CHAPTER 8 HEAT, GAS LAW and THERMODYNAMICS

CHAPTER 8
Heat, Gas Law and Thermodynamics

[12 Hours]

8.1 Heat
8.2 ideal gas equation
8.3 Thermodynamics

8.2 HEAT

At the end of this topic, students should be
able to :
a) Define heat conduction.
b) Solve problems related to rate of heat

transfer, dQ  kA dT through a cross-
sectional ardet a. *onlydoxne material
c) Discuss graphs of temperature distance
T-x for heat conduction through
insulated. * one material & lagged material

a) Heat

 is the energy transferred
from one object to
another because of a
difference in
temperatures between the
two objects .

 Heat flows from higher
temperature to lower
temperature.

 Unit of heat is Joule (J) ,
kg m2 s–2

b) Thermal Conduction

a process whereby heat is transferred through a solid from a region of high
temperature to a region of lower temperature (Temperature difference, T)

• Heat can be transferred by • A lot of free electrons are available in a
three mechanisms, metal
• Conduction
• Convection • When one end (A) of the metal rod is
• Radiation heated, the free electrons will gain
additional thermal energy due to the
AB heating

Figure 1 • As a result, they move faster than before
and some of them collide with colder
• Thermal conduction in a molecules at B
insulator rod (non metal)
• This collision cause the cold end (B) of
• Consider a non-metal rod is the metal rod become hot
heated at its one end as
shown in Figure 1. • The transfer of energy by free electrons
is found to be faster compared to the
transfer of energy by vibrating
molecules in the lattice

• It is because electrons are lighter and
move faster

Rate of Heat Flow

Insulator

T1 A T2 T1  T2

Figure 2 x

• Consider: Heat conduction through a lagged rod which has
cross sectional area A and length dx as shown in Figure 2

• Assumption: No heat is lost to the surroundings,

• Effect: Heat can only flow through the cross sectional area
(From higher temperature region, T1 to lower temperature
region, T2)

• When the rod in steady condition, the rate of heat flows is
constant along the rod

Rate of Heat Flow

• A relationship between the where dQ : rate of heat flow
rate of heat flow with cross dt
sectional area &
temperature gradient: A: cross sectional area

dQ   A dT dT : temperature gradient
dt x x
k : thermal conductivity

dQ  kA dT OR • The negative sign indicates that
dt x heat always in the direction of
decreasing temperature
dQ  kA T2  T1
• It is a scalar quantity
dt x
• Unit: J s1 or watt (W)

Example 1

A glass window of cross-sectional ANSWER:

area 1.50 m2 and thickness 0.20 cm (i) Rate of heat flow:
is closed in winter. The
temperatures of the inner and dQ  kA dT  9450 W
outer surfaces of the window are
dt x

15˚C and 0˚C. (ii) To reduce the amount of heat
loss to the surroundings
(i) Calculate the rate of heat

flow through the window. through this window:

(ii) Suggest how you would 1. Increase the thickness of the

reduce the amount of heat glass, and

loss to the surroundings 2. Decrease the surface area of

through this window. the window

(kglass = 0.84 W m-1 K-1) 3. Cover the glass with an
insulator

Thermal Conductivity, k

• is defined as a rate of heat flows perpendicularly through the
unit cross sectional area of a solid, per unit temperature
gradient along the direction of heat flow

OR  dQ 
k    dt 
A dT 
 x

• Thermal conductivity is a property of conducting material
where good conductors will have higher values of k
compared to poor conductors

• It is a scalar quantity, Unit : W m1 K1 or W m1 C1.

• Table 1 shows value of k for various substances.

Substance k ( W m1 K1)
Silver
Copper 406.0
Steel 385.0
Glass 50.2
Wood
0.8
0.08

Table 1

Example 2

The rate of heat conduction is 80 From ANSWER:

kJ in an hour at a wall with an dQ  kA dT
area of 140 cm2. If the dt x

temperature gradient across the OR
wall is 15 0C m-1, calculate the
 dQ 
thermal conductivity of the wall. k    dt 

[105.82 W m-1 0C-1] A dT 
x

c) Temperature gradient in the lagged metal bar

insulator Y  If the metal bar XY was
completely covered
X
T1 T2 with a good insulator,
there will be no heat

insulator loss to the surroundings

Temperature,T T1  T2 along the bar

T1  This cause the

temperature gradient

will be constant along

the bar as shown in

Figures 3 and

T2 length, x dQ  constant
dt
0 along metal bar XY

Figure 3

8.2 IDEAL GAS EQUATIONS

a) State Gas’s Law
b) Sketch the following graphs of an ideal gas:

i. p-V graph at constant temperature
ii. V-T graph at constant pressure
iii. p-T graph at constant volume.
c) Explain the following graphs of an ideal gas:
i. p-V graph at constant temperature
ii. V-T graph at constant pressure
iii. p-T graph at constant volume.
d) State ideal gas equation
e) Use the ideal gas equation, PV  nRT

• An ideal gas is defined
as a perfect gas which
obeys the three gas laws
(Boyle’s, Charles’s and
Gay-Lussac’s) exactly

• It is a relationship
between MACROSCOPIC
behaviour of gas

P

Boyle’s Law Charles’ Gay-Lussac
P  1/V Law Law
VT
PT
IDEAL
GAS
LAW

Sketch and explain graphs
Boyle’s Law
For a fixed amount of gas at a constant
temperature, gas pressure is inversely
proportional to gas volume.

