Kinetic Theory of Gases MCQs for NEET — Physics Questions with Answers

(d) Root mean square velocity at T₁ temperature,

              U₁ = $\sqrt{\frac{3R{T}_1}{M}}=\sqrt{\frac{3R(27+273)}{M}}$            .....(i)

    Root mean square velocity at T₂ temperature,

             U₂ = $\sqrt{\frac{3R{T}_2}{M}}=\sqrt{\frac{3R(927+273)}{M}}$           ......(ii)

      Eq. (i) divided by Eq. (ii)

             U₁/U₂ = $\sqrt{\frac{27 +273}{927 + 273}}$= $\sqrt{\frac{300}{1200}}$=1/2

                  U₂ = 2U₁

(b) Initial volume, V₁ = 500 mL

Initial temperature,

              T₁ = 27$°$C = 27 + 273 = 300 K

Final temperature,

             T₂ = -5 + 273 = 268 K

             V₂ = ?

             V₁/T₁ = V₂/T₂

             V₂ = V₁T₂/T₁ = (500 x 268)/300 = 446.66 mL

(c) H₂ and He possess minimum molar mass among all gases.

(a) Increase of pressure decreases volume and molecules come closer to each other.

(c) For ideal gases PV = nRT, .^.. Z =1; because Z = PV/nRT

(a) V$\propto$1/P or density $\propto$P (^..^. d$\propto$1/V),Boyle's law 

(c) b = 4 x N x volume of one molecule in rest.

(a) Average kinetic energy, E = (3/2)RT

 u_rms = $\sqrt{\frac{3RT}{M}}$ = $\sqrt{\frac{2E}{M}}$

(a) According to Boyle's law,

           p₁V₁ = p₂V₂

          750 x 600 =p₂ x 500

                    p₂ = (750 x 600)/500 = 900 mm

   So, the increase in pressure = 900-750 =150 mm