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JEE MainPhysicsKinetic Theory of Gases

A closed container holds a mixture of Helium gas (molar mass = 4 g mol ⁻¹ ) and Oxygen gas (molar mass = 32 g mol ⁻¹ ) in thermal equilibrium. Let R_E be the ratio of the average translational kinetic energy per molecule of Helium to that of Oxygen, and R_v be the ratio of the root-mean-square speed of Helium molecules to that of Oxygen molecules. The values of R_E and R_v are respectively:

Options

  1. AR_E = 1:1 and R_v = 1:2 2
  2. BR_E = 8:1 and R_v = 2 2 :1
  3. CR_E = 1:1 and R_v = 2 2 :1
  4. DR_E = 1:8 and R_v = 1:2 2

Correct answer

C. R_E = 1:1 and R_v = 2 2 :1

Step-by-step solution

Since both gases are in the same container, they are in thermal equilibrium and share the same absolute temperature T . The average translational kinetic energy per molecule for any ideal gas is given by 3 2 k_B T . This value depends only on temperature and is independent of the nature or mass of the gas. Thus, the ratio of their average translational kinetic energies is: R_E = 1:1 The root-mean-square speed of gas molecules is given by v_ rms = 3RT M , where M is the molar mass. For Helium and Oxygen at the same

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