At constant temperature, the volume of a given quantity of gas is inversely proportional to its pressure : V ∝ ^{1}/_{P}

So at constant temperature, if the volume of a gas is doubled, its pressure is halved.
OR
At constant temperature for a given quantity of gas, the product of its volume and its pressure is a constant : PV = constant, PV = k

At constant temperature for a given quantity of gas : P_{i}V_{i} = P_{f}V_{f}

where P_{i} is the initial (original) pressure, V_{i} is its initial (original) volume, P_{f} is its final pressure, V_{f} is its final volume
P_{i} and P_{f} must be in the same units of measurement (eg, both in atmospheres), V_{i} and V_{f} must be in the same units of measurement (eg, both in litres).

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

A hypothetical gas which obeys Boyle's Law at all temperatures and pressures is called an Ideal Gas.
A Real Gas is one which approaches Boyle's Law behaviour as the temperature is raised or the pressure lowered.

Graphical Representations

Compression of Hydrogen gas at 25^{o}C

Pressure (mm Hg)*

Volume (mL)

P x V

Graph

760

23

1.75 x 10^{4}

912

19.2

1.75 x 10^{4}

1064

16.4

1.75 x 10^{4}

1216

14.4

1.75 x 10^{4}

1368

12.8

1.75 x 10^{4}

1520

11.5

1.75 x 10^{4}

Volume (mL)

1/Pressure (1/mm Hg)*

Graph

11.5

6.6 x 10^{-4}

12.8

7.3 x 10^{-4}

14.4

8.2 x 10^{-4}

16.4

9.4 x 10^{-4}

19.2

1.1 x 10^{-3}

23

1.3 x 10^{-3}

* A pressure of 760 mm Hg is equal to 1 atmosphere (atm) or 101.3 kilopascals (kPa)

Calculations : P_{i}V_{i} = P_{f}V_{f}

A certain mass of gas occupies a volume of 2.5L at 90 kPa pressure. What pressure would the gas exert if it were placed in a 10.0 L container at the same temperature?
P_{i} = 90 kPa V_{i} = 2.5 L
P_{f} = ? V_{f} = 10.0 L
P_{i}V_{i} = P_{f}V_{f} 90 x 2.5 = P_{f} x 10.0
225 = P_{f} x 10.0
225 ÷ 10.0 = P_{f} P_{f} = 22.5 kPa

4.5L of gas at 125 kPa is expanded at constant temperature until the pressure is 75 kPa. What is the final volume of the gas?
P_{i} = 125 kPa V_{i} = 4.5 L
P_{f} = 75 kPa V_{f} = ?
P_{i}V_{i} = P_{f}V_{f} 125 x 4.5 = 75 x V_{f} 562.5 = 75 x V_{f} V_{f} 562.5 ÷ 75 = 7.5 L

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