Class 11th

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

5.1. Initial Pressure, P₁ = 1 bar                           Final Pressure, P₂ =?      

               Initial Volume, V₁= 500 dm³                       Final Volume, V₂=200 dm³
As the value of temperature constant (=30°C)
                  P₁V₁=P₂V₂
1 bar x 500 dm³ = P₂ x 200 dm³ 

Or                   P₂=500/200 bar

                           =2.5 bar

= mRT/MV

Replacing m/V by d (i.e. density), we get

                            P = dRT/M

    P? d

Hence, at a given temperature, the density of a gas is directly proportional to its pressure.