WBCHSE Class 11 Chemistry Notes For Boyle’s Law – Definition, Equation and Examples

WBCHSE Class 11 Chemistry Notes On Boyle’s Law Definition And Equation

Boyle’s Law

Robert Boyle, an Irish physicist, carried out a set of experiments to study the effect of pressure on the volume of a fixed mass of air. He devised a very simple apparatus, using a bent tube (J-tube), some mercury, and a measuring scale.

Boyle performed his experiments in a room where the temperature remained fairly constant. He observed how the length of the air trapped above the mercury in the closed limb of the tube varied with the pressure applied to it and found a definite quantitative relationship between the two.

“WBCHSE Class 11 Chemistry, Boyle’s law, definition, equation, and examples”

He increased the pressure on the trapped air by adding more mercury in the open limb of the tube. These observations showed that the length of the trapped air (consequently, the volume of air as volume is proportional to the length) varied inversely with the pressure applied on it.

Boyle worked with air, but it was found later that all gases behave the way the air trapped in Boyle’s tube did (at constant temperature). This relationship between the volume and pressure of a fixed amount of gas at constant temperature can be expressed mathematically as follows.

Read and Learn More WBCHSE For Class11 Basic Chemistry Notes

⇒ \(p \propto \frac{1}{V} or V \propto \frac{1}{p}(n, T constant)\)

V = \(\frac{k}{p}\)

or pV = k, where k is a constant.

Boyle’s Law Class 11

The value of the constant depends upon the amount of gas (n) and its temperature (T).

This quantitative relationship is called Boyle’s law and can be stated as follows. For a fixed amount of gas, the pressure is inversely proportional to the volume if the temperature remains constant. In other words, the product of the pressure and volume of a given mass of gas is constant, provided the temperature remains constant.

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“Boyle’s law, WBCHSE Class 11, chemistry notes, with derivation and formula”

Suppose the initial pressure and volume of a given mass of gas are p₁ and V₁ respectively at temperature T. If the pressure is then changed to p₂ while keeping the temperature constant and the volume becomes V₂ then according to Boyle’s law,

∴ \(p_1 V_1=p_2 V_2 \quad \text { or } \quad \frac{p_1}{p_2}=\frac{V_2}{V_1}\)

This implies that if the volume of the fixed amount of gas is doubled at a constant temperature, its pressure will reduce to half. This way of expressing Boyle’s law is useful while solving numerical problems.

Experimental verification of Boyle’s law: Boyle’s law can be verified experimentally by measuring the volumes of a given mass of gas at different pressures while keeping the temperature constant.

1. A plot of p against V at constant temperature is a hyperbola and it is called an isotherm. pV = constant, and the value of the constant depends upon n and T. Therefore, at every temperature for a fixed amount of gas, there is a separate pressure-volume curve that shows such curves at different temperatures for a fixed amount of gas.
2. A plot of pV against p at constant temperature is a straight line parallel to the x-axis which indicates that pV remains constant with changing pressure.
3. A graph of V against 1/p gives a straight line through the origin, which shows that volume increases uniformly with an increase in 1 /p or volume is inversely proportional to pressure.

Implications of Boyle’s law: The quantitative relationship discovered by Boyle between the pressure and volume of gas showed that a gas is compressible. When a given mass of a gas is compressed, the number of molecules it has does not change. They come closer and occupy less space. In other words, the gas becomes denser.

“WBCHSE Class 11, chemistry notes, on Boyle’s law, pressure-volume relationship”

This is why mountain air is rarer than the air at sea level. The air at sea level is denser because it is compressed by the mass of air above it. As one climbs higher and higher up a mountain, the pressure decreases and the density of air decreases.

This is why mountaineers have to carry a supply of oxygen with them. For the same reason, the size of a weather balloon increases as it ascends to higher altitudes.

A relationship can be obtained between density and pressure by using Boyle’s law. You already know that density (d) = \(\frac{\text { mass }}{\text { volume }} \frac{(m)}{(V)}\)

But pV = k (from Boyle’s law).

∴ V = \(\frac{k}{p} \quad \text { or } \quad d=\frac{m}{k / p}=\left(\frac{m}{k}\right) p\)

This shows that the pressure of a gas is directly proportional to its density.

Boyle’s Law Applications And Examples In WBCHSE Class 11 Chemistry

Example 1. 100 mL of CO₂ was collected at 27°C and 1 bar pressure. What would be the volume of the gas if the pressure changed to 0.96 bar at the same temperature?
Solution:

Given

100 mL of CO₂ was collected at 27°C and 1 bar pressure.

V₁ = 100 ml V₂ = ?

p₁ = 1 bar p₂ = 0.96 bar

“Boyle’s law, definition, mathematical equation, and real-life applications, WBCHSE Chemistry”

p₁V₁ = p₂V₂ (at constant temperature)

∴ \(V_2=\frac{p_1 V_1}{p_2}=\frac{1 \times 100}{0.96}=105.5 \mathrm{~mL}\)

The volume of carbon dioxide = 105.5 mL.

Example 2. A gas occupies a volume of 2.0 L tif 745 mmHg pressure. Calculate the additional pressure required to decrease the volume of the gas to 1.5 L.
Solution:

Given

A gas occupies a volume of 2.0 L tif 745 mmHg pressure

p₁ = 745 mm p₂ = ?

V₁ = 2.0L V₂ = 1.5 L

From Boyle’s law, pV = pV

∴ p₂ = \(\frac{p_1 V_1}{V_2}=\frac{745 \times 2}{1.5}=993.3 \mathrm{mmHg}\)

The additional pressure required = 993.3 – 745 = 248.3 mmHg.