04 First Law of Thermodynamics

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2011 Imam Siswanto Kimia Fisika I

KIMIA FISIKA I


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First Law of Thermodynamicsuniversitas airlangga

excellence with morality

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This law is merely the law of conservation of energy, namely, that energycan be neither created nor destroyed.

Matematically :

DE + w = q

or

DE = q - w

Since internal energy depends only on the state of the system, then thechange in the energy , involved in going from a state where the energy isE1 to another state where the energy isE2 must be given by

DE = E2E1


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3

D2

1

V

V

pdVqE

Under special conditions these equation may take on special forms as follows :

1. Constant volume, dV = 0, dw = 0 and DE = q

2. Opposing pressure zero,p = 0, dw = 0 DE = q

3. Opposing pressure constant, w = p(V2V1) andDE = qp(V2V1)

4. Opposing pressure variable,p must be known as a function ofVfor the given

situation.

It must be clearly understood that, except under the condition to be pointed out

below, the pressure which determines the amount of work done is not the

pressure of the gasP, but the pressure against which the gas is working, namely,

p.


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Reversibility and Maximum Work

The maximum work is obtainable from a system when any change takingplace in it is entirely reversible.

Example :

1. Find the work done when 2 moles of hydrogen expand isothermally from15 to 25 liters against a constant pressure of 1 atm at 250 C.

2. Calculate the work performed when 2 moles of hydrogen expandisothermally and reversibly at 250 C from 15 to 25 liters.


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The Enthalpy af a System

Thermal changes at constant pressure are most conveniently expressed interms of another function,H, called the enthalpy o the relationr heat contentof a system.

This function is defined by the relation :

H = E + PVWhen the pressure remains constant throughout the process, then

DH = DE + PDV


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Heat Capacity

The amount of heat required to raise the temperature of the system 1 degreeis

and Cis thus the heat capacity of the system.

While dq = de + pdv, therefore

When the volume is held constant

It tell us that Cv is the rate of the change of the internal energy withtemperature at constant volume.

dqC

dT

dE pdvC

dT

v

v

EC

T


7/33 2011 Imam Siswanto Kimia Fisika I



7

Heat Capacity

However, when the heat absorption occurs reversibly at constant pressure, whenp = P and

But, ifH = E + PV is differentiated with respect to T at constant P we get

Consequently,

It tell us that Cp is the rate of the change of the internal energy with temperatureat constant pressure.

p

P P

E VC P

T T

P P P

H E VP

T T T

p

P

HC

T





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Heat Capacity

From the previous definition of Cv and Cp we have

ButH = E + PV. Differentiating this equation with respect to temperature atconstant pressure we obtain

Then

The difference between the two heat capacities

p v

P v

H EC C

T T

P P P

H E VP

T T T

p v

P P v

E V EC C P

T T T





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Heat Capacity

The internal energy,E, in general, will be a function of any two of threevariables (P, V, T). If we take Tand Vas our independent variables, then

E = f(V,T)

And

Dividing both sides of the equation by dTand imposing the condition ofconstant pressure, we get


V T

E EdE dT dV T V

P V T P

E E E V

T T V T





10

Heat Capacity

Finally,


p v

V T P P v

E E V V EC C P

T V T T T

T P P

E V VPV T T

Perfectly general equation





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Dependence of State Function on Variables

If we start with the fact thatE = f(T,V), then

According to the second law of thermodynamics :

Then

V T

E EdE dT dV

T V

T V

E PT P

V T

v

V

PdE C dT T P dV

T





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Similar consideration of H as a function of T and P gives

According to the second law of thermodynamics :

Then

P T

H HdH dT dP

T P

p

P

VdH C dT V T dP

T

T P

H VV T

P T





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From

And

We get

Effect of volume changes on Cv

v

v

EC

T

T V

E PT P

V T

2

2

v

T V

C PTV T





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From

And

We get

Effect of pressure changes on Cp

2

2

p

T P

C VTP T

p

P

HC

T

T P

H VV T

P T





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The Thermodynamic Behavior of Ideal Gases

We have

Form PV = nRT

Thus

Consequently, the internal energy of an ideal gas is independent of thevolume and depends only on the temperature.

T V

E PT P

V T

V

P nR

T V

0

T

E nRTP

V V





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We have

Form PV = nRT

Thus

P

V nR

T P

0

T

H nRTV

P P

T P

H VV TP T

Consequently, the enthalpy of an ideal gas is independent of the pressure and

depends only on the temperature.





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We have

Form PV = nRT

Thus

P

V nR

T P

0

T

H nRTV

P P

T P

H VV TP T

Consequently, the enthalpy of an ideal gas is independent of the pressure and

depends only on the temperature.





