IIT-JEE 2007 Assertion-Reason Type MCQ on Kinetic Energy of Gas Molecules

Assertion-Reason Type Multiple Choice Questions are often interesting to answer, as is the case with the following one which appeared in IIT-JEE 2007 question paper:

STATEMENT-1

The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume

because

STATEMENT-2

The molecules of a gas collide with each other and the velocities of the molecules change due to the collision

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

Statement-1 is true. To prove this you may use the expression for the translational kinetic energy (E) of a gas molecule, given by

E = (3/2) kT

where ‘k’ is Boltzman’s constant and ‘T’ is the absolute (Kelvin) temperature.

[Remember that the translational kinetic energy of any type of molecule is (3/2)kT since the energy per degree of freedom is ½ kT and in our three dimensional space, there can be three translational degrees of freedom only]

The total translational kinetic energy of all the molecules of the given mass of gas is (3/2) kT×nV where ‘V’ is the volume of the gas and ‘n’ is the number of molecules per unit volume.

But, the pressure (P) of the gas is given by

P = nkT .

Therefore, total translational kinetic energy = (3/2)PV = 1.5 PV.

Statement -2 is true. In fact this follows from the postulates of kinetic theory. But, the pressure exerted by a gas is not because of the collision between the gas molecules, but is because of the collision of gas molecules with the walls of the container. So, statement-2 is not a correct explanation for statement-1.

The correct option therefore is (B).

Two Questions on Internal Energy of a Gas

The molar heat capacity (molar specific heat) of a gas at constant volume is given by

C_v = (n/2)R where ‘n’ is the number of degrees of freedom of the gas molecule and ‘R’ is the universal gas constant.

The molar heat capacity at constant pressure is given by

C_p = C_v + R = (n/2)R + R = [(n+2)/2]R, in accordance with Meyer’s relation.

The number of degrees of freedom of a mono atomic gas molecule is 3. Therefore, in the case of a mono atomic gas, C_v = (3/2)R and C_p = (5/2)R.

The number of degrees of freedom of diatomic gas molecules is 5 and that of triatomic and polyatomic gas molecules is 6 if we do not consider the possible vibrational modes. Therefore, in the case of a diatomic gas, C_v = (5/2)R and C_p = (7/2)R where as in the case of triatomic and polyatomic gases, C_v = (6/2)R = 3R and C_p = 4R.

Now, consider the following MCQ:

When an ideal diatomic gas without vibrational modes is heated at constant pressure, the fraction of heat energy supplied which increases the internal energy of the gas is

(a) 5/7 (b) 3/5 (c) 2/5 (d) 2/3 (e) 1/3

In the case of a diatomic gas, the molar specific heats at constant volume and constant pressure are respectively (5/2)R and (7/2)R. This means that the increase in internal energy of one mole of the gas when we supply (7/2)R joule of heat at constant pressure is (5/2)R joule. The remaining R joule of heat is spent on doing work during the expansion of the gas.

The fraction of the total energy supplied which increases the internal energy of the gas is therefore (5/2)R/(7/2)R = 5/7 [Option(a)].

[Note that this fraction in the case of a mono atomic gas will be 3/5].

Let us consider another question:

Two moles of an ideal mono atomic gas is heated from 20°C to 120°C. The increase in internal energy of the gas is nearly (Universal gas constant = 8.3 J mol^–1 K^–1).

(a) 1660 J (b) 2490 J (c) 3320 (d) 4150 J (e) 4980 J

Some of you may be inclined to wonder which molar specific heat you have to use here (C_p or C_v?). Don’t have any doubt. The increase in internal energy depends on the temperature rise and not on the conditions of constant volume or constant pressure. When you heat a gas at constant volume, the entire heat supplied is used to increase the internal energy. When you heat the gas at constant pressure, a part of the heat supplied is used for doing external work since the gas has to expand to keep the pressure constant. In both cases, the increase in the internal energy of the gas is the same if the temperature rise is the same. You have to use the molar specific heat at constant volume, which is equal to (3/2)R in the case of a mono atomic gas, for calculating the increase in internal energy.

So, increase in internal energy = No. of moles×C_v×Temperature rise = 2× (3/2)R(120–20) = 2490 J, on substituting the value 8.3 for the universal gas constant R.

Kinetic energy of gas molecules

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).

You can find more multiple choice questions at physicsplus: Questions on Kinetic Theory of Gases