Most of you might be remembering the expression for the total average kinetic energy of a gas molecule: E = (n/2)kT where ‘n’ is the number of degrees of freedom, ‘k’ is Boltzman’s constant and T is the absolute temperature. The values for ‘n’ are 3 for mono atomic, 5 for diatomic and 6 for tri and polyatomic gas molecules. [Note that we have not considered the possible vibrational modes of diatomic and polyatomic gas molecules here].
One important thing you have to note is that the average translational kinetic energy of all types of molecules is (3/2)kT since we have a universe with three dimensional space (n=3).
If you consider one mole of gas molecules, the average total kinetic energy (for one mole) is (n/2)RT and the average translational kinetic energy (for one mole) is (3/2)RT where ‘R’ is universal gas constant.
Now, consider the following M.C.Q.:
The average translational kinetic energy of oxygen molecule at a temperature ‘T’ is 6×10-21 J. The average translational kinetic energy of helium atom at a temperature 2T will be
(a) 3.6×10-21 J (b) 7.2×10-21 J (c) 2.88×10-20 J (d) 6×10-21 J (e) 1.2×10-20 J
Since the translational kinetic energy is (3/2)kT for all types of gas molecules, it depends only on temperature and the answer is 1.2×10-20 J [Option (e)].
The following MCQ high lights the type of the molecule in addition to the temperature you will have to consider while calculating the kinetic energy:
The ratio of the total kinetic energy of all the molecules in one mole of hydrogen at temperature ‘T’ to the total kinetic energy of all the molecules in two moles of helium at temperature 2T is
(a) 5:6 (b) 5:12 (c) 5:3 (d)1:2 (e) 1:4
The total kinetic energies per mole in the case of hydrogen and helium are, respectively, (5/2)RT and (3/2)RT since hydrogen is diatomic and helium is mono atomic. Hence the required ratio is [(5/2)RT] : [2×(3/2)R×2T] = 5:12 [Option (b)].
The following question appeared in the I.I.T.1997 test paper:
The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21×10-21J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)
(a) 12.42×10-21J, 968 m/s (b) 8.78×10-21J, 684 m/s (c) 6.21×10-21J, 968 m/s (d) 12.42×10-21J, 684 m/s
Average translational energy per mole = (3/2)RT . When the temperature changes from 300 K to 600 K, the energy is doubled. The rms speed of gas molecules = √(3kT/m) where ‘k’ is Boltzman’s constant and ‘m’ is the molecular mass. So rms speed becomes √2 times when the temperature changes from 300 K to 600 K. The correct option is (d).
One important thing you have to note is that the average translational kinetic energy of all types of molecules is (3/2)kT since we have a universe with three dimensional space (n=3).
If you consider one mole of gas molecules, the average total kinetic energy (for one mole) is (n/2)RT and the average translational kinetic energy (for one mole) is (3/2)RT where ‘R’ is universal gas constant.
Now, consider the following M.C.Q.:
The average translational kinetic energy of oxygen molecule at a temperature ‘T’ is 6×10-21 J. The average translational kinetic energy of helium atom at a temperature 2T will be
(a) 3.6×10-21 J (b) 7.2×10-21 J (c) 2.88×10-20 J (d) 6×10-21 J (e) 1.2×10-20 J
Since the translational kinetic energy is (3/2)kT for all types of gas molecules, it depends only on temperature and the answer is 1.2×10-20 J [Option (e)].
The following MCQ high lights the type of the molecule in addition to the temperature you will have to consider while calculating the kinetic energy:
The ratio of the total kinetic energy of all the molecules in one mole of hydrogen at temperature ‘T’ to the total kinetic energy of all the molecules in two moles of helium at temperature 2T is
(a) 5:6 (b) 5:12 (c) 5:3 (d)1:2 (e) 1:4
The total kinetic energies per mole in the case of hydrogen and helium are, respectively, (5/2)RT and (3/2)RT since hydrogen is diatomic and helium is mono atomic. Hence the required ratio is [(5/2)RT] : [2×(3/2)R×2T] = 5:12 [Option (b)].
The following question appeared in the I.I.T.1997 test paper:
The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21×10-21J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)
(a) 12.42×10-21J, 968 m/s (b) 8.78×10-21J, 684 m/s (c) 6.21×10-21J, 968 m/s (d) 12.42×10-21J, 684 m/s
Average translational energy per mole = (3/2)RT . When the temperature changes from 300 K to 600 K, the energy is doubled. The rms speed of gas molecules = √(3kT/m) where ‘k’ is Boltzman’s constant and ‘m’ is the molecular mass. So rms speed becomes √2 times when the temperature changes from 300 K to 600 K. The correct option is (d).
You can find more multiple choice questions at physicsplus: Questions on Kinetic Theory of Gases
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