Gas Laws

INTRODUCTION

Matter can be classified into three categories depending upon its physical state namely solid, liquid and gaseous states. Solids have a definite volume and shape; liquids also have a definite volume but no definite shape; gases have neither a definite volume nor a definite shape.

DISTINCTION BETWEEN THREE STATES OF MATTER

MEASURABLE PROPERTIES OF GASES

Mass, volume, temperature are the important measurable properties of gases.

$\bullet$Mass: The mass of the gas is related to the number of moles as

n = $\frac{w}{M}$

Where n = number of moles

w = mass of gas in grams

M = molecular mass of the gas

$\bullet$Volume: Since gases occupy the entire space available to them, therefore the gas volume means the volume of the container in which the gas is enclosed.

Units of Volume: Volume is generally expressed in litre or cm³ or dm³ 1m³ = 10³ litre

= 10³ dm³ = 10⁶ cm³.

$\bullet$Pressure: The force exerted by the gas per unit area on the walls of the container is equal to its pressure.

Units of Pressure: The pressure of a gas is expressed in atm, Pa, Nm^–2, bar or,

lb/In² (psi).

760 mm = 1 atm = 10132.5 KP_­a = 101325 P_a = 101325 Nm^–2

760 of Hg = 1.01325 bar = 1013.25 milli bar = 14.7 lb/2n² (psi)

$\bullet$Temperature: Temperature is defined as the degree of hotness. The SI unit of temperature is Kelvin. On the Celsius scale water freezes at 0°C and boils at 100°C where as in the Kelvin scale water freezes at 273 K and boils at 373 K.

GAS LAWS

The state of a sample of gas is defined by 4 variables i.e. P, V, n & T. Gas laws are the simple relationships between any two of these variables when the other two are kept constant.

Boyle's Law: The changes in the volume of a gas by varying pressure at a constant temperature of a fixed amount of gas was quantified by Robert Boyle in 1662. The law was named after his name as Boyle's law. It states that:

The volume of a given mass of a gas is inversely proportional to its pressure at a constant temperature.

Mathematically

P $\propto$$\frac{1}{V}$ (n, T constant)

V $\propto$ $\frac{1}{P}$ (n, T constant)

i.e. P = $\frac{K}{V}$ (where K is the constant proportionality)

or PV = K (constant)

Let V₁ be the volume of a given mass of the gas having pressure P₁ at temperature T. Now, if the pressure is changed to P₂ at the same temperature, let the volume changes to V₂. The quantitative relationship between the four variables P₁, V₁, P₂ and V₂ is:

P₁V₁ = P₂V₂ (temperature and mass constant)

Graphical Representation of Boyle's Law

$\bullet$Fig. (a) show the plot of V vs P at a particular temperature. It shows that P increases

V decreases.

$\bullet$Plot (b) shows the plot of PV vs P at particular temperature. It indicates that PV value remains constant inspite of regular increase in P.

Illustration 1. A sample of gas occupies 100 litres at 1 atm pressure and at 0°C. If the volume of the gas is to be reduced to 5 litres at the same temperature, what additional pressure must be applied?

Solution: Here P₁ = 1 atm P₂ = ?

V₁ = 100 litre V₂ = 5 litre

T₁ = 273 K T₂ = 273 K

As temperature is constant

Then from Boyle's law

P₁V₁ = P₂V₂

or 1 x 100 = P₂ x 5

$\therefore$ P₂ = 20 atm

Charle's Law: The French Scientist, Jacques Charles in 1787 found that for a fixed amount of a gas at constant pressure, the gas expands as temperature increases. The law can be stated as the volume of a given mass of a gas increases or, decrease by 1/273 of its volume at 0°C for each degree rise or, fall of temperature, provided pressure is kept constant.

Charles also found that for a given mass of a gas if pressure is kept constant, the volume increases linearly with temperature.

V = V₀ (1 + $\alpha$t) or V – V₀ = V₀ $\propto$ t

If the temperature is measured in the Celsius scale and V₀ is the volume at 0°C, it is found that a = 1/273. The volume at temperature T is then;

V_t = V₀ $\left(1+\frac{t}{273}\right)$ = V₀ $\left(\frac{273+t}{273}\right)$ V_T = $\frac{V_0}{273}T$

Where T = 273 + t is the temperature on the Kelvin scale, which has – 273°C as its zero point.

