Unit3

BASIC THERMODYNAMICS J2006/3/1

UNIT 3

BASIC THERMODYNAMICS

OBJECTIVES

General Objective : To understand the laws of thermodynamics and its constants.

Specific Objectives : At the end of the unit you will be able to:

 define the definitions of Boyle’s Law, Charles’ Law and
Universal Gases Law

 define and show the application of the specific heat capacity at
constant pressure

 define and apply the specific heat capacity at constant volume


INPUT

3.0 Definition Of Perfect Gases

Did you know, one important type of fluid that has many applications in
thermodynamics is the type in which the working temperature of the fluid remains
well above the critical temperature of the fluid? In this case, the fluid cannot be
liquefied by an isothermal compression, i.e. if it is required to condense the fluid,
then cooling of the fluid must first be carried out. In the simple treatment of such
fluids, their behavior is likened to that a perfect gas. Although, strictly speaking, a
perfect gas is an ideal which can never be realized in practice. The behavior of many
‘permanent’ gases, e.g. hydrogen, oxygen, air etc is very similar to the behavior of a
perfect gas to a first approximation.

A perfect gas is a collection of particles that:
 are in constant, random motion,

 have no intermolecular attractions (which leads to elastic collisions in which no

energy is exchanged or lost),
 are considered to be volume-less points.

You are more familiar with the term ‘ideal’ gas. There is actually a distinction
between these two terms but for our purposes, you may consider them
interchangeable. The principle properties used to define the state of a gaseous system
are pressure (P), volume (V) and temperature (T). SI units (Systems International)
for these properties are Pascal (Pa) for pressure, m3 for volume (although liters and
cm3 are often substituted), and the absolute scale of temperature or Kelvin (K).

Two of the laws describing the behavior of a perfect gas are Boyle’s Law and
Charles’ Law.


3.1 Boyle’s Law

The Boyle’s Law may be stated as follows:
Provided the temperature T of a perfect gas remains constant, then volume, V of a
given mass of gas is inversely proportional to the pressure P of the gas, i.e. P ∝ 1/V
(as shown in Fig. 3.1-1), or P x V = constant if temperature remains constant.

P

P ∝ 1/V

1/V

Figure 3.1-1 Graph P ∝ 1/V

If a gas changes from state 1 to state 2 during an isothermal process, then

P1 V1 = P2 V2 = constant (3.1)

If the process is represented on a graph having axes of pressure P and volume V, the
results will be as shown in Fig. 3.1-2. The curve is known as a rectangular
hyperbola, having the mathematical equation xy = constant.

P

P1 1

P2 2
3
P3

V1 V2 V3 V
PV = constant
Figure 3.1-2 P-V graph for constant temperature


Example 3.1
A quantity of a certain perfect gas is heated at a constant temperature from an
initial state of 0.22 m3 and 325 kN/m2 to a final state of 170 kN/m2. Calculate
the final pressure of the gas.

Solution to Example 3.1

From equation P1V1 = P2V2

 325 kN/m 2 
= ( 0.22 m 3 )
P1
∴V2 = V1 x  170 kN/m 2  = 0.421 m 3

P2  

3.2 Charles’ Law

The Charles’s Law may be stated as follows:
Provided the pressure P of a given mass of gas remains constant, then the volume V
of the gas will be directly proportional to the absolute temperature T of the gas, i.e.
V ∝ T, or V = constant x T. Therefore V/T = constant, for constant pressure P.

If gas changes from state 1 to state 2 during a constant pressure process, then

V1 V2
= = constant (3.2)
T1 T2

If the process is represented on a P – V diagram as before, the result will be as shown
in Fig. 3.2.

P

1 2

0 V
V1 V2
Figure 3.2 P-V graph for constant pressure process


Example 3.2
A quantity of gas at 0.54 m3 and 345 oC undergoes a constant pressure process
that causes the volume of the gas to decreases to 0.32 m3. Calculate the
temperature of the gas at the end of the process.


