# CHAPTER 16 : THERMODYNAMICS

Mechanics

Oct 27, 2013 (4 years and 7 months ago)

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CHAPTER 16 :
THERMODYNAMICS

16.1 Work

Heat

Form of energy which is transferred from one
body to another body at lower temperature, by
virtue of the temperature difference between the
bodies.

Internal energy

Total energy content of a system. It is the sum of
all forms of energy possessed by the atoms and
molecules of the system.

16.1 Work

System

The collection of matter within prescribed
and identifiable boundaries.

Surroundings

Everything else in the environment.

Thermodynamics system

System that can interact with its
surroundings or environment in at least
two ways, one of which is heat transfer.

16.1 Work

Work

Assume a system in a piston that contains
fluid, the work done by the system as its
volume changes.

pA

dx

A

16.1 Work

When the piston moves out an infinitesimal
distance dx, the work dW done by this force is

A : cross
-
sectional area of the cylinder

p : pressure exerted by the system at the

surface of the piston.

F : total force exerted by the system on

the piston

16.1 Work

dW=pdV.

The work done in a
volume change from
V
1

to V
2
,

Graph p versus V

p

p

(a) p vs V at changes pressure

V

(a) p vs V at constant pressure

V

16.2 First Law of
Thermodynamics

The internal energy of
a system changes
from an initial value
U1 to a final value U2
due to heat, Q and
work, W:

Cyclic process

The change in internal energy depends on the
gas temperature.

It also depends on the initial and final state.

Cyclic process :

Initial state = Final state

U1 = U2

Q = W

Example:

Two moles of the monatomic argon gas
expand isothermally at 298 K, from an
initial volume of 0.025 m3 to a final volume
of 0.050 m3 (Argon is an ideal gas), find
(a) work done by the gas

(b) the heat supplied to the gas.

(a) W=+3400 J,

(b) Q =+3400 J

16.3 Thermodynamic
Processes

Isothermal process

Isochoric process

Isobaric process

Graph
Isothermal process

The work done by the system is the area
under the curve of a pV
-
diagram as a
graph shown below..

p

V

V
i

V
f

Work,W

p
i

p
f

Isothermal process

Isochoric process

a constant
-
volume process, where there is
no work done in the system, W = 0.

The first law of thermodynamic will be,

U2
-
U1 = ΔU = Q.

Graph
Isochoric process

The graph pV of an isochoric process is
shown below.

p

v

Isothermal process

a constant
-
temperature process.

Heat transfer and work in the system must not
change and the internal energy will be zero,
ΔU=0.

The changes in pressure,p and volume,V will
take place.

Isobaric process

Isobaric process is a constant
-
pressure
process.

W = p(V2

V1)

The first law of thermodynamic will be,

Q= ΔU +W = ΔU + p( V2
-
V1)

Graph Isobaric process

p

V

Isobaric process carried out at a constant pressure

a process with no
heat transfer to or
from a system, Q=0.

ΔU =
-
W

Example

The temperature of three moles of a monatomic
ideal gas is reduced from Ti = 540 K to Tf = 350
K by two different methods. In the first method
5500 J of heat flows into the gas, while in the
second method, 1500 J of heat flows into it. In
each case, find (a) the change in the internal
energy of the gas and (b) the work done by the
gas.

16.4 Molar Specific Heat Capacities At Constant
Pressure And Volume

Molar specific heat capacities of constant pressure, Cp

When an n mole of gas undergoes a heating process,

the equation that relates Q and the temperature changes,
dT

is written as ;

Q = nCdT ……………..(1)

Whereas Q = heat

n = number of mole

C = molar specific heat capacities

dT = temperature changes

When the gas is heated at constant
pressure, the (1) equation will be;

Q = nCpdT

Or Cp =

Where Cp = molar specific heat capacities at
constant

pressure.

Molar specific heat capacities of constant
volume, Cv

Hence, when an n mole of gas is heated at
constant volume, the (1) equation will be;

Q= nCvdT

Or

Where Cv = molar specific heat capacities at
constant volume

The universal gas constant, R

the substraction between Cp and Cv that
is;

Cp
-
Cv = R

Where R = 8314 J mol
-
1 K

Example

A certain perfect gas has specific heat
capacities as follows:

Cp = 0.84 kJ / mol K and

Cv = 0.657 kJ / mol K

Calculate the gas constant.

16.5 Specific Heat At Constant Pressure And
Volume

Example

A 2.00 mol sample of an ideal gas with γ =