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Charles' Law

Key Concepts

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 : ^V_i/_{T_i} = ^V_f/_{T_f}
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^oC 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 (^oC)	Temperature (K)	^V/_{T (K)}
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 : ^V_i/_{T_i} = ^V_f/_{T_f}

A sample of gas at 101.3kPa had a volume of 1.2L at 100^oC. What would its volume be at 0^oC at the same pressure?
V_i = 1.2L V_f = ?
T_i = 100^oC = 100 + 273 = 373K T_f = 0^oC = 0 + 273 =273K
1.2/373 = ^V_f/273
3.22 x 10^-3 = ^V_f/273
V_f = 3.22 x 10^-3 x 273 = 0.88L (880mL)
A balloon had a volume of 75L at 25^oC. To what does the temperature need to raised in order for the balloon to have a volume of 100L at the same pressure?
V_i = 75L V_f = 100L
T_i = 25^oC = 25 + 273 = 298K T_f = ? (K)
^V_i/_{T_i} = ^V_f/_{T_f}
75/298 = 100/T_f
0.2517 = 100/T_f
T_f = 100/0.2517 = 397K (397-273 = 124^oC)