From the question
V1 = 0.54 m3
T1 = 345 + 273 K = 618 K
V2 = 0.32 m3

V1 V2
=
T1 T2

V2
∴ T2 = T1 x
V1

 0.32 m 3 
= ( 618 K ) 
 0.54 m 3 

 
= 366 K

3.3 Universal Gases Law

Charles’ Law gives us the change in volume of a gas with temperature when the
pressure remains constant. Boyle’s Law gives us the change in volume of a gas with
pressure if the temperature remains constant.

The relation which gives the volume of a gas when both temperature and the
pressure are changed is stated as equation 3.3 below.

PV
= constant = R (3.3)
T

i.e. P1V1 P2V2 (3.4)
=
T1 T2


No gases in practice obey this law rigidly, but many gases tend towards it. An
PV
imaginary ideal that obeys the law is called a perfect gas, and the equation =R
T
is called the characteristic equation of state of a perfect gas.

The constant, R, is called the gas constant. The unit of R is Nm/kg K or J/kg K.
Each perfect gas has a different gas constant.

The characteristic equation is usually written
PV = RT (3.5)

or for m kg, occupying V m3,
PV = mRT (3.6)

Another form of the characteristic equation can be derived using the kilogram-mole
as a unit. The kilogram-mole is defined as a quantity of a gas equivalent to m kg of
the gas, where M is the molecular weight of the gas (e.g. since the molecular weight
of oxygen is 32, then 1 kg mole of oxygen is equivalent to 32 kg of oxygen).

From the definition of the kilogram-mole, for m kg of a gas we have,
m = nM (3.7)

(where n is the number of moles).
Note: Since the standard of mass is the kg, kilogram-mole will be written simply as
mole.

Substituting for m from equation 3.7 in equation 3.6,

PV
PV = nMRT or MR = (3.8)
nT


Now Avogadro’s hypothesis states that the volume of 1 mole of any gas is the same
as the volume of 1 mole of any other gas, when the gases are at the same temperature
and pressure. Therefore V/n is the same for all gases at the same value of P and T.
That is the quantity PV/nT is constant for all gases. This constant is called the
universal gas constant, and is given the symbol Ro.

i.e. PV (3.9)
MR = Ro = or PV = nRo T
nT

or since MR = Ro then,
Ro
R= (3.10)
M

Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 1 oC
is approximately 22.71 m3. Therefore from equation 3.8

PV 1 x 10 5 x 22.71
R0 = = = 8314.4 J/mole K
nT 1 x 273.15

From equation 3.10 the gas constant for any gas can be found when the molecular
weight is known, e.g. for oxygen of molecular weight 32, the gas constant is

Ro 8314.4
R= = = 259.8 J/kg K
M 32

Example 3.3
0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2
and a temperature of 45 oC. The gas is compressed until the pressure reaches
1.27 MN/m2 and the temperature is 83oC. If the gas is assumed to be a perfect
gas, determine:

a) the mass of gas (kg)
b) the final volume of gas (m3)

Given:
R = 0.29 kJ/kg K



From the question
V1 = 0.046 m3
P1 = 300 kN/m2
T1 = 45 + 273 K = 318 K
P2 = 1.27 MN/m2 = 1.27 x 103 kN/m2
T2 = 83 + 273 K = 356 K
R = 0.29 kJ/kg K

From equation 3.6
PV = mRT

P1V1 300 x 0.046
m= = = 0.1496 kg
RT1 0.29 x 318

From equation 3.4, the constant volume process i.e. V1 = V2

P1 P2
=
T1 T2

P   1.27 x 10 3 
T2 = ( T1 )  2
P  = ( 318) 
  300  = 1346 K

 1   


Activity 3A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!

