The ratio of average translational kinetic en | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For a diatomic molecule,the translational degrees of freedom are $f_t = 3$ and the rotational degrees of freedom are $f_r = 2$.According to the eq

Vedclass · Accepted Answer

$3/2$

Vedclass · Answer

(D) For a diatomic molecule,the translational degrees of freedom are $f_t = 3$ and the rotational degrees of freedom are $f_r = 2$.According to the equipartition theorem,the average energy associated with each degree of freedom is $\frac{1}{2} k_B T$.Average translational kinetic energy $E_t = f_t 	imes \frac{1}{2} k_B T = 3 	imes \frac{1}{2} k_B T = \frac{3}{2} k_B T$.Average rotational kinetic energy $E_r = f_r 	imes \frac{1}{2} k_B T = 2 	imes \frac{1}{2} k_B T = 1 k_B T$.The ratio of average translational kinetic energy to rotational kinetic energy is $\frac{E_t}{E_r} = \frac{\frac{3}{2} k_B T}{k_B T} = \frac{3}{2}$.

The ratio of average translational kinetic energy to rotational kinetic energy of a diatomic molecule at temperature $T$ is

Similar Questions

The mean rotational kinetic energy of a diatomic molecule at temperature $T$ is

When $x$ amount of heat is given to a gas at constant pressure,it performs $x/3$ amount of work. The average number of degrees of freedom per molecule is

If the degrees of freedom of a gas molecule is $6$,then the total internal energy of the gas molecule at a temperature of $47^{\circ} C$ (in $eV$) is (Boltzmann constant $= 1.38 \times 10^{-23} \ J \ K^{-1}$)

The number of rotational degrees of freedom of a monatomic molecule is

What will be the average value of energy for a monoatomic gas in thermal equilibrium at temperature $T$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module