Mechanical Engineering Principles

IDEAL GAS LAWS 261

Initial pressurepR=200 kPa, and the pressure of
gasesRandQtogether,p=pR+pQ=320 kPa.

By Dalton’s law of partial pressure, the pressure of
gasQalone ispQ=p−pR= 320 − 200 =120 kPa

Now try the following exercise

Exercise 122 A further problem on Dal-

ton’s law of partial pressure

1. A gasAin a container exerts a pressure of
   120 kPa at a temperature of 20°C. GasB
   is added to the container and the pressure
   increases to 300 kPa at the same temper-
   ature. Determine the pressure that gasB
   alone exerts at the same temperature.
   [180 kPa]

23.6 Characteristic gas equation

Frequently, when a gas is undergoing some change,
the pressure, temperature and volume all vary simul-
taneously. Provided there is no change in the mass
of a gas, the above gas laws can be combined, giving

p 1 V 1

T 1

=

p 2 V 2

T 2

=k where k is a constant.

For an ideal gas, constantk=mR, wheremis the
mass of the gas in kg, andRis thecharacteristic
gas constant,

i.e.

pV

T

=mR

or pV=mRT

This is called thecharacteristic gas equation.In
this equation,p = absolute pressure in Pascal’s,
V=volume in m^3 ,m=mass in kg,
R = characteristic gas constant in J/(kg K), and
T=thermodynamic temperature in Kelvin.
Some typical values of the characteristic gas
constant R include: air, 287 J/(kg K), hydrogen
4160 J/(kg K), oxygen 260 J/(kg K) and carbon
dioxide 184 J/(kg K).
Standard temperature and pressure(i.e.STP)
refers to a temperature of 0°C, i.e. 273 K, and
normal atmospheric pressure of 101.325 kPa.

23.7 Worked problems on the

characteristic gas equation

Problem 8. A gas occupies a volume of

2.0 m^3 when at a pressure of 100 kPa and a

temperature of 120°C. Determine the volume

of the gas at 15°C if the pressure is increased

to 250 kPa.

Using the combined gas law:

p 1 V 1

T 1

=

p 2 V 2

T 2

whereV 1 = 2 .0m^3 ,p 1 =100 kPa,p 2 =250 kPa,

T 1 =( 120 + 273 )K=393 K and

T 2 =( 15 + 273 )K=288 K, gives:

( 100 )( 2. 0 )

393

=

( 250 )V 2

288

from which,volume at 15°C,

V 2 =

( 100 )( 2. 0 )( 288 )

( 393 )( 250 )

= 0 .586 m^3

Problem 9. 20 000 mm^3 of air initially at a

pressure of 600 kPa and temperature 180°C

is expanded to a volume of 70 000 mm^3 at a

pressure of 120 kPa. Determine the final

temperature of the air, assuming no losses

during the process.

Using the combined gas law:

p 1 V 1

T 1

=

p 2 V 2

T 2

where V 1 =20 000 mm^3 ,V 2 =70 000 mm^3 ,

p 1 =600 kPa,p 2 =120 kPa, and

T 1 =( 180 + 273 )K=453 K, gives:

( 600 )(20 000)

453

=

( 120 )(70 000)

T 2

from which,final temperature,

T 2 =

( 120 )(70 000)( 453 )

( 600 )(20 000)

=317 Kor 44 °C