The study of behaviuor of gases has led to the formulation of a few important generalisations called as Gas Laws.
Boyle's Law
Robert Boyle proposed this law in the year 1662, giving the relationship between pressure and volume of given mass of a gas at constant temperature. This law states that volume (V) of a given mass of gas is inversely proportional to the pressure (P) at constant temperature.
Mathematically it can be expressed as,
V a (1/P) (constant T) ……. (i)
or PV = k = constant
At a given temperature, when the pressure of the gas is changed from
P1 to P2 the relation becomes
P1 V1 = P2 V2 = constant
where, V2 is the new volume of the gas. The product of volume and pressure for a given mass of a gas at constant temperature is constant. This aspect can be experimentally verified by taking the pressure volume data for a gas like 10g of oxygen at 25°C. It is observed that with the increase in pressure the volume decreases and the product 'PV' remains constant. This data when plotted with 'P' along the x-axis and 'V' long the y-axis gives a curve.
Fig: - Variation of P and V at constant T
The curve shows the inverse relation of 'P' and 'V'. When the pressure is P1, the volume is V1 and when the pressure is increased to P2, the volume V2 is smaller than V1. If the graph is plotted between 'P' and
1 / V, a straight line passing through the origin is obtained. On plotting the product 'PV' along y-axis and Pressure 'P' along x-axis, a horizontal line is obtained, indicating 'PV' to be constant even if we change pressure.
Fig: - Plot of P versus 1/V
The P-V curve for a given gas is different at different temperatures. The plot of 'PV' against 'P' at different temperature is known as Isotherms. The higher curve corresponds to higher temperature. Boyle's law expresses the compressible nature of gas, which gives a measure of its increased density.
Problem
1. A gas occupies a volume of 250 mL at 745 mm Hg and 25°C. What additional pressure is required to reduce the gas volume to 200mL at the same concentration?
Solution
P1 = 745 mm Hg V1 = 250 mL
P2 = ? V2 = 200mL
P1V1 = P2V2
P2 = (1V1 / V2) = {(745mm Hg x (250 ml) / (200 ml)} = 931.25 mm Hg
The additional pressure required is 931.25 - 745 = 186.25 mm.
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