If we combine the observations of Boyle, Charles, and Avogadro we find a single expression with four variables and one unified constant that we write as PV = nRT

In chemistry we use the following units for these quantities

P = pressure in atmospheres

V = volume in liters

n = moles

T = temperature in Kelvins

R = 0.821 l atm/mol K

**Illustration : **Calculate the pressure exerted on the walls of a 3 litre flask when 7 gms of N_{2} are introduced into the same at 27°C.

**Solution: **$\large P = \frac{W R T}{M V}$

$\large P = \frac{7 \times 0.082 (273+27)}{28 \times 3}$

P = 2.05 atm

**Illustration :** An open vessel at 27°C is heated until 3/5th of the air in it has been expelled. Assuming that the volume of the vessel remains constant find

(a) the temperature at which vessel was heated?

(b) the air escaped out if vessel is heated to 900 K?

(c) temperature at which half of the air escapes out?

**Solution:** One should clearly note the fact that on heating a gas in a vessel there are the number of moles of gas which go out, the volume of vessel remains constant.

Let initial moles of gas at 300 K be ‘ n ‘ . On heating 3/5 moles of air are escaped out at temperature T.

moles of air left at temperature T = (n −3n/5) = 2n/5

(a) Under similar conditions of P and V

n_{1}T_{1} = n_{2}T_{2}

n × 300 = (2n/5) × T

T = 750 K

(b) On heating vessel to 900 K , let n1 moles be left again n_{1}T_{1} = n_{2}T_{2}

n_{1} × 900 = 300 × n

n_{1} = n/3

moles escaped out = n − n/3 = 2n/3 moles

(c) Let n/2 moles are escaped out at temperature T then

n_{1}T_{1} = n_{2}T_{2}

= n × 300

T = 600 K

**Exercise : **(i) An open bulb containing air at 19°C was cooled to a certain temperature at which the no. of moles of gaseous molecules increased by 25%. What is the final temperature ?

(ii) A long rectangular box is filled with Cl_{2} (atomic weight 35.45) which is known to contain only ^{35}Cl and ^{37}Cl. If the box could be divided by a partition and the two types of Cl_{2} molecules put in the two compartments respectively, calculate where should the partition be made if pressure on both sides are to be equal. Is this pressure the same as the original pressure?

### Relation between Molecular Mass and Gas Densities :

** Actual density: **Density and Molecular weight

From the ideal gas equation

$\large P = \frac{n RT}{V}$

$\large P = \frac{W }{M \times V} R T = \frac{d R T}{M}$

$\large M = \frac{d}{P} R T$

Hence from a knowledge of d/P density under unit pressure molecular weight can be determined. At low pressure below 1 atm d/p has been found to decrease linearly with pressure. Extrapolation of the line to zero pressure

P = 0 gives the ideal value of (d/P) at P → 0

$\large M = (\frac{d}{P} )_{P \rightarrow 0}R T$

** Vapour Density : **For gases another term which is often used is vapour-density. Vapour density of a gas is defined as the ratio of the mass of the gas occupying a certain volume at a certain temperature and pressure to the mass of hydrogen occupying the same volume at the same temperature and pressure i.e.

$\large W(gas) = \frac{P V M}{R T}$

$\large W_{H_2} = \frac{P V \times 2}{R T}$ ; ( Since , mol. wt.of Hydrogen is 2)

$\large \frac{W_{gas}}{W_{H_2}} = \frac{M}{2}$ = Vapour density of gas

Vapour density of a gas is same at any temperature, pressure and volume.

**Molecular Weight and Effective Molecular Weight : **Molecular wt. of a gas mixture or effective molecular weight:- Suppose we have to find the molecular weight of air and we are told that air contains 79% nitrogen and 21% oxygen (by mole or volume).

First of all let us understand what is meant by molecular wt. of a gas. Molecular wt. of a gas is the weight in gms of one mole of the gas (hence the unit gm/mole). Now if we take one mole of air it would contain 79/100 moles of Nitrogen and 21/100 moles of oxygen. The weight of one mole of air would be 0.79 × 28 + 0.21 × 2 = 22.54 gm/mole.

$\large M_{effective} = \Sigma X_i M_i$

$\large = \Sigma mole \;fraction \;of \;a \;gas \times mol. \;wt. \;of \;a\; gas$

X_{i} (Mole fraction) of a gas is $\large \frac{n_i}{\Sigma n_i} = \frac{moles \;of \;a\; gas}{Total \;moles \;of \;gas}$

### Dalton’s Law of Partial Pressures :

“The total pressure of a mixture of non-reacting gases is equal to the sum of their partial pressures”. By the partial pressure of a gas in a mixture is meant, the pressure that the gas will exert if it occupies alone the total volume of the mixture at the same temperature.

Consider n_{1}, moles of gas 1 and n_{2} moles of gas 2 occupying a vessel of volume V at temperature T K and exerting a total pressure P.

Let n_{1} moles of gas 1 alone occupy the same vessel of capacity V at the same temperature TK and exert a pressure P_{1}. Then by definition, P_{1} is partial pressure of gas 1 in the mixture.

Let n_{2} moles of gas 2 alone occupy the same volume V at the same temperature TK and exert a pressure P_{2}. The P_{2} is the partial pressure of gas 2 in the mixture.

By Dalton’s Law , $\large P = P_1 + P_2 + P_3 + ….$

Derivation : $\large n = n_1 + n_2 + n_3 + …$

$\large \frac{P V}{R T} = \frac{P_1 V}{R T} + \frac{P_2 V}{R T} + \frac{P_3 V}{R T} + …$

$\large P = P_1 + P_2 + P_3 + ….$

*Relationship between partial pressures and number of moles:*

(i) $\large P_1 = \frac{n_1}{n_1 + n_2} P = X_1 P $ ; where x_{1} = mole fraction of gas

(ii) Partial pressure of a gas in the mixture $\large = \frac{Volume \; of \; the \; gas}{Total \; volume} \times P$

**Partial pressure and aqueous tension:**

Dalton’s law is used to calculate the pressure of a dry gas when it is collected over water at atmospheric pressure. By Dalton’s law.

Pressure of dry gas = atmospheric pressure – aqueous tension

Aqueous tension depends on temperature. It increases with temperature and becomes 760 mm at 100°C.

** Illustration : **A mixture containing 1.6 g of O

_{2}and 1.4 g of N

_{2}and 0.4 g of He occupies a volume of 10 litre at 27

^{o}C. Calculate the total pressure of the mixture and partial pressure of each component.

** Solution: ** $\large P’_{He} = \frac{0.4}{4} \times \frac{1}{10} \times 0.0821 \times 300 = 0.2463 atm$

$\large P’_{O_2} = \frac{1.6}{32} \times \frac{1}{10} \times 0.0821 \times 300 = 0.123 atm$

$\large P’_{N_2} = \frac{1.4}{28} \times \frac{1}{10} \times 0.0821 \times 300 = 0.123 atm$

$\large P_{Total} = P’_{He} + P’_{O_2} + P’_{N_2}$

P_{total} = 0.4925 atm