WBCHSE Class 11 Physics Expansion Of Gases Boyles Law Notes

Expansion Of Gases Boyles Law

Boyle’s Law Notes for Class 11 WBCHSE

Boyle’s law: The relation between the volume and the pressure of a gas, at constant temperature, was first proposed by the British scientist Robert Boyle in 1660 AD, and that relation is called Boyle’s law.

Statement: At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.

Mathematically, if V is the volume of a gas of certain mass, and p is its pressure, then according to Boyle’s law,

⇒ \(V \propto \frac{1}{p}\), when temperature remains constant.

or, \(V=\frac{k}{p}\) or, pV = k = constant ….(1)

The value of k depends on the mass and the temperature of the gas. Hence, the product of the pressure and the volume of a fixed mass of gas at a fixed temperature remains constant.

Thus at constant temperature, if V₁, V₂, V₃ ….. are volumes of a fixed mass of a gas at pressures p₁, p₂, p₃ …., then according to this law,

∴ \(p_1 V_1=p_2 V_2=p_3 V_3 \cdots\) =k (a constant)

Graphical representation of Boyle’s law: Boyle’s law can be expressed through different graphs:

1. p- V graph: Plotting volumes V of a fixed mass of gas, at constant temperature, along the horizontal axis and the corresponding pressures along the vertical axis, the obtained graph is a rectangular hyperbola, Such a graph is called the isothermal of the gas. At different values of fixed temperatures, the isothermals are different rectangular hyperbolas.

2. \(\frac{1}{V}-p\) graph: For a fixed mass of a gas at a constant temperature, the graph of pressure p plotted along the horizontal axis and reciprocal I of volume V plotted along the vertical axis, is a straight line passing through the origin. At different values of fixed temperatures, \(\frac{1}{V}-p\) graphs will be different straigth lines.

A V-\(\frac{1}{p}\) graph is also of the same nature as that of a p-^ graph.

Understanding Boyle’s Law in Gas Expansion

3. pV-p graph: Graph obtained, by plotting pressures p of a fixed mass of gas at constant temperature along the horizontal axis and the corresponding products pV along the vertical axis, is a straight line parallel to the p axis, At different fixed temperatures, pV-p graphs will be different parallel lines.

At constant temperature, the pV- V graph is also similar.

These graphs, obtained experimentally, verify Boyle’s law.

Expansion Of Gases Boyles Law Numerical Examples

Example 1. The volume of a fixed mass of gas at STP is 500 cm³. What will its volume be at 700 mmHg pressure If its temperature remains constant?
Solution:

The volume of a fixed mass of gas at STP is 500 cm³.

As temperature remains constant, Boyle’s law is applicable.

Given, p₁ = 76 cmHg, V₁ = 500 cm³, p₂ = 70 cmHg.

∴ p₁V₁= p₂V₂

∴ 76 x 500 = 70 x V₂

or, \(V_2=\frac{76 \times 500}{70}=542.86 \mathrm{~cm}^3\)

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Practice Questions on Boyle’s Law for Class 11

Example 2. While tabulating the pressures and volumes for a fixed mass of a gas at a fixed temperature, a student forgets to record a few observations, as shown below. Fill in the blanks.

Solution:

Since the temperature is fixed, Boyle’s law is applicable here, i.e., pV = constant.

From reading 1 and 3, p₁V₁ = 100 x 80 = 8000 and p₃V₃ = 200 x 40 = 8000. Hence the value of the constant in this case is 8000.

∴ For reading (2), \(125 \times V_2=8000 \text { or, } V_2=\frac{8000}{125}=64 \mathrm{~cm}^3\)

and for reading (4) \(p_4 \times 32=8000 \text { or, } p_4=\frac{8000}{32}=250 \mathrm{mmHg} \text {. }\)

Example 3. The volume of a gas at 1 standard atmosphere is compressed to 1/6 th of its value at constant temperature. What will be its final pressure?
Solution:

Given

The volume of a gas at 1 standard atmosphere is compressed to 1/6 th of its value at constant temperature.

Let the initial volume at 1 standard atmosphere be x cm³ and the final pressure be p.

The temperature remains constant; so applying Boyle’s law, we get, 1 x x = p x x/6 or, p = 6 standard atmospheres.

Examples of Boyle’s Law in Everyday Life

Question 4. A 100 cm long vertical cylinder, closed at the bot¬tom end, has a movable, frictionless, air-tight disc attached at its other end. An ideal gas is confined within the cylinder. Initially, when the disc between the confined gas and atmosphere is in equilibrium, the gas column length is 90 cm. Mercury is poured slowly on the disc. When the disc descends by 32 cm, mercury over it is just about to overflow. Find the atmospheric pressure if the operation took place at a constant temperature of the gas. Neglect the weight or thickness of the disc.
Solution:

Given

A 100 cm long vertical cylinder, closed at the bot¬tom end, has a movable, frictionless, air-tight disc attached at its other end. An ideal gas is confined within the cylinder. Initially, when the disc between the confined gas and atmosphere is in equilibrium, the gas column length is 90 cm. Mercury is poured slowly on the disc. When the disc descends by 32 cm, mercury over it is just about to overflow.

Let the atmospheric pressure be p cmHg and α = area of the cross-section of the cylinder. Initially in equilibrium, the pressure of confined gas = atmospheric pressure p.

Now, the volume of enclosed gas V = 90α cm³.

When mercury just overflows from the cylinder, pressure of the gas p₁ = (p + 42) cm and volume V₁

= (90 – 32)α = 58α cm³.

As the temperature is constant, according to Boyle’s law, \(p V=p_1 V_1\)

or, \(p \times 90 \alpha=(p+42) \times 58 \alpha\)

90p = \(58 p+58 \times 42\)

or, \(32 p=58 \times 42\)

or, p = \(\frac {58 \times 42}{32}=76.125 \mathrm{cmHg}\)