Mechanical Engineering Principles

IDEAL GAS LAWS 259

Hence,force on the piston
=( 750 × 103 Pa)( 300 × 10 −^4 m^2 )= 22 .5kN

Now try the following exercise

Exercise 119 Further problems on Boyle’s

law

1. The pressure of a mass of gas is increased
   from 150 kPa to 750 kPa at constant tem-
   perature. Determine the final volume of
   the gas, if its initial volume is 1.5 m^3.
   [0.3 m^3 ]


2. In an isothermal process, a mass of gas
   has its volume reduced from 50 cm^3 to
   32 cm^3. If the initial pressure of the gas
   is 80 kPa, determine its final pressure.
   [125 kPa]


3. The piston of an air compressor com-
   presses air to^14 of its original volume
   during its stroke. Determine the final pres-
   sure of the air if the original pressure is
   100 kPa, assuming an isothermal change.
   [400 kPa]


4. A quantity of gas in a cylinder occupies a
   volume of 2 m^3 at a pressure of 300 kPa.
   A piston slides in the cylinder and com-
   presses the gas, according to Boyle’s law,
   until the volume is 0.5 m^3. If the area of
   the piston is 0.02 m^2 , calculate the force
   on the piston when the gas is compressed.
   [24 kN]

23.3 Charles’ law

Charles’ lawstates:

for a given mass of gas at constant pressure, the vol-

ume V is directly proportional to its thermodynamic

temperature T,

i.e.V∝TorV=kT or

V

T

=k

at constant pressure, where

T=thermodynamic temperature in Kelvin (K).

A process that takes place at constant pressure is

called anisobaricprocess. The relationship between

the Celsius scale of temperature and the thermody-

namic or absolute scale is given by:

kelvin =degrees Celsius+ 273

i.e. K=°C+^273 or°C=K−^273

(as stated in Chapter 19).

If a given mass of gas at a constant pressure

occupies a volumeV 1 at a temperatureT 1 and a

volumeV 2 at temperatureT 2 ,then

V 1

T 1

=

V 2

T 2

Problem 4. A gas occupies a volume of

1.2 litres at 20°C. Determine the volume it

occupies at 130°C if the pressure is kept

constant.

Since the change occurs at constant pressure (i.e. an

isobaric process), Charles’ law applies,

i.e.

V 1

T 1

=

V 2

T 2

whereV 1 = 1 .2l,T 1 = 20 °C=( 20 + 273 )K=

293 K andT 2 =( 130 + 273 )K=403 K.

Hence,

1. 2

293

=

V 2

403

from which,volume at 130°C,V 2 =

( 1. 2 )( 403 )

293

= 1 .65 litres

Problem 5. Gas at a temperature of 150°C

has its volume reduced by one-third in an

isobaric process. Calculate the final

temperature of the gas.

Since the process is isobaric it takes place at constant

pressure and hence Charles’ law applies,

i.e.

V 1

T 1

=

V 2

T 2

whereT 1 =( 150 + 273 )K=423 K andV 2 =^23 V 1.