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

An ideal gas is made of polyatomic molecules. Each of the molecules has three translational, three rotational and $f$ number of vibrational modes. If the ratio of heat capacities $C_p/C_v$ of the gas is $8/7$, then the value of $f$ is:

$A$ monoatomic gas molecule has

The kinetic energy per gram mole for a diatomic gas at room temperature is:

For an ideal gas,$R = \frac{2}{3} C_v$. This suggests that the gas consists of molecules which are: $[R = \text{universal gas constant}]$

$A$ polyatomic gas with $n$ degrees of freedom has a mean energy per molecule given by (where $k$ is Boltzmann's constant and $T$ is temperature).

Vedclass Test Series

Exam Paper Generator

Online Exam Module