egr 334 thermodynamics chapter 3: section 12-14 lecture 10: ideal gas law quiz today?

EGR 334 ThermodynamicsChapter 3: Section 12-14

Lecture 10: Ideal Gas Law

Quiz Today?

Today’s main concepts:• Be able to explain the Ideal Gas Law• Be able to explain when it is appropriate to use the Ideal

Gas Law• Be able to use the Ideal Gas Law to determine State

Properties• Be able to apply the Ideal Gas Law to the solution of

1st Law problems.

Reading Assignment:

Homework Assignment:

• Read Chap 3: Sections 15

From Chap 3: 102, 107,115, 125

3

When the general compressibility factor, Z 1,then the following relationship between pressure, temperature, and volume of a gas applies.

Ideal Gas Law

pv RT

which can also be written

absolute pressure

absolute temperature

specific volume

p

T

v

mol

f mol

8.314 kJ/kmol K

1.986 Btu/lb

1545 ft lb /lb

o

o

R R

R

or

pV mRT

pV nRT

R may be found on Table 3.1m = massn = number of moles

4Sec 3.13 : Ideal gases and u, h, cv, cp

If a gas behaves as an ideal gas, then its specific internal energy, u, depends only on temperature.

Enthalpy, h, was defined as:

pv RTFor an ideal gas, since then

( )u u T

h u pv

h u RT

( ) ( )h h T u T RT ( )u u TSince and

Therefore: If a gas can be treated as an ideal gas, its intensive properties of specific energy and enthalpy are entirely functions of temperature.


If u and h are only functions of temperature, then the specific heats may be used to determine relations between temperature change and energy levels.

For an ideal gas, the expressions for u and h can be simplified since u = f(T) and h = f(T)

VV

uc

T

P

P

hc

T

2

1

T

VTdu c dT

2

1

T

PTdh c dT

For many cases, the specific heats will be treated as constant values over a limited temperature range and these integrals will be approximated as:

2 1 2 1( )Vu u c T T 2 1 2 1( )ph h c T T


Another important ideal gas equation may be written as

h T u T RT

When the specific heat ratio is used, this equationmay also be written as V

P

c

ck

1P

kRc T

k

For monotonic gases (Ar, Ne, He) with k = 1.4 _

5

2P mono gasesc R

and

dh duR

dT dT

P Vc T c T R

1V

Rc T

k


Temperature Dependence:Specific heats cv and cp are functions of temperature.

V

P

c

ck

An alternative method is to use a formula to represent cp based on

If possible, you should look uptheir values from tables which give the specific heat at the temperature indicated.(see Tables A-20 and A-20E)

2 3 4pcT T T T

R

where Table A-21 has values of , , , , for different gases

2

1

( )T

VTTdu c dT

2

1

( )T

PTTdh c dT


where if cV and cP are treated as constants, then

2 1 2 1Vu T u T c T T

It best to use an actual function of the specific heats to evaluate u and h, by integration

2

1

T

VTdu c dT

2

1

T

PTdh c dT

But, often this is simplified, evaluating cV and cP at an average T

1 2( ) ( )

2ave

v vV

c T c Tc

either and 1 2( ) ( )

2ave

p pp

c T c Tc

1 2

2ave

T TV vc c

or the specific heat at the average temperature may be used.

1 2

2ave

T Tp pc c

2 1 2 1Ph T h T c T T

9

1) Decide if a substance can be treated as an ideal gas…if yes, then

Sec 3.13 : Ideal gases and u, h, cv, cp

2 1 _ 2 1V aveu u c T T

Summary:

2

1

T

VTdu c dT

2

1

T

PTdh c dT

pv RT2) To evaluate changes in internal energy, u, and enthalpy, h:

2 1 2 1Ph h c T T

i) Integrate with cv and cp as function of T (see Table A-21)

ii) Use value of cv or cp at an average temperature (see Table A-20)

iv) Look up temperature dependent values of u and h on property tables.

Table A-22 has property values for AirTable A-23 has property values for CO2, C0, H20, O2, and N2.

2 1( ) ( )u u T u T 2 1( ) ( )h h T h T

iii) Use k and R to define and then

2 1 2 1Vu u c T T 1P

kRc

k

1v

Rc

k

2 1 _ 2 1P aveh h c T T

pV mRT

or

or

or

pV nRTor

10

Example: (3.111) A piston cylinder assembly contains air at 2 bar, 300K, and a volume of 2 cubic meters. the air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is pV = constant. Assuming ideal gas behavior, determine a) the mass of the air, b) the work, and c) the heat transfer.


11

Example: (3.111) A piston cylinder assembly contains air at 2 bar, 300K, and a volume of 2 cubic meters. the air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is pV = constant. Assuming ideal gas behavior, determine a) the mass of the air, b) the work, and c) the heat transfer.