3.1 Study the statements in the table below. Mark the answers as TRUE or
FALSE.

STATEMENT TRUE or FALSE
i. Charles’ Law gives us the change in volume
of a gas with temperature when the
temperature remains constant.
ii. Boyle’s Law gives us the change in volume of
a gas with pressure if the pressure remains
constant.
iii. The characteristic equation of state of a
perfect gas is PV = R .
T
iv. Ro is the symbol for universal gas constant.
v. The constant R is called the gas constant.
vi. The unit of R is Nm/kg or J/kg.

3.2 0.04 kg of a certain perfect gas occupies a volume of 0.0072 m3 at a pressure
6.76 bar and a temperature of 127 oC. Calculate the molecular weight of the
gas (M). When the gas is allowed to expand until the pressure is 2.12 bar the
final volume is 0.065 m3. Calculate the final temperature.


Feedback To Activity 3A

3.1 i. False
ii. False
iii. True
iv. True
v. True
vi. False

3.2 From the question,
m = 0.04 kg
V1 = 0.072 m3 V2 = 0.072 m3
P1 = 6.76 bar = 6.76 x 102 kN/m2 P2 = 2.12 bar = 2.12 x 102 kN/m2
T1 = 127 + 273 K = 400 K

From equation 3.6
P1V1 = mRT1

P1V1 6.76 x 10 2 x 0.0072
∴R = = = 0.3042 kJ/kg K
mT1 0.04 x 400
Then from equation 3.10
R
R= o
M

Ro 8.3144
∴M = = = 27 kg/kmol
R 0.3042

i.e. Molecular weight = 27


From equation 3.6
P2V2 = mRT2

P2V2 2.12 x 10 2 x 0.065
∴ T2 = = = 1132.5 K
mR 0.04 x 0.3042

i.e. Final temperature = 1132.5 – 273 = 859.5 oC.

CONGRATULATIONS, IF YOUR ANSWERS ARE CORRECT YOU CAN
PROCEED TO THE NEXT INPUT…..


INPUT

3.4 Specific Heat Capacity at Constant Volume (Cv)

The specific heat capacities of any substance is defined as the amount of heat energy
required to raise the unit mass through one degree temperature raise. In
thermodynamics, two specified conditions are used, those of constant volume and
constant pressure. The two specific heat capacities do not have the same value and it
is essential to distinguish them.

If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the
temperature of the gas by 1 degree whilst the volume of the gas remains constant,
then the amount of heat energy supplied is known as the specific heat capacity at
constant volume, and is denoted by Cv. The unit of Cv is J/kg K or kJ/kg K.

For a reversible non-flow process at constant volume, we have
dQ = mCvdT (3.11)

For a perfect gas the values of Cv are constant for any one gas at all pressures and
temperatures. Equations (3.11) can then be expanded as follows :
Heat flow in a constant volume process, Q12 = mCv(T2 – T1) (3.12)

Also, from the non-flow energy equation
Q – W = (U2 – U1)
mcv(T2 – T1) – 0 = (U2 – U1)
∴ (U2 – U1) = mCv(T2 – T1) (3.13)

i.e. dU = Q

Note:
In a reversible constant volume process, no work energy transfer can take
place since the piston will be unable to move i.e. W = 0.


The reversible constant volume process is shown on a P-V diagram in Fig. 3.4.

P

P2 2

P1 1

V
V1 = V2
Figure 3.4 P-V diagram for reversible constant volume process

Example 3.4

3.4 kg of gas is heated at a constant volume of 0.92 m 3 and temperature 17 oC
until the temperature rose to 147 oC. If the gas is assumed to be a perfect gas,
determine:

c) the heat flow during the process
d) the beginning pressure of gas
e) the final pressure of gas

