Charles’ law

It is evident from Boyle’s law that, for a given gas, the product pV could be used as an indication of its temperature and, in fact, this is the basis of a scale of temperature. It can

Be shown that for an ideal gas, at constant pressure, the volume is related to the temperature in a linear fashion. Experimental results support this, and reference to Figure 2.3 shows just how this could be so. Suppose that experimental results allow a straight line to be drawn between two points A and B, as a graph of volume against temperature. If the line is extended to cut the abscissa at a point P, having a temperature of-273.15°C, it is clear that shifting the origin of the co-ordinate system to the left by 273.15°C will give an equation for the straight line, of the form

-273.15° fa 0° tb

Temperature °C (a)

-273.15° fa 0° tb
temperature °c (a)

Temperature in kelvin (b)

Fig. 2.3 Charles’ law and absolute temperature.

temperature in kelvin (b)
fig. 2.3 charles’ law and absolute temperature.
V=aT, <2-6)

2.6 The general gas law 9

Where T is the temperature on the new scale and a is a constant representing the slope of the line.


T = 273.15° + t (2.7)

This graphical representation of Charles’ law shows that a direct proportionality exists between the volume of a gas and its temperature, as expressed on the new abscissa scale. It also shows that a new scale of temperature may be used. This new scale is an absolute one, so termed since it is possible to argue that all molecular movement has ceased at its zero, hence the internal energy of the gas is zero and, hence also, its temperature is at an absolute zero. Absolute temperature is expressed in kelvin, denoted by K, and the symbol T is used instead of t, to distinguish it from relative temperature on the Celsius scale.


15 m3 s-1 of air at a temperature of 27°C passes over a cooler coil which reduces its temperature to 13°C. The air is then handled by a fan, blown over a reheater, which increases its temperature to 18°C, and is finally supplied to a room.

Calculate the amount of air handled by the fan and the quantity supplied to the room.


According to Charles’ law:

V= aT, that is to say,

Charles’ law

Hence, the air quantity handled by the fan

(273 + 13)

(273 + 27)

= 14.3 m3 s“1

And the air quantity supplied to the room

(273 + 18)

(273 + 27)

= 14.55 m3 s“1

One further comment, it is clearly fallacious to suppose that the volume of a gas is directly proportional to its temperature right down to absolute zero; the gas liquefies before this temperature is attained.

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