State 1: p1 = 2 bar T1 = 300 K V1 = 2 m3

State 2: p2 = 1 bar T2 = ? V2 = ?

For Ideal Gas: pV mRT

For constant pV:

1 1 2 2pV p V

3 5 21 1

1

(2 )(2 ) 10 / 14.65

(0.2870 / )(300 ) 1 1000

pV bar m N m kJ Jm kg

RT kJ kg K K bar J N m

3 312 1

2

22 4

1

p barV V m m

p bar

12

Example: (3.111) continued...


State 1: p1 = 2 bar T1 = 300 K V1 = 2 m3

State 2: p2 = 1 bar T2 = ? V2 = 4 m3

find T2:

1st Law of Thermodynamics:

1 1 2 2pV p V C

2 2 1 1

2 1

p V pVR

T T

32 2

2 1 21 1

1 4(300 ) 300

2 2

p V bar mT T K K

pV bar m

U Q W 2 1( ) 0vU c T T

CW pdV dV

V

2 21 1

1 1

ln lnV V

W C pVV V

3 5 2

33

4 10 / 1(2 )(2 ) ln 277.3

2 1 1000

m N m J kJbar m kJ

m bar N m J

0 277.3Q U W kJ

13

Example: (3.124) Two kilograms (2 kg) of air, initially at 5 bar, 350 K and 4 kg of CO initially at 2 bar, 450 K are confined to opposite sides of a rigid, well-insulated container by a partition. The partition is free to move and allows conduction from one gas to the other without energy storage in the partition itself. The air and CO each behave as ideal gases with constant specific heat ratio, k = 1.395. Determine at equilibrium (a) the temperature in K, (b) the pressure, in bar, and (c) the volume occupied by each gas, in m3.


14

Example: (3.124) k = 1.395.

system system systemU Q W


0CO airU U

air COU U

CO State 1: Air: State 1:

_1

_1

450

2

4

CO

CO

CO

T K

p bar

m kg

_1

_1

350

5

2

air

air

air

T K

p bar

m kg

State 2:

_ 2 _ 2 2CO airT T T

1st Law of Thermo:

_1 _ 2 2CO airp p p CO air totalm m m

since system is isolated

or

15

Example: (3.124) continued…

2 1 U m u u


CO State 1: Air: State 1: _1

_1

450

2

4

CO

CO

CO

T K

p bar

m kg

_1

_1

350

5

2

air

air

air

T K

p bar

m kg

2 1Vmc T T

For CO: 2 _1COCO CO V COU m c T T

,

0.2968 /0.7514

1 1.395 1CO

V CO

kJ kg KR kJc

k kg K

2(4 )(0.7514 / ) 450COU kg kJ kg K T K

For Air: 2 _1airair air V airU m c T T

_

0.2870 /0.7266

1 1.395 1air

V air

kJ kg KR kJc

k kg K

2(2 )(0.7266 / ) 350airU kg kJ kg K T K

16



CO State 1: Air: State 1: _1

_1

450

2

4

CO

CO

CO

T K

p bar

m kg

_1

_1

350

5

2

air

air

air

T K

p bar

m kg

2 2(4 )(0.7514 / ) 450 (2 )(0.7266 / ) 350 0kg kJ kg K T K kg kJ kg K T K

0CO airU U

2 2(3.006 / ) 450 (1.453 / ) 350 0kJ K T K kJ K T K

2 2(3.006 / ) 1352.7 (1.453 / ) 508.55 0kJ K T kJ kJ K T kJ

2(4.459 / ) 1858.3kJ K T kJ

2 416.8T K

17

Example: (3.124) continued

T CO AirV V V

Find Vtotal

Then find pfinal , ,CO f Air f

ftotal total

mRT mRTmRTp

V V


25 5

CO,i Air,i

4 0.2968 450 2 0.2870 350 ( )( / )( ) 1000

/2 10 5 10

kg kJ kg K K N m

N m kJ

CO,i Air,i

mRT mRT

P P

32.67 0.40 = 3.07 m

3 5

4 0.2968 416.8 2 0.2870 416.8 ( )( / )( ) 1000 12.39

3.07 10 /

kg kJ kg K K J N m barbar

m kJ J N m

18


CO,fCO

mRTV

P

The final volumes are then,


325

Air,f

2 0.2870 416.8 ( )( / )( ) 1000 1.001 m

/2.39 10Air

kg kJ kg K K N mV

N m kJ

Air,f

Air

mRTV

P

325

CO,f

4 0.2968 416.8 ( )( / )( ) 10002.079

/2.39 10CO

kg kJ kg K K N mV m

N m kJ

19

Solution using IT:

Note: The results are slightly different as the cv and cp values that IT pulled out slightly different.

20

End of Slides for Lecture 10

egr 334 thermodynamics chapter 3: section 12-14 lecture 10: ideal gas law quiz today?

Documents