Given
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K


From the question
m = 3.4 kg
V1 = V2 = 0.92 m3
T1 = 17 + 273 K = 290 K
T2 = 147 + 273 K = 420 K
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K


a) From equation 3.13,
Q12 = mCv(T2 – T1)
= 3.4 x 0.72(420 – 290)
= 318.24 kJ

b) From equation 3.6,
PV = mRT

Hence for state 1,
P1V1 = mRT1
mRT1 3.4 kg x 0.287 kJ/kgK x 290 K
P1 = = 3
= 307.6 kN/m 2
V1 0.92 m

c) For state 2,
P2V2 = mRT2
mRT2 3.4 kg x 0.287 kJ/kgK x 420 K
P2 = = 3
= 445.5 kN/m 2
V2 0.92 m

3.5 Specific Heat Capacity at Constant Pressure (Cp)

If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the
temperature of the gas by 1 degree whilst the pressure of the gas remains constant,
then the amount of heat energy supplied is known as the specific heat capacity at
constant pressure, and is denoted by Cp. The unit of Cp is J/kg K or kJ/kg K.

For a reversible non-flow process at constant pressure, we have
dQ = mCpdT (3.14)

For a perfect gas the values of Cp are constant for any one gas at all pressures and
temperatures. Equation (3.14) can then be expanded as follows:

Heat flow in a reversible constant pressure process Q = mCp(T2 – T1) (3.15)


3.6 Relationship Between The Specific Heats

Let a perfect gas be heated at constant pressure from T1 to T2. With reference to the
non-flow equation Q = U2 – U1 + W, and the equation for a perfect gas
U2 – U1 = mCv(T2 – T1), hence,
Q = mCv(T2 – T1) + W

In a constant pressure process, the work done by the fluid is given by the pressure
times the change in volume, i.e. W = P(V2 – V1). Then using equation PV = mRT,
we have
W = mR(T2 – T1)

Therefore substituting,
Q = mCv(T2 – T1) + mR(T2 – T1) = m(Cv + R)(T2 – T1)

But for a constant pressure process from equation 3.15,
Q = mCp(T2 – T1)

Hence, by equating the two expressions for the heat flow Q, we have
mCp(T2 – T1) = m(Cv + R)(T2 – T1)
∴Cp = Cv + R
Alternatively, it is usually written as

R = Cp - C v 3.16

3.7 Specific Heat Ratio (γ)

The ratio of the specific heat at constant pressure to the specific heat at constant
volume is given the symbol γ (gamma),

Cp
i.e. γ= (3.17)
Cv

Note that since Cp - Cv= R, from equation 3.16, it is clear that Cp must be greater than
Cv for any perfect gas. It follows therefore that the ratio Cp/Cv = γ , is always greater
than unity. In general, γ is about 1.4 for diatomic gases such as carbon monoxide
(CO), hydrogen (H2), nitrogen (N2), and oxygen (O2). For monatomic gases such as


argon (A), and helium (He), γ is about 1.6, and for triatomic gases such as carbon
dioxide (CO2), and sulphur dioxide (SO2), γ is about 1.3. For some hydro-carbons the
value of γ is quite low (e.g. for ethane (C2H6), γ = 1.22, and for iso-butane (C4H10), γ
= 1.11.

Some useful relationships between Cp , Cv , R, and γ can be derived.
From equation 3.17

Cp - Cv= R

Dividing through by Cv
Cp R
−1 =
Cv Cv
Cp
Therefore using equation 3.17, γ = , then,
Cv

R
γ −1 =
Cv

R
Cv = 3.18
(γ − 1)
Also from equation 3.17, Cp = γCv hence substituting in equation 3.18,

γR
Cp = γ Cv =
(γ − 1)
γR
Cp = 3.19
(γ − 1)


Example 3.5

A certain perfect gas has specific heat as follows

Cp = 0.846 kJ/kg K and Cv = 0.657 kJ/kg K

Find the gas constant and the molecular weight of the gas.


From equation 3.16

R = Cp - C v

i.e. R = 0.846 – 0.657 = 0.189 kJ/kg K
or R = 189 Nm/kg K

From equation 3.10

R0
M=
R

8314
i.e. M= = 44
189


Activity 3B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!

