Class 11th

(a) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours is now distributed over a larger space.

(b) On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.

(c) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

5.7. Applying PV = nRT

                      We get P = nRT / V

Given:               n_CH4 = 3.2 / 16 mol = 0.2 mol

              n_CO2 = 4.4 /44 mol = 0.1 mol

So,

P_CH4 = (0.2 mol) (0.0821 dm³atm K^-1 mol^-1) (300 K) / (9 dm³)

        = 0.55 atm

 P_CO2= (0.1 mol) (0.0821 dm³atm K^-1 mol^-1) (300 K) / (9 dm³)

         = 0.27 atm

P_total = P_CH4 + P_CO2 = 0.55 + 0.27

         = 0.82 atm

         = 8.314 x 10⁴ Pa

Hence, the pressure exerted by a mixture is 8.314 x 10⁴ Pa

5.6. The chemical equation for the reaction is
                                    2 Al + 2 NaOH + H₂O? 2 NaAlO₂ + 3H₂ 
i.e. 2 moles of Al give 3 moles of H₂ gas.

Moles of aluminium = 270.15g = 5.56*10^?3 moles

Moles of H₂ produced= 23*5.56*10^?3

           = 8.33*10^?3 moles

           Volume of H₂ = nRT / P

                                   = 8.33*10^?3 x 0.0831*293

                      = 0.203 litres

Therefore, volume of dihydrogen released is 0.203 litres.

Answer: According to ideal gas equation
                        PV = nRT

Or                        P= nRT/V

Replacing n by m/M, we get

                            P M₁ x 2 = 28 x 5 (Molecular mass of N₂ = 28 g/mol)
or                                             M₁ = 70 g/mol