Thermodynamics and Kinetics PDF

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Thermodynamics and Kinetics with diagrams and derived examples...

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5.60 Fall 2017

Lectures #3-5

Isothermal Gas Expansion

page 1

(DT = 0)

gas (p1, V1, T ) = gas (p2, V2, T ) Irreversibly (many ways possible) (1)

Set p ext = 0 Gas v2

w (1) = - òV pextdV = 0 1

(2)

Set p ext = p 2 p2 T

p2

T p 2,V2

p 1 ,V1

v

w (2) = - ò 2 p2dV = -p2 (V2 -V1 ) V1

p p1

p2 V1

-w(2)

V2

Note, work is negative: system expands against surroundings

5.60 Fall 2017

Lectures #3-5

(3)

page 2

Carry out change in two steps

gas (p1, V 1, T ) = gas (p3, V3, T ) = gas (p2, V2, T )

p1 > p3 > p2 p2

p3 p3

T

T

T

p2,V2

p3,V3

p1,V1 v3

v2

V1

V3

w(3) = - ò p3 dV - ò p2 dV = - p3 (V3 - V1 ) - p2 (V2 - V3 ) p p1 p3

More work delivered to surroundings in this case.

p2 V1 V3

V2

-w (3) (4)

Reversible change

p

p = pext throughout V

wrev = -ò 2 pdV

p1

V1

p2 V1

-

V2 rev

Maximum work delivered to surroundings for isothermal gas expansion is obtained using a reversible path

For ideal gas: V V p wrev = - ò 2 nRT dV = -nRT ln 2 = nRT ln 2 V1 V V1 p1

5.60 Fall 2017

Lectures #3-5

page 3

The Internal Energy U

dU = d-q +d-w

(First Law)

dU = C pathdT - pextdV And U (T ,V

)

Þ

æ ¶U ö æ ¶U ö dU = ç ÷ dT + ç ÷ dV è ¶T øV è ¶V øT

Some frequent constraints: •

Reversible

Þ

dU = d-qrev +d-wrev = d-qrev – pdV ( p = pext )

•

Isolated

Þ

d- q = d-w = 0

•

Adiabatic

Þ

d-q = 0

Þ dU = d-w

•

Constant V

Þ

w=0

Þ dU = d-qV

reversible

=

-pdV

Constant V

æ ¶U ö æ ¶U ö dU = ç ÷ dV ÷ dT + ç è ¶V øT è ¶T øV but also æ ¶U ö Þ d-qV = ç ÷ dT è ¶T øV d-qV = CV dT

So

Þ

æ ¶U ö ç ÷ = CV è ¶T øV

very important result!!

æ ¶U ö ÷ è ¶V øT

dU = CV dT + ç

dV what is this?

5.60 Fall 2017

Lectures #3-5

(to get ç ¶ ÷ ) è ¶V øT æ

Joule Free Expansion of a Gas

gas

page 4

U ö

Adiabatic

q=0

Expansion into Vac. (pext=0)

w =0

vac

gas (p1, T 1, V1) = gas (p 2, T 2, V2) Since q = w = 0 Recall

Þ

dU or D U = 0

æ ¶U ö dU = CV dT + ç ÷ dV = 0 è ¶V øT æ ¶U ö ç ÷ dVU = -CV dTU è ¶V øT æ ¶U ö æ ¶T ö ç ÷ = -CV ç ÷ è ¶V øT è ¶V øU

Joule did this.

•

Constant U

measure in Joule exp't!

æ DT ö ç ÷ è DV øU

æ DT ö æ ¶T ö \ dU = CV dT -CV hJ dV ÷ =ç ÷ º hJ è DV øU è ¶V øU Joule coefficient

lim ç DV ®0

For Ideal gas

Þ

hJ = 0

exactly

dU = CV dT

Always for ideal gas

U(T)

only depends on T

The internal energy of an ideal gas depends only on temperature Consequences

Þ

Þ

DU = 0

DU = òCV dT

For all isothermal expansions or compressions of ideal gases For any ideal gas change in state

5.60 Fall 2017

Lectures #3-5

page 5

For an isothermal reversible expansion of an ideal gas: gas (p1, V1, T ) = gas (p2, V2, T ) Now we know: DT = 0

p 1 > p 2, V1 < V 2

Þ DU = 0 V2 V -wrev = ò nRT dV = nRT ln 2

p p1

V1

V

V1

V w rev = −nRT ln 2 < 0 V1

p2 V1

-

V2 rev

What is q rev? First Law Þ DU = q +w = 0

V qrev = -w rev =nRT ln 2 > 0 V1 Heat flows from surroundings to system (to replace energy used by the system doing work on surroundings).

5.60 Fall 2017

Lectures #3-5

H(T,p)

Enthalpy

page 6

H º 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, T 1, V 1)

=

const .p

gas ( p, T 2, V2)

U1

U2

DU = q + w = q p - p DV define as H

DU + p DV = q p D U + D ( pV ) = q p

H º U + pV Choose

æ ¶H ö ç¶ ÷ è T øp

for a reversible constant p process æ ¶H ö æ ¶H ö ÷ dp ÷ dT + ç p ¶ è ¶T ø p è øT

Þ

H (T , p ) è ¶T øp

•

DH = q p

Þ

¶H ö What are æç ÷

D (U + pV ) = q p

Þ

dH = ç

æ¶ H ö and ç ÷ ? è ¶ p øT

for a reversible process at constant p ( dp = 0)

