 Chemistry Ideal Gas Equation and Dalton's Law of Partial Pressures

### Topics Covered :

● Ideal Gas Equation
● Density and Molar Mass of a Gaseous Substance
● Dalton's Law of Partial Pressures
● Partial Pressure in Terms of Mole Fraction

### Ideal Gas Equation :

text(Definition :) The three laws which we have learnt till now can be combined together in a single equation which is known as ideal gas equation.

● At constant T and n; V prop 1/p Boyle’s Law

● At constant p and n; V ∝ T Charles’ Law

● At constant p and T ; V ∝ n Avogadro Law

● Thus V prop (nT)/p ...........(15)

=> V = R (nT)/p ...........(16)

where R is proportionality constant.

● On rearranging the equation (16) we obtain

pV = n RT .............(17)

=> R = (p V)/(n T) ..................(18)

● R is called gas constant.

● It is same for all gases.

● Therefore, it is also called text(Universal Gas Constant).

● Equation (17) is called text(ideal gas equation).

=> Equation (18) shows that the value of R depends upon units in which p, V and T are measured.

=> If three variables in this equation are known, fourth can be calculated.

● From this equation we can see that at constant temperature and pressure, n moles of any gas will have the same volume because

V = ( n RT)/p

and n, R, T and p are constant.

● This equation will be applicable to any gas, under those conditions when behaviour of the gas approaches ideal behaviour.

● Volume of one mole of an ideal gas under STP conditions (273.15 K and 1 bar pressure) is 22.710981 L mol^(-1).

● Value of R for one mole of an ideal gas can be calculated under these conditions as follows :

R = ((10^5 pa) (22.71 ×10^(–3)m^3))/((1 mol) (273.15 K))

= 8.314 Pa m^3 K^(–1) mol^(–1)

= 8.314 × 10^(–2) bar L K^(–1) mol^(–1)

= 8.314 J K^(–1) mol^(–1)

● At STP conditions used earlier (0 °C and 1 atm pressure), value of R is 8.20578 × 10^(–2) L atm K^(–1) mol^(–1).

● Ideal gas equation is a relation between four variables and it describes the state of any gas, therefore, it is also called text(equation of state).

● Ideal gas equation is the relationship for the simultaneous variation of the variables.

=> If temperature, volume and pressure of a fixed amount of gas vary from T_1, V_1 and p_1 to T_2, V_2 and p_2 then we can write

(p_1 V_1)/T_1 = nR and (p_2 V_2)/T_2 = nR

=> (p_1V_1)/T_1 = (p_2 V_2)/T_2 .........(19)

● Equation (19) is a very useful equation.

● If out of six, values of five variables are known, the value of unknown variable can be calculated from the equation (19).

● This equation is also known as text(Combined gas law).
Q 3056223174 At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 mL. Solution:

p_1 = 760 mm Hg , V_1 = 600 mL

T_1 = 25+273 = 298 K

V_2 = 640 mL and T_2 = 10+273 = 283K

According to Combined gas law

(p_1 V_1)/T_1 = (p_2 V_2)/T_2

=> p_2 = ( p_1 V_1 T_2)/(T_1 V_2)

=> p_2 = (760 mm xx600 mL xx 283 K)/(298K xx640 mL)

 = 676.6 mm Hg

### Density and Molar Mass of a Gaseous Substance :

Ideal gas equation can be rearranged as follows :

n/V = p/(RT)

Replacing n by m/M we get

m/(MV) = p/(RT) ................(20)

d/M = p/(RT) (where d is the density) ......................... (21)

On rearranging equation (21), we get the relationship for calculating molar mass of a gas.

M = (d RT)/p ...........(22)

### Dalton’s Law of Partial Pressures :

=> The law was formulated by John Dalton in 1801.

text(Statement :) The total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases i.e., the pressures which these gases would exert if they were enclosed separately in the same volume and under the same conditions of temperature.

● In a mixture of gases, the pressure exerted by the individual gas is called text(partial pressure).

● Mathematically,

p_text(Total) = p_1+p_2+p_3..... ( at constant T, V) ............(23)

where p_text(Total) is the total pressure exerted by the mixture of gases and p_1, p_2, p_3 etc. are partial pressures of gases.

=> Gases are generally collected over water and therefore are moist.

● Pressure of dry gas can be calculated by subtracting vapour pressure of water from the total pressure of the moist gas which contains water vapours also.

● Pressure exerted by saturated water vapour is called text(aqueous tension).

● Aqueous tension of water at different temperatures is given in Table.

p_text(dry gas) = p_text(Total) - Aqueous tension ...........(24)

### Partial pressure in terms of mole fraction :

● Suppose at the temperature T, three gases, enclosed in the volume V, exert partial pressure p_1, p_2 and p_3 respectively.

Then,

p_1 = (n_1 RT)/V .............(25)

p_2 = (n_2 RT)/V .........(26)

p_3 = (n_3 RT)/V .............(27)

where n_1 n_2 and n_3 are number of moles of these gases.

● Thus, expression for total pressure will be

p_text(Total) = p_1 + p_2 + p_3

= n_1 (RT)/V + n_2 (RT)/V + n_3 (RT)/V

= ( n_1 + n_2 + n_3) (RT)/V .........(28)

● On dividing p_1 by p_text(total) we get

p_1/p_text(total) = (n_1/(n_1+n_2+n_3))

 = n_1/(n_1+n_2+n_3)

where n = n_1 + n_2 + n_3

x_1 is called mole fraction of first gas.

Thus p_1 = x_1 p_text(total)

Similarly, for other two gases we can write

p_2 = x_2 p_text(total) and p_3 = x_3 p_text(total)

Thus, a general equation can be written as p_1 = x_1 p_text(total) ........(29)

where p_i and x_i are partial pressure and mole fraction of ith gas respectively. If total pressure of a mixture of gases is known, the equation (29) can be used to find out pressure exerted by individual gases.
Q 3016323279 A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture ? Solution:

Number of moles of dioxygen

 = (70.6 g)/{32 g mol^(-1)}

= 2.21 mol

Number of moles of neon

 = (167.5 g)/{20 g mol^(-1)}
 = 8.375 mol

Mole fraction of dioxygen

 = (2.21)/(2.21+8.375)

= (2.21)/(10.585)

 = 0.21

Mole fraction of neon  = (8.375)/(2.21+8.375)

= 0.79

Partial pressure of a gas = mole fraction × total pressure

⇒ Partial pressure of oxygen = 0.21 × (25 bar) = = 5.25 bar

Partial pressure of neon = 0.79 × (25 bar) = 19.75 bar 