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Thermodynamics

Session 6

Prepared by Vinod Kallur, RVCE, Bangalore.

Guidelines to represent processes on p

V diagram:

In the figure above there are two isotherms shown. For these isotherms, .12 TT There is one

isotherm passing through every point on p V diagram. Higher the temperature higher will

be location of the isotherm. As they go downwards they converge as shown. No two

isotherms will ever intersect.

Similarly every point has an adiabatic curve passing through it. The adiabatic curve follows

the relation constpV . Thus every point is an intersection of an isotherm and an adiabatic

curve as shown below. The adiabatic curve is steeper.

p

V

T1

T2

p

V

Isotherm

Adiabatic curve

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No two adiabatic curves will ever intersect.

The mathematical relation between pressure, volume and temperature applicable all ranges of

values has not been possible. However, there are equations which are valid over certain

ranges.

Conditions for any equation to be equation of state:

1. The equation should reduce to

RTpV

As p 0.

2. p-Vcurve should have a point of inflexion for isothermal at critical point C,

0

CV

pand 0

2

2

CV

p

Van der Waals equation of state:

RTbVV

ap

2

Here a and b are constants that depend upon the gas. If we apply the partial derivativeconditions, we get,

C

C

C

C

p

RTb

p

TRa

8;

64

2722

The Van der Waals equation is cubic in V. Therefore it should have three roots for a given

set of values of pressure and temperature and for a given substance. For any temperature

higher than critical temperature, there is one positive root. For critical temperature and

critical pressure, all the three roots will be equal which gives critical volume. For temperature

and pressure less than critical values, there will be three roots, least one gives the molar

volume of saturated liquid and highest gives that of saturated vapor which are volumes given

by the two ends of horizontal section of the isotherm.

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Redlich Kwong Equation:

TbVV

a

bV

RTp

)(

Applying the partial derivative condition, we get

C

C

C

C

p

RTb

p

TRa

0867.0;

4278.02/52

Note that the above equations reduce to the equation ,RTpV as p 0.

Other equations of state are Peng Robinson and Virial equations.

Numerical problem:

Air is compressed from an initial state of 1 bar and 298 K to a final state of 5 bar and 298 K

through two mechanically reversible processes in a closed system. First heating at constant

volume followed by cooling at constant pressure. Calculate W, Q, change in enthalpy and

internal energy. Use2

5,

2

7 RC

RC VP .

The path is shown in the figure. 1 - a represents heating at constant volume and a - 2

represents cooling at constant pressure. Note that initial and final states lie on a single

isotherm.

Let us do the calculations for one mol of gas.

For the path 1 a, since volume is constant, work done is zero.

dQdU

a

1

2

V

p

Isotherm at 298 K

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dTCdU V

aT

T

Va dTCU

1

1

To use this we need to know temperature at a. Since it is closed system and volume is same

at 1 and a,

11TpTp aa

Since pressure at a is same as that at 2,

Kp

Tp

p

TpT

a

a 14901

)298(5

2

1111

aT

T

Va dTCU

1

1

kJ

J

R

dTR

775.24

7.24775

)2981490(2

)314.8(5

)2981490(2

5

2

51490

298

Thus Q1-a = U1-a = 24.775 kJ.

kJJdTCH Pa 686.3434686)2981490(2

)314.8(71490

298

1

You can use the formula

)()( 111111 VpVpUpVUH aaaaaa

to calculate change in enthalpy.

For the step a 2,

dWdQdU

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2

2

T

T

Va

a

dTCU

kJ

J

R

dTR

775.24

7.24775

)1490298(2

5

25

298

1490

kJJdTCH Pa 686.3434686)1490298(2

)314.8(7298

1490

2

pdVdW

2

2

a

a pdVW

Along the path a 2, since pressure is constant and equal to 5 bar,

2

2

a

a pdVW

)(105 25

aVVx

We should know the volume at a.

From gas law, at the initial condition, 351

02477.0101

)198(314.8m

xV

Note that we have taken one mol.

1

11

2

22

T

Vp

T

Vp

The temperatures are equal,

2

112

p

Vp

V

3004954.0

5

)02477.0(1

m

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Similarly,

3

51245.0

101

)1498(314.8m

xp

RTV

a

aa

kJJxVVpW aa 773.5959773)1245.0004954.0(105)(5

22

Since the volume is decreasing, work is done on the system.

kJQ

Q

WQU

a

a

aaa

548.84

)773.59(775.24

2

2

222

For the entire process,

kJQQQ aa 773.59)548.84(775.242121

0686.34686.34

0775.24775.24

2121

2121

aa

aa

HHH

UUU

kJWWW aa 773.59)773.59(02121

Since there are no changes in internal energy and enthalpy, work done on the system is given

out as heat.