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From the previous deduction, it must follow also that Cv and Cp are functionsof T only and are independent of volume and pressure. Therefore

Furthermore, since for an ideal gas (E/V)T

= 0, then

0v

T

C

V

0p

T

C

P

p v

P

VC C P

T





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But for an ideal gas

And so we find that

And per mole

P

VP nR

T

p vC C nR

p vC C R

This conclusion is the same as obtained from the kinetic theory of gases


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Isothermal and Adiabatic Processes

Isothermal : any process conducted in a manner such that the temperatureremains constant during the entire operation.

Adiabatic : process in which no heat is absorbed or evolved by the system.

i f ii i i l


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For isothermal process

For isothermal and reversible process,

w = wm ,p = P = nRT/V

or

Isothermal Processes in Ideal Gases

2

1

V

V

q w pdV

2

1

2

1ln

V

m

V

dV V

w nRT nRT V V 1

2

lnm

Pw nRT

P

Fi t L f Th d ii it i l


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For adiabatic process DE = -w.

Any work in an adiabatic process is done at the expense of the internalenergy. As work is performed, the internal energy of the system willdecreases, and consequently the temperature drops.

From the above equation, therefore :

Since P = nRT/V,

Finally,

Adiabatic Processes in Ideal Gases

vPdV dE nC dT

vnRTdV nC dT

V

/ /

1 1 2 2 1

Cv R Cv RV T V T C

Fi t L f Th d ii it i l


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A very common form is one involving P and V :

where =Cp/Cv

or

Adiabatic Processes in Ideal Gases

1 1 2 2PV P V

1 2

2 1

P V

P V

Fi t L f Th d iuniversitasairlangga


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Exercises

1. A gas expands against a variable opposing pressure given byp = 10/Vatm, where Vis the volume of the gas at each stage of the expansion.Further, in expanding from 10 to 100 liters, the gas undergoes a changein internal energy of DE= 100 cal. How much heat is absorbed by the gasduring the process ?

2. Two liters of N2 at 00 C and 5 atm pressure are expanded isothermallyagainst a constant pressure of 1 atm until the pressure of the gas is also1 atm. Assuming the gas to be ideal, what are the values of w, DE, DH,and q for the process

3. Calculate the work done by 5 moles of an ideal gas during expansionfrom 5 atm at 250 C to 2 atm at 500 C against a constant pressure of 0.5atm. If for the gas Cp = 5.0 cal mole

-1 degree-1 , find also DE, DH, and q forthe process.



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Exercises

4. Assuming CO2 to be an ideal gas, calculate the work done by 10 g of CO2in expanding isothermally and reversibly from a volume of 5 liters to 10liters at 270 C. What are DE, DH, and q for the process.

5. Two liters of N2 at 00 C and 5 atm pressure are expanded isothermally

and reversibly until the confining pressure is 1 atm. Assuming the gas tobe ideal, calculate of w, DE, DH, and q for the expansion.



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The Joule-Thomson Effect

V1 V2

Porous plug

The situation with real gases

The work done on the system at the left piston is -P1V1 , the work done by

the system at the right piston is P2V2, and hence the net workdoneby the system is

w = P2V2P1V1



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Since the process was conducted adiabatically,

DE = - w

E2E1 = - (P2V2P1V1)

E2

+ P2

V2

= E1

+ P1

V1

H2 = H1

DH = 0

Thus, the process was conducted at constant enthalpy.

The Joule-Thomson coefficient, , is defined as

H

T

P

First La of Thermod namicsuniversitasairlangga


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The Joule-Thomson coefficient, , may be thought as of the number ofdegrees temperature change produced per atmosphere drop in pressureunder conditions of constant enthalpy.

For a coolingis positive, while for an observed heating is negative.

Example : N2, at a pressure such as 200 atm undergoes heating effect at -1500 C, cooling at -100 to 2000 C.

First Law of Thermodynamicsuniversitasairlangga


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The Joule-Thomson coefficient can be related to other thermodynamicquantities as follow :

and

p

T

HC

P

p

P

VC T V

T



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The Carnot Cycle

DE1 = q2w1

DE2 =w2

DE3 =q1w3

DE4 =w4



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The Carnot Cycle

The total change in internal energy for the complete cycle must be

DE = DE1 + DE2 + DE3 + DE4

= (q2q1)wm

Since the system is back to its initial state, we must have DE = 0, hence

wm = (q2q1)

and on division of both sides by q2,

2 1

2 2

mw q qq q



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The Carnot Cycle

Or in another form2 1 2 1

2 2 2

mw q q T T

q q T



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Ki i Fi ik I

First Law of Thermodynamicsgg


Exercise

One mole of an ideal gas monoatomic gas is carried through the cycle of

below, consisting of steps A, B, and C and involving states 1, 2, and 3. Fill inthe following table. Assume all of the processes are reversible.

A

BC

1 2

3

273 546T, 0 K

22,4

44,8

V,

l/mol

State P, Pa V, liter T, K

1 22,4 273

2 22,4 546

3 44,8 546

Step q, J w, J E, J

A

B

C

Cycle 0

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04 First Law of Thermodynamics

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