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
- AR_E = 1:1 and R_v = 1:2 2
- BR_E = 8:1 and R_v = 2 2 :1
- CR_E = 1:1 and R_v = 2 2 :1
- 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