

Comment on State Functions : P

V

00

≠Δ=Δ

QS

Problem:

(a) Calculate the entropy increase of an ideal gas in an isothermal process.(b) Calculate the entropy increase of an ideal gas in an isochoric process.

( ) ( ) 2/,, NfNUVNfNVU =Ω

You should be able to do this using (a) Sackur-Tetrode eq. and (b) TQSd δ=

⎟⎠⎞

⎜⎝⎛ +=+= dV

VTdTfNkPdVdUQ B 2

δ ⎟⎠⎞

⎜⎝⎛ +==

VdV

TdTfNk

TQdS B 2δ

( )[ ]2/ln fB VTNgNkS =

⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ =

i

fBconstT V

VNkS ln ⎟⎟

⎠

⎞⎜⎜⎝

⎛=Δ =

i

fBconstV T

TNkfS ln

2

(all the processes are quasi-static)

Let’s verify that we get the same result with approaches a) and b) (e.g., for T=const):

Since ΔU = 0, ⎟⎟⎠

⎞⎜⎜⎝

⎛==−= ∫

i

fB

V

V

B

VV

TkNdVVTkNWQ

f

lnδδ ⇒TQS δ=Δ

(Pr. 2.34)

Problem:A body of mass M with heat capacity (per unit mass) C, initially at temperature T0+ΔT,

is brought into thermal contact with a heat bath at temperature T0..

(a) Show that if ΔT

Problem (cont.)

(b) Ω for the (non-equilibrium) state with Tbacteria = 300.03K is greater than Ωin the equilibrium state with Tbacteria = 300K by a factor of

630145023

20

10/1038.1

/102expexp0

0 ≈≈⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅

=⎟⎟⎠

⎞⎜⎜⎝

⎛ Δ=

Ω

Ω−

−

Δ+

eKJKJ

kS

B

total

TT

T

The number of “1ps” trials over the lifetime of the Universe: 3012

18

101010

=−

Thus, the probability of the event happening in 1030 trials:

( )( ) 01010# 63030 ⇒×= −event an of occurrence ofy probabilitevents

Pr. 3.32. A non-quasistatic compression. A cylinder with air (V = 10-3 m3, T = 300K, P =105 Pa) is compressed (very fast, non-quasistatic) by a piston (S = 0.01 m2, F = 2000N, Δx = 10-3m). Calculate δW, δQ, ΔU, and ΔS.

VPSTU Δ−Δ=Δ

WQU δδ +=Δholds for all processes, energy conservationquasistatic, T and P are well-defined for any intermediate state

quasistatic adiabatic ≡ isentropic non-quasistatic adiabatic

[ ]

JmPa 11010101

111

11

111

1)(

2335

1

1

11

=××⋅=Δ

≈

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛ Δ+

−=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−⎟

⎟⎠

⎞⎜⎜⎝

⎛

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

−=

====

−−

−

−

−−

∫∫

xxVP

xxVP

VVVP

VVVP

dVV

VPconstPVdVVPW

ii

ii

f

iii

if

ii

V

Vii

V

V

f

i

f

i

γ

γ

γγ

γ

γγγ

γ

γγ

δ[ ]fi

V

V

VVPdVPWf

i

−=−= ∫δ

J2m10m10Pa102 23225

=××⋅= −−

The non-quasistatic process results in a higher T and a greater entropy of the final state.

S = const along the isentropic

line

P

V ViVf

1

2

2*

δQ = 0 for both

Caution: for non-quasistatic adiabatic processes, ΔS might be non-zero!!!An example of a non-quasistatic adiabatic process

( ) ( )NfkUkNfVkNNVUS BBB lnln2ln,, ++=Direct approach:

JWU 2==Δ210−=ΔVV

J/K K

10m 10Pa 10 2-33-5

3001

300

2502

2510

2

ln2

ln

2

=××

=

⎥⎦⎤

⎢⎣⎡ ×+−=⎥

⎦

⎤⎢⎣

⎡ −+

−≈

+=Δ

−

i

ii

i

if

i

ifB

i

fB

i

fB

TVP

UUUf

VVV

kN

UU

kNfVV

kNS

J 250m 10Pa 1025

25

233-5 =××=== iiiBi VPTkN

fU

adiabatic quasistatic ≡ isentropic 0=Qδ WU δ=Δ VPTkNf B Δ−=Δ2

VVTkNTkNf BB Δ−=Δ2

constTV f =2/ f

f

i

i

f

VV

TT

/2

⎟⎟⎠

⎞⎜⎜⎝

⎛=

0ln22

ln =−=Δi

fB

i

fB V

Vf

kNfVV

kNS

adiabatic non-quasistatic

K/J300

2K300

2J=≈=

Δ=Δ

TQ

TUS

The entropy is created because it is an irreversible, non-quasistatic compression.

P

V ViVf

To calculate ΔS, we can consider any quasistaticprocess that would bring the gas into the final state (S is a state function). For example, along the red line that coincides with the adiabat and then shoots straight up. Let’s neglect small variations of T along this path (Δ U

The inverse process, sudden expansion of an ideal gas (2 – 3) also generates entropy (adiabatic but not quasistatic). Neither heat nor work is transferred: δW = δQ = 0 (we assume the whole process occurs rapidly enough so that no heat flows in through the walls).

i

fBrev V

VTkNW ln=δ

The work done on the gas is negative, the gas does positive work on the piston in an amount equal to the heat transfer into the system

J/K300

1ln0 =Δ≈=−==Δ>−=xx

TVP

VVkN

TW

TQSWQ

i

ii

f

iB

revrevrevrev

P

V ViVf

1

2

3 Because U is unchanged, T of the ideal gas is unchanged. The final state is identical with the state that results from a reversible isothermal expansion with the gas in thermal equilibrium with a reservoir. The work done on the gas in the reversible expansion from volume Vf to Vi:

Thus, by going 1 → 2 → 3 , we will increase the gas entropy by J/K300

2321 =Δ →→S

Lecture 6. Entropy of an Ideal Gas (Ch. 3)Multiplicity for a Single particleMultiplicity of a Monatomic Ideal Gas (simplified)More Accurate Calculation of pEntropy of an Ideal GasProblemEntropy of MixingGibbs ParadoxProblemAn Ideal Gas: from S(N,V,U) - to U(N,V,T)Partial Derivatives of the EntropyThermodynamic Identity for dU(S,V,N)The Exact Differential of S(U,V,N)Mechanical Equilibrium and PressureThe “Pressure” Equation of State for an Ideal GasQuasi-Static ProcessesProblem:Problem:Problem (cont.)An example of a non-quasistatic adiabatic process