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MHT CET202411 May 2024Evening ShiftPhysicsKinetic Theory of GasesActual

' N ' molecules of gas A , each having mass ' m ' and ' 2 N ' molecules of gas B , each of mass ' 2 m ' are contained in the same vessel which is at constant temperature ' T '. The mean square velocity of B is V^2 and mean square of x -component of A is ^2 . The value of ^2 V^2 is

Options

  1. A3: 2
  2. B2: 3
  3. C1: 2
  4. D2: 1

Correct answer

B. 2: 3

Step-by-step solution

Mean square velocity of molecule = 3 kT m For gas A, x component of mean square velocity of molecule = ^2 Mean square velocity =3 ^2= 3 kT m ...(i) For gas B, Mean square velocity = V ^2= 3 kT 2 ~m ...(ii) From (i) and (ii) 3 ^2 ~V ^2 = 3 kT m 2 ~m 3 kT ^2 ~V ^2 = 2 3

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