Ideal gas Equation:-

 Any Gas which follows all the gas laws is known as Ideal Gas.

PV = nRT

Ideal gas equation where P → pressure of Gas

V → volume of gas

N → number of mole

R → Gas constant

T → Temperature

Derivation:-

V ∝ (1/P) at constant Temperature (Boyles law)

V ∝ T at constant Pressure              (Charles law)

V ∝ n                                                 (Avogadro’ law)

At const T & P

Combining all these

V ∝ (nT/P) or V = (RnT/P) or PV =

Where R → Gas constant

Also (P₁V₁/T₁) = P₂V₂/T₂

Relation between pressure and density of gas

As PV = nRT (ideal gas equation)

Also PV = (w/M) RT (because n = (w/M) = wt./molar mass

Or PM = (w/V) RT (As (w/V) = density of gas)

Or PM = dRT or d = PM/RT

Values of Gas constant →

R = 0.0821 L atm k^-1 mol^-1

R = 0.083 bar dm³ k^-1 mol^-1

R = 8.314 J k^-1 mol^-1

R = 8.314 k Pa dm³ k^-1 mol^-1

R = 2 cal k^-1 mol^-1

1 cal = 4.184 J

1 J = 10⁷ erg

(1) Numerical → 8 g of methane is placed is 5 litre container at 27^°C .Find Boyle’s constant .

Solution → PV = Boyle’s constant

PV = (w/M) RT = 8/16 x 0.0821 x 300 = 12.31 L atm.

(2) Numerical: The density of a gas is 3.80 g litre^-1 at STP. Calculate its density at 27^°C and 700 loss pressure

Solution → d = (PM/RT) for the same gas

Or (3.80/d₂) = (760/700) x (300/273) (d₁/d₂) = (P₁/P₂) x (T₂/T₁)

Or d₂ = 3.185 g litre^-1

 

Daltons Law OF PARTIAL PRESSURE → It states that “when 2 or more gases which don’t react chemically are enclosed in vessel, the total pressure exerted by gaseous mixture is equal to the sum of all the partial pressures that each gas would exert when present alone in same vessel at the same temperature.”

Let P₁, P₂, P₃, are partial pressures of gases 1, 2, 3 reply then total pressure, P = P₁ + P₂ + P₃

Applications of the law → (i) In determination of dry gas:- whenever a gas is collected over water. Its moist them.

P_{dry gas} = P_{moist gas}– Aqueous tension (at t^°C)

Where aq. tension is the pressure due to water vapours

(ii) In calculation of partial pressures →

As PV = nRT or P = (nRT/V) then

P = P_A + P_B + P_C +………. = (RT/V) (n_A + n_B + n_C +……)

Divide P_A by P

Or (P_A/P) = n_A/n_A + n_B + n_C +…. = x_A

Or P_A = x_A * P where x_A → mole fraction of gas A.

So partial pressure = mole fraction * total pressure

Numerical → A neon –O₂ mix contains 70.6 g dioxygen and 167.5 g neon of P_T = 25 bar. What is& P_Ne in mix. (At. Mass of Ne = 20)

Solution:-

= 0.21 x 25 = 5.25 bar & P_Ne = 0.79 x 25 = 19.7 bar