Let V₁ be the volume of a certain mass of a gas at temperature T₁ and at pressure P. If temperature is changed to T₂ keeping pressure constant, the volume changes to V₂. The relationship between for variables V₁, T₁, V₂ and T₂ is:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$ (Pressure and Mass Constant)

Graphical Representation of Charle's Law $\therefore$Additional pressure that should be applied = P₂ – P₁ = 20 – 1 = 19 atm

COMBINED GAS EQUATION

The Boyle's and Charles' law can be combined to give a relationship between the three variables P, V and T. Let a certain amount of a gas in a vessel have a volume V₁, pressure P₁ and temperature T₁. On changing the temperature and pressure to T₂ and P₂ respectively, the gas occupies a volume V₂.

Then we can write

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

The above relation is called the combined gas law.

Avogadro's Law: The Avogadro's law states that at a given temperature and pressure, the volume of a gas is directly proportional to the amount of gas i.e.

V $\propto$ n (P and T constant)

or V = constant x n

Where n is the amount of the substance

It was said that 1 mol of any gas at 0°C and under 1 atm pressure occupies 22.4$\times$10^–3 m³ or 22.4 litre.

Avogadro further generalised the statement that a mole of any substance contains 6.022$\times$10²³ particles (molecules, atoms or any other entities).

IDEAL GAS EQUATION

A gas that would obey Boyle's and Charle's law under the conditions of temperature and pressure is called an ideal gas.

Here, we combine four measurable variables P, V, T and n to give a single equation.

V $\propto$ n [P, T constant] Avogadro's law

V $\propto$ T [n, P constant] Charle's law

V $\propto$ $\frac{1}{P}$[n, T constant] Boyle's law

The combined gas law can be written as $V\propto \frac{nT}{P}$ or PV $\propto$ nT

PV = nRT

this is called ideal gas equation

where R is the constant of proportionality or universal gas constant

The value of R was found out to be

R = 8.314 J mol^–1 K^–1

R = 0.0821 litre atm K^–1 mol^–1

R = 2 cal K^–1 mol^–1

Relation between molar mass and density (From ideal gas equation)

$n=\frac{W}{M}$

$P=\frac{nRT}{V}$

$P=\frac{W}{V}\frac{RT}{M}$

$P=d\frac{RT}{M}$ $\Rightarrow$ $d=\frac{PM}{RT}$

Illustration 2. What is the increase in volume when the temperature of 600 ml of air increases from 27°C to 47°C under constant pressure?

Solution: Charle's law is applicable as the pressure and amount remains constant.

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$or

V_­1 = $\frac{T_1}{T_2}\times V_2$

V₁ = $\frac{320}{300}\times 600$ = 640 ml

Increase in volume of air = 640 – 600 = 40

Gay Lussac's Law (temperature pressure law)

It state that pressure of the given of mass a gas is directly proportional to the Kelvin temperature at constant volume

$P\propto T$ (n.v. are constant)

$\frac{P}{T}=K$ $\Rightarrow$ $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

PAY-LOAD

When a balloon is buoyant, it can take along some weight into the upper atmosphere. The maximum weight a balloon can carry along is called its pay-load.

Pay load = weight of the air displaced - (weight of the balloon + weight of the gas it contains)

Illustration 3. A flask of 2 dm³ capacity contains O₂ at 101.325 kPa and 300 K. The gas pressure is reduced to 0.1 Pa. Assuming ideal behaviour, answer the following:

(i) What will be the volume of the gas which is left behind?

(ii) What amount of O₂ and the corresponding number of molecules are left behind in the flask?

(iii) If now 2g of N₂ is introduced, what will be the pressure of the flask?

Solution: Given that

V₁ = 2dm³ P_t = 101.325 kPa

P₂ = 0.1 Pa T = 300 K

We have the following results.

(i) The volume of O₂ left behind will be the same, i.e. 2dm³

(ii) The amount of O₂ left behind is given by

n = P₂ $\frac{V_1}{RT}=\frac{10^{-4}kpa\times 2dm}{\left(8.314kpad{m}^3{k}^{-1}mo{l}^{-1}\right)\left(300K\right)}$

$=8.019\times 10^{-8}mol$

(iii) 2g of N₂ = 1/14 mol

Total amount of gases in flask, 1/ 14 mol + 8.125$\times$10^–8 mol $\approx$ 1/14 mol.

Thus, the pressure of the flask is given by

P = $\frac{nRT}{V}$ = $\frac{\left(\frac{1}{14\overrightarrow{\leftarrow }mol}\right)\times \left(8.315kpad{m}^3{k}^{-1}mo{l}^1\times 300k\right)}{\left(2d{m}^3\right)}$ = 89.08 kPa.