P 1
V

or P V = constant
or P1 V1 = P2 V2



• Graphs of the Boyle’s law.

a. P (kPa) b. PV

0 T2 0 T2
c. P (kPa) T1 d. PV T1
V(m3)
T2 P

T1 T2
T1
1 (m-3)
V
0 V0

Charles’s Law
For a fixed amount of gas at a constant
pressure, gas volume is directly proportional
to its absolute temperature.

V T

or V  constant
T

or V 1  V 2
T1 T2



-- the Charles’s Law can be represented
graphically as shown :

V (m3) V (m3)

-273.15°C

0 T (K) 0 T (°C)

Gay- Lussac / Pressure Law
For a fixed amount of gas at a constant
volume, gas pressure is directly
proportional to its absolute
temperature.

PT

or P  constant
T

or P1  P2
T1 T2



-- the relationship between
pressure & temperature
does not depends on the
type of gas being
considered.

-- the pressure law can be represented graphically
as shown :

P (Pa) P (Pa)

-273.15°C

0 T (K) 0 T (°C)

Ideal Gas Equation

 For n mole of an ideal gas, the  R  PVm  101.3103 0.0224
equation of state is written as T 273.15
R  8.31 J K 1 mol1
PV  nRT
R: molar gas constant (constant)
n: the number of mole gas

n m m: mass of gas  If the Boltzmann constant, k is
M M: molar mass of gas defined as

n N N: no. of molecules k  R  1.381023 J K 1
NA NA: Avogadro’s constant NA
= 6.02 x 1023 mol-1
 then the equation of state
 Consider 1 mole of gas at becomes

standard temperature and PV  NkT

pressure (S.T.P.), T = 273.15 K, P
= 101.3 kPa and Vm = 0.0224 m3

Example 3

What is the approximate number of atoms in a cubic
meter of an ideal monoatomic gas at a temperature of
27 °C and a pressure of 1 x105 Pa ?

Solution

P = 1 x105 Pa PV  nRT , n  N
T = ( 27 + 273.15 ) = 300.15 K NA
V = 1 m3 (cubic metre)
NA = 6.02x1023 mol-1 PV  ( N ) RT
R = 8.31 J K-1 mol-1 NA
number of atoms , N = ?
N  2.41x1025

Example 4 Solution: V  3V ; m  m;
A BA
The volume of vessel A is three T  300 K;T  500 K
times of the volume vessel B. The 0A 0B
vessels are filled with an ideal gas
and are at a steady state. The Since the vessels A and B are
temperature of vessel A and vessel
B are at 300 K and 500 K connected by a narrow tube thus
respectively as shown below. the pressure for both vessels is
same, finally i.e.
A B
PA  PB  P
(300 K) (500 K)
The system is in the steady state,
If the mass of the gas in the vessel
A is m, obtain the mass of the gas in thus T  T  300 K;
the vessel B in terms of m. 0A A
T  T  500 K
0B B

By applying the equation of state Vessel B:

for an ideal gas, m 500 mB R
M  M 
PV  nRT and n  PVB  (2)

PV   m RT
M 

Therefore, Vessel A : By equating the eqs. (1) and (2)

hence

PAVA   mA  RTA 100 m R  500 mB R
 M  M  M 

P3VB    m R300 mB  m
 M 5


PVB  100 m R (1)
 M 

Example 5
A small bubble rises from the bottom of a lake,
where the temperature and pressure are 8 °C and
6.4 atm to the water’s surface where the
temperature is 25 °C and the pressure is 1.0 atm.
Calculate the final volume (in m3) of the bubble if
its initial volume was 2.1 cm3.

Solution

initial
P1= 6.4 atm
V1= 2.1 cm 3 = 2.1x10 -6 m3
T1=(8 + 273.15)=281.15 K

final P2= 1.0 atm * n1 = n2
V2= ? m 3
T2=(25+273.15)=298.15 K

P1V 1  P2 V 2  V 2  P1V 1 x T 2
T1 T2 T1 P2

V 2  (6.4 atm) (2.1x106 m3 ) x 298.15K
281.15K 1.0 atm

V 2  1.425x105 m3

8.3 THERMODYNAMICS

a) State the first law of thermodynamics.
b) Solve problem related to first law of thermodynamics.
c) Define the thermodynamics processes

i. Isothermal
ii. Isochoric
iii. Isobaric
iv. Adiabatic
c) Interpret p-V graph for all the thermodynamics processes.

