CHAPTER 8 HEAT, GAS LAW and THERMODYNAMICS

Isovolumetric / Isochoric Process

- is defined as a process that occurs at
constant volume

No work is done since the area under such a
curve is zero. ( No change in volume )

W 0

All of the heat added to the gas goes
completely into increasing the gas’s internal
energy, U and therefore its temperature.

In terms of 1st Law of Thermodynamic:

QUW
Q  U  0

QUU2U1

Work done in Isovolumetric process

Since the volume of the system in
isovolumetric process remains unchanged, thus

dV  0

Therefore the work done in the isovolumetric
process is

WPdV0

Gay-Lussac’s law is obeyed during
isovolumetric (isochoric) process.

P1  P2
T1 T2

Adiabatic Process
- is defined as a process that occurs without
heat transfer into or out of a system

 500 K
 250 K

Since no heat is added to or removed from the
system ; thus

Q0

From the 1st Law of Thermodynamic :

QUW
0UW
W  U

During expansion, W is done by the gas at the
expense of its internal energy.

* Gas use it own internal energy to do work.

∆U negative – internal energy and thus
temperature, both decrease. Such an adiabatic
expansion is a cooling process.

Similarly, an adiabatic compression is a
warming process (temperature increase).

In practice to carry out an adiabatic
process :

1. Gas must be held in a container with
thick walls made of good insulator.

2. Process carried out fast – no enough
time for heat to flow through the walls of
the container.

Adiabatic change does NOT obey Boyle’s law. The
gas temperature is no longer constant.

QT- isothermal Adiabatic curve is steeper
QS - adiabatic than isothermal curve
because in adiabatic there
is a changes in internal
energy ( temperature
change) while in isothermal
internal energy, U remain
unchanged ( temperature
constant).

1300K

700K



Example 8

For the P-V graph below, name the following
processes:

(a) P to Q (b) Q to R [ 4 marks ]
(c) Q to T (d) Q to S

PQ – Isobaric
QR – Isovolumetric
QT – Isothermal
QS - Adiabatic

Example 9

On a P-T graph, sketch the general paths for the
following reversible processes for an ideal gas : (a)
Isothermal (b) Isobaric (c) Isometric

Solution (a)Isothermal

P (b)Isobaric

P2 (c)Isometric

(b)

(a) (c)

P1

T
T1 T2

Example 10

A vessel contains an ideal gas in condition A as
shown in figure.

Isovolumetric V constant
P decreases, T decreases
∆ UAB = - 2.49 kJ ; WAB = 0

P

Isobaric P constant
- Gas expands
V increases, T increases
∆ UBC = + 3.74 kJ

V1 V2

When the condition of the gas changes from
that of A to B, there is a change in internal
energy of 2.49 kJ. When the condition of the
gas changes from B to C, there is a change in
internal energy of 3.74 kJ.

Determine the total amount of heat transfer in
the process ABC. State whether heat has been
transferred into or out of the gas at the end of
the process ABC.

Solution
Process AB : Isovolumetric – WAB = 0
- Gas is being cooled, P ↓ ; T↓ , ∆U -ve

Apply 1st Law of Thermodynamics

QABUABWAB
QABUAB2.4k9J

Process BC : Isobaric
- Gas expands, V ↑ ; T ↑, ∆U +ve

QBCUBCWBC
UBCP(V2V1)

QBC3.713410153(301020212)0

3.741301.50130

QBC  5.24 kJ

The total amount of heat transfer in ABC

QABCQABQBC

2.491305.24130

QABC  2.75 kJ

2.75 kJ heat is transferred into the gas.