DALTON'S LAW OF PARTIAL PRESSURE

The relation between the pressures of the mixture of non-reacting gases enclosed in a vessel to their individual pressure is described in the law. The law was given by John Dalton in 1807. It states that.

At constant temperature, the pressure exerted by a mixture of two or more non-reacting gases enclosed in a definite volume, is equal to the sum of the individual pressures which each gas would exert if present alone in the same volume.

The individual pressures of gases are known as partial pressures.

If P is the total pressure of the mixture of non-reacting gases at temperature T and volume V, and P₁, P₂, P₃ …. represent the partial pressures of the gases, then

P = P₁ + P₂ + P₃+ ……… (T, V are constant)

Partial Pressure in terms of Mole Fraction

Mole fraction $\chi$ defines the amount of a substance in a mixture as a fraction of total amount of all substances. If n=1 moles of way substance is present in n moles of the mixture, then mole fraction of the substance, $\chi$₁ = $\frac{n_1}{n}$

If $P_{N_2}$ is the partial pressure of nitrogen the mixture of SO₂ and N_­2. Then

$P_{N_2}=n_{N_2}\times \frac{RT}{V}$ (($n_{N_2}$= no. of moles of N₂)

and P_mixture = $\left(n_{N_2}+n_{S{O}_2}\right)\frac{RT}{V}$

dividing we get,

$\frac{P_{N_2}}{P_{mixture}}=\left(\frac{n_{N_2}}{n_{N_2}+n_{S{O}_2}}\right)={\chi }_{N_2}$

or, $p_{N_2}={\chi }_{N_2}\times P_{mixture}$

Illustration 4. A 2.5 litre flask contains 0.25 mole each of sulphur dioxide and nitrogen gas at 27°C. Calculate the partial pressure exerted by each gas and also the total pressure.

Solution: Partial pressure of SO₂

$P_{S{O}_2}=\frac{nRT}{V}=\frac{0.25\times 8.314\times 300}{2.5\times {10}^{-3}}$

= 2.49$\times$10⁵ Nm^–2 = 2.49$\times$10⁵ P_a

Similarly $P_{N_2}$. = 2.49$\times$10⁵ P_a

Following Dalton's Law

P_Total = $P_{N_2}$ + $P_{S{O}_2}$

= 2.49$\times$10⁵ P_a + 2.49$\times$10⁵ P_a = 4.98$\times$10⁵ P_a

GRAHAM'S LAW OF DIFFUSION/EFFUSION

The ability of a gas to spread and occupy the whole available volume irrespective of other gases present in the container is called diffusion.

Effusion is the process by which a gas escapes from one chamber of a vessel through a small opening or an orifice. $\text{r = }\frac{\text{volume diffused }}{timetaken}=\frac{V}{T}$

Thomas Graham in 1831 proposed the law of gaseous diffusion. The law states under similar conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional to the square roots of their densities.

$r\propto \frac{1}{\sqrt{d}}$where is the rate of diffusion and d is the density of the gas.

Now, if there are two gases A and B having r₁ and r₂ as their rates of diffusion and d₁ and d₂ their densities respectively. Then

$r_1\propto \frac{1}{\sqrt{d_1}}$and $r_2\propto \frac{1}{\sqrt{d_2}}$ or, $\frac{r_1}{r_2}=\sqrt{\frac{d_2}{d_1}}$(at same T and P)

$\frac{r_1}{r_2}=\sqrt{\frac{\frac{M_2}{2}}{\frac{M_1}{2}}}=\sqrt{\frac{M_2}{M_1}}=\frac{V_1}{V_2}\frac{T_1}{T_2}$

Here M₁ and M_­2 are the molecular masses of the gases having densities d₁ and d₂ respectively.

Graham's law of diffusion also holds good for effusion.

Effect of pressure on state of diffusion

The rate of diffusion (r) of a gas at constant temperature is directly preoperational to its pressure

$\frac{r_1}{r_2}=\frac{P_1}{P_2}\times \sqrt{\frac{M_2}{M_1}}$

Illustration 5. Which of the two gases ammonia and hydrogen chloride will diffuse faster and by what factor?

Solution: $\frac{r_{N{H}_3}}{r_{HCl}}={\left(\frac{M_{HCl}}{M_{N{H}_3}}\right)}^{1/2}$

= ${\left(\frac{36.5}{17}\right)}^{1/2}$ = 1.46 or $r_{N{H}_3}$ = 1.46 r_HCl

Thus ammonia will diffuse 1.46 times faster than hydrogen chloride gas