Introduction

Thermodynamics is the study of how energy can be
transferred between a gas system & its surroundings.

Effect of such energy transfer is the
change of internal energy of the
system.

This will affect the macroscopic
properties of the system like
pressure (P), volume (V) and
temperature (T).

a)&b) First Law of Thermodynamics

• States: “The heat (Q) • It is a statement of the
supplied to a system is equal conservation of energy
to the increase in the for a thermodynamic
internal energy, U of the system
system plus the work done,
W by the system on its Heat Q, energy provided
surroundings.” to a system not disappear

Q  U W but is transformed into
the internal energy of the
Q: heat added/ remove from the system system and the work the
∆U: change in internal energy of the system
W: work done by the system (on the system performed

surrounding )

(+)ve (–)ve

Q System absorb heat/ System loses heat/
Heat out of the system (Gas
Heat into the system (Gas expansion)
compression)
U Increase (T)
Decrease (T)
W Work done by the system
Work done on the system
+Q -Q
system

-W +W

Figure 5

• The change in the Figure 5
internal energy (U) of
a system during any
thermodynamic process
is independent of path

• For example a
thermodynamics system
goes from state 1 to
state 2 as shown in
Figure 5

U12  U142  U132

• The internal energy depends • The change in the
only on temperature of the internal energy also zero
system in the cyclic
thermodynamic process
• If the initial and final (repeated process)
temperature (state) of the because the initial and
system is the same, hence final state of the system
is the same
U  U2 U1  0
• The internal energy is
because U  f nRT not depend on the
2 volume of the system

thus U 2  U1

Example 6 Solution:
Q  U W
Calculate the change in
internal energy refer to U  Q W
the system below? U  150  (100)
U  50 J
system
∆U > 0 due to increase
Q = 150 J W= 100 J in internal energy

Example 7

Figure shows a p-V diagram of an ideal gas. Change in
internal energy from K to M is + 400 J, while the work done
when the gas undergoes a change following a path KLM is +
100 J. Calculate the heat flows into the system that follow
the path of KLM.

Answer 7

Calculate the heat flows into the system that follow the
path of KLM.

Knowing that ∆UKLM = ∆UKM = +400J

From the first law of thermodynamics,

QKLM = WKLM + ∆UKLM
QKLM = 100 J + 400 J
QKLM = 500 J

c)&d) Thermodynamics Processes

A thermodynamic process is process where the
parameters of a system such as pressure,
volume & temperature undergo changes.

This can happen due to either :

(a) Exchange of heat between system &
surrounding.

(b) Mechanical work being done on or by
the system ( ex: gas expand )

(c) Both of (a) & (b)

4 ways in which thermodynamic process can be
carried out in a controlled manner :

(a) Isothermal : a process that occurs at
constant temperature, T

(b) Isobaric : a process that occurs at
constant pressure, P

(c) Isochoric/ Isovolumetric : a process
that occurs at constant volume,V

(d) Adiabatic : a process involving no heat
flow in or out of the system

Isothermal Process
- a process that occurs at constant temperature

T2 300K
T1 100K

Since T remain the same after the process,
thus the internal energy, U of the gas does not
change.

U  0

In terms of 1st Law of Thermodynamic:

QUW
Q  0W
Q W

PVconstant

Gas that undergoes isothermal changes obey
Boyle’s law.

P1V 1  P2V 2

All the heat added to the gas goes into doing
work (expand or compress).

If the gas expand isothermally, thus V2>V1
- (W= positive)

If the gas compress isothermally, thus V2<V1
- (W= negative)

Work done in Isothermal process
 From the equation of state for an ideal gas,

PVnRTthen P  nRT

V

 Therefore the work done in the isothermal
process which change of volume from V1 to V2,
is given by

W V2 PdV
V1

WV2nRdTV
V1  V

WnRV2T1dV
V1V

 WnRlTnVVV12

WnRlnVT2lnV1

WnRlTnVV12

For isothermal process, the temperature of the

system remains unchanged, thus
V2  P1
P1V1 P2V2 V1 P2

The equation can be expressed as

WnRlTnPP12

45

The magnitude of the work done on the gas is
equal to the area under the curve.

W  nRT [ln V 2 ]
V1

or

W  nRT [ln P1 ]
P2

Isobaric Process
- is defined as a process that occurs at constant pressure

As the heated gas expands, its temperature
increase & gas crosses to higher temperature.

Temperature, T increase means internal
energy, U of the gas increases.

Work done in isobaric process
The work done during the isobaric process which
change of volume from V1 to V2 is given by

W V2 PdVand Pconstant
V1

WP V2 dV
V1

WPV2V1

OR

W PV

W  P(V 2 V 1) AurenadgerrPa-Vtphhe

Heat added to the gas goes into :
(1) work done by the gas
(2) changing the internal energy of the gas.

In terms of 1st Law of Thermodynamic:

QUW
Q  U  P(V 2 V 1)

Changes in volume at constant pressure obey
Charles’ Law.

V1  V2
T1 T2