At constant pressure, the volume of a given quantity of gas is directly proportional to the absolute temperature : V ∝ T (in Kelvin)

So at constant pressure, if the temperature (K) is doubled, the volume of gas is also doubled.

OR

At constant pressure for a given quantity of gas, the ratio of its volume and the absolute temperature is a constant : ^{V}/_{T} = constant, ^{V}/_{T} = k

At constant pressure for a given quantity of gas : ^{Vi}/_{Ti} = ^{Vf}/_{Tf}

where V_{i} is the initial (original) volume, T_{i} is its initial (original) temperature (in Kelvin), V_{f} is its final volme, T_{f} is its final tempeature (in Kelvin)

V_{i} and V_{f} must be in the same units of measurement (eg, both in litres), T_{i} and T_{f} must be in Kelvin NOT celsius.

temperature in kelvin = temperature in celsius + 273 (approximately)

All gases approximate Charles' Law at high temperatures and low pressures.

A hypothetical gas which obeys Charles' Law at all temperatures and pressures is called an Ideal Gas.

A Real Gas is one which approaches Charles' Law as the temperature is raised or the pressure lowered.

As a Real Gas is cooled at constant pressure from a point well above its condensation point, its volume begins to increase linearly.
As the temperature approaches the gases condensation point, the line begins to curve (usually downward) so there is a marked deviation from Ideal Gas behaviour close to the condensation point.
Once the gas condenses to a liquid it is no longer a gas and so does not obey Charles' Law at all.
Absolute zero (0K, -273^{o}C approximately) is the temperature at which the volume of a gas would become zero if it did not condense and if it behaved ideally down to that temperature.

Graphical Representations

Expansion of Hydrogen gas at constant pressure

Volume (mL)

Temperature (^{o}C)

Temperature (K)

^{V}/_{T (K)}

Graph

25

-23

250

0.1

30

27

300

0.1

35

77

350

0.1

40

127.5

400.5

0.1

45

177

450

0.1

Calculations : ^{Vi}/_{Ti} = ^{Vf}/_{Tf}

A sample of gas at 101.3 kPa had a volume of 1.2 L at 100^{o}C. What would its volume be at 0^{o}C at the same pressure?

V_{i} = 1.2L V_{f} = ?

T_{i} = 100^{o}C = 100 + 273 = 373 K T_{f} = 0^{o}C = 0 + 273 =273 K

1.2/373 = ^{Vf}/273

3.22 x 10^{-3} = ^{Vf}/273

V_{f} = 3.22 x 10^{-3} x 273 = 0.88 L (880 mL)

A balloon had a volume of 75 L at 25^{o}C. To what does the temperature need to raised in order for the balloon to have a volume of 100 L at the same pressure?

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