3.3 Two kilograms of a gas receive 200 kJ as heat at constant volume process. If
the temperature of the gas increases by 100 oC, determine the Cv of the
process.

3.4 A perfect gas is contained in a rigid vessel at 3 bar and 315 oC. The gas is
then cooled until the pressure falls to 1.5 bar. Calculate the heat rejected per
kg of gas.
Given:
M = 26 kg/kmol and γ = 1.26.

3.5 A mass of 0.18 kg gas is at a temperature of 15 oC and pressure 130 kN/m2.
If the gas has a value of Cv = 720 J/kg K, calculate the:

i. gas constant
ii. molecular weight
iii. specific heat at constant pressure
iv. specific heat ratio


Feedback To Activity 3B

m = 2 kg
Q = 200 kJ
(T2 – T1) = 100 oC = 373 K

Q = mCv(T2 – T1)
Q 200
Cv = = = 0.268 kJ/kgK
m( T2 − T1 ) 2(373)

P1 = 3 bar
T1 = 315 oC = 588 K
P2 = 1.5 bar
M = 26 kg/kmol
γ = 1.26

From equation 3.10,
R 8314
R= o = = 319.8 J/kg K
M 26

From equation 3.18,
R 319.8
Cv = = = 1230 J/kg K = 1.230 kJ/kg K
(γ − 1) 1.26 − 1

During the process, the volume remains constant (i.e. rigid vessel) for the
mass of gas present, and from equation 3.4,


P1V1 P2V2
=
T1 T2

Therefore since V1 = V2,
P 1.5
T2 = T1 2 = 588 x = 294 K
P1 3
Then from equation 3.12,
Heat rejected per kg gas, Q = Cv(T2 – T1)
= 1.230(588 – 294)
= 361.6 kJ/kg

3.5 From the question
m = 0.18 kg
T = 15 oC = 288 K
V = 0.17 m3
Cv = 720 J/kg K = 0.720 kJ/kg K

i. From equation 3.6,
PV = mRT
PV 130 x 0.17
R= = = 0.426 kJ/kgK
mT 0.18 x 288

ii. From equation 3.10,
R
R= o
M
R 8.3144
M = o = = 19.52 kg/kmol
R 0.426

iii. From equation 3.16,
R = Cp - C v
Cp = R + Cv = 0.426 + 0.720 = 1.146 kJ/kg K

iv. From equation 3.17,
C p 1.146
γ = = = 1.59
C v 0.720


SELF-ASSESSMENT

You are approaching success. Try all the questions in this self-assessment
section and check your answers with those given in the Feedback to Self-
Assessment on the next page. If you face any problem, discuss it with your lecturer.
Good luck.

1. 1 m3 of air at 8 bar and 120 oC is cooled at constant pressure process until the
temperature drops to 27 oC.
Given R = 0.287 kJ/kg K and Cp = 1.005 kJ/kg K, calculate the:

i. mass of air
ii. heat rejected in the process
iii. volume of the air after cooling.

2. A system undergoes a process in which 42 kJ of heat is rejected. If the
pressure is kept constant at 125 kN/m2 while the volume changes from
0.20 m3 to 0.006 m3, determine the work done and the change in internal
energy.

3. Heat is supplied to a gas in a rigid container.The mass of the container is 1 kg
and the volume of gas is 0.6 m3. 100 kJ is added as heat. If gas has
Cv = 0.7186 kJ/kg K during a process, determine the:

i. change in temperature
ii. change in internal energy


Feedback To Self-Assessment

Have you tried the questions????? If “YES”, check your answers now.

1. i. m = 7.093 kg
ii. Q = 663 kJ
iii. V2 = 0.763 m3

2. W = -24.25 kJ
(U2 – U1) = -17.75 kJ

3. i. (T2 – T1) = 139.2 K
ii. (U2 – U1) = 100 kJ

CONGRATULATIONS!!!!
…..May success be with
you always….

Unit3

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