Test: Law of Equipartition of Energy - JEE MCQ

A triatomic nonlinear gaseous atom has 6 degrees of freedom, that are 3 in all transrational directions and three rotational barriers in all the three axises.

Let us consider 1 mole of an ideal gas at kelvin temperature T. It has N molecules (Avogadro's number). The internal energy of an ideal gas is entirely kinetic. The average KE per molecule of a ideal gas is ½ ​nkT (k is boltzman constant), where n is degree of freedom. Therefore, the internal energy of one mole of a gas would be
 
E=N(1/2​nKT)=1/2​nRT             (∵k=R/N​)
 
Now, C_v​=dE/dT​=n/2 ​R
 
and Cp​=n/2 ​R+R=(n/2​+1)R
 
​C_p/ C_v ​​=​(n/2​+1)R/n/2​=(1+2/n​)

Oxygen molecule is a diatomic molecule and  has 3 translational and 2 rotational degrees of freedom, 

The total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to ½ k_BT. This is known as the law of equipartition of energy.

For monoatomic gas degrees of freedom are 3, while for diatomic gas is it 5, and for triatomic gas it is 7.

1 calorie = 4.184 joules. That is the conversion rate, no explanation.

Degrees of freedom are the ways in which a molecule of the gas can execute motion.
So in case of triatomic gas molecule:
1. It can translate (move) in all 3 dimensions, which accounts for 3 degrees of freedom (since there are 3 dimensions in which it could translate (move)).
2. This molecule can also revolve with Moment of Inertia ≠ 0 around all three axes, x, y, and z, which accounts for another 3 degrees of freedom (since there are 3 axes of rotation).

The average K.E. of gas molecules depends only on the absolute temperature of the gas and is directly proportional to it.

Definition of Mean Path.

The mean free path is the average distance travelled by a molecule between two successive collisions.