Þ

dH = đ qp Þ

æ ¶H ö and dH = ç ÷ dT è ¶T ø p æ ¶H ö

đq p = ç ÷ dT è ¶T øp \

but

æ ¶H ö = Cp ç ÷ è ¶T ø p

đq p = C pdT also

5.60 Fall 2017

•

Lectures #3-5

æ ¶H ö ç ÷ è ¶p øT

Þ

page 7

Joule-Thomson expansion

adiabatic, q = 0 porous partition (throttle)

gas (p, T 1)

w = pV - pV 1 1 2 2

Þ \

=

gas (p , T 2)

DU = q + w = pV - pV = -D (pV 1 1 2 2 DU + D ( pV ) = 0 Þ \

) D (U + pV ) =

0

DH = 0

Joule-Thomson is a constant Enthalpy process. æ ¶H ö ÷ dp è ¶p øT

dH = C pdT + ç

æ ¶H ö ÷ dp H è ¶p øT

C pdT = - ç

æ ¶H ö æ ¶T ö ç ÷ = - Cp ç ÷ ¬ can measure this è ¶p øT è ¶p ø H

Þ

Define

\

Þ

æ DT ö ç ÷ è Dp ø H

æ DT ö æ ¶T ö lim ç = ç ÷ ÷ º µJT ¬ Joule-Thomson Coefficient D p® 0 D p è øH è ¶ p øH

æ¶H ö çç ÷÷ = - Cp µJT ¶ p è øT

and

dH = Cp dT - Cp µJT dp

5.60 Fall 2017

Lectures #3-5

U(T),

For an ideal gas:

page 8

pV=nRT

H ºU (T ) + pV = U (T ) + nRT

()

⇒

HT

only depends on T, no p dependence ⎛ ∂H ⎞ ⎜ ⎟ = µJT = 0 for an ideal gas ⎝ ∂ p ⎠T

For most real gases µJT < 0 at room temperature and below Þ

Use J-T expansion to liquefy gases

dU = CV dT

And Þ

dH = C pdT

for any ideal gas process

Proof that C p = C V + R for an ideal gas using a thermodynamic cycle

will get us C V

g( p,V ,T ) (2)

will get us Cp

constant V

=

(1)

constant p, reversible

g( p + dp,V ,T + dT ) (3) constant T

close cycle here, isothermally

g (p ,V + dV ,T +dT ) If we work with state functions, it doesn’t matter which path we take, e.g. constant V or constant p then constant T .

5.60 Fall 2017

(1) Constant V: dU1 = đq + đw

Lectures #3-5

constant V

= CV dT

(2) Constant p : æRö dU2 = đq + đw = C pdT - pdVp = C pdT - p çç ÷÷dT = (C p - R )dT èpø (3) Constant T : dU3 = 0 (isothermal, ideal gas)

U is a state function Þ Both paths must have same dU dU1 = dU2 + dU3 Þ

CV = C p - R

page 9

5.60 Fall 2017

•

Lectures #3-5

page 10

Reversible Adiabatic Expansion (or compression) of an Ideal Gas 1 mole gas (V1,T1) = 1 mole gas (V2,T 2) đq = 0 Ideal gas Þ

adiabatic Þ

From 1st Law

\

C V dT = − pdV

CV ∫

T2

T1

⇒ p =RT V

V dV dT = −R ∫ 2 V1 T V

Define

γ ≡

Cp CV

dU = -p dV Þ CV

dT T

= −R

dV V

For monatomic ideal gas:

Cvd T = -pd V along path

(There is one mole of gas)

⎛T ⎞ ⎛ V ⎞ ⇒ ⎜ 2 ⎟ = ⎜ 1⎟ ⎜⎝ T ⎟⎠ ⎜⎝ V ⎟⎠ 1 2

⇒

đw = -p dV

Reversible Þ dU = CvdT

Cp

R CV

⎛T ⎞ ⎛ V ⎞ ⎜ 2⎟ =⎜ 1⎟ ⎜⎝ T ⎟⎠ ⎜⎝ V ⎟⎠ 1 2

C p − CV = R for i.g.

⎯⎯⎯⎯⎯⎯→

⎛ T ⎞ ⎛V ⎞ CV ⎜ 2 ⎟ = ⎜ 1⎟ ⎜⎝ T ⎟⎠ ⎜⎝V ⎟⎠ 1 2

γ −1

3 CV = R 2 5 Cp = R 2

⎫ ⎪ 5 ⎪ ( > 1 generally) ⎬ γ = 3 ⎪ ⎪⎭

In an adiabatic expansion (V2 > V1), the gas cools ( T2 > T1). And in an adiabatic compression (V 2 < V 1), the gas heats up. For an ideal gas (one mole)

pV T = R

Þ

æ p2 ç è p1

g

ö æ V1 ö ÷= ç ÷ ø èV2 ø

pV g is constant along a reversible adiabat

g g Þ pV 1 1 = pV 2 2

−1

5.60 Fall 2017

Lectures #3-5

page 11

For an isothermal process

T = constant

Þ

pV = constant

p V2adiabat < V2isotherm because the gas cools during reversible adiabatic expansion

p1

p2 V2ad V2iso

V1

• Irreversible Adiabatic Expansion of an ideal gas against a constant external pressure 1 mol gas (p 1,T1) = 1 mol gas (p2,T2) adiabatic Constant p ext = p2 Ideal gas 1st Law \ Integrating:

Þ Þ Þ Þ

đq = 0 đ w = - p2dV dU = C vd T dU = - p 2d V

C vd T = - p2d V C v ( T 2 - T1) = - p 2 ( V2 - V1) æ

Using pV = RT

Note

(pext=p2)

p2 < p1

T2 (CV + R ) = T1 çCV + è

Þ

Note also (-w rev) > (-wirrev)

T2 < T 1

p2 ö R÷ p1 ø

Again, expansion cools

Less work is recovered through an irreversible process...