A diatomic gas consisting of rigid molecules | vedclass

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

Question

(A) According to the equipartition theorem,the rotational kinetic energy of a rigid diatomic molecule is given by $\frac{1}{2} I \omega^2 = k_B T$,whe

Vedclass · Accepted Answer

$6 	imes 10^{12} 	ext{ rad} \cdot s^{-1}$

Vedclass · Answer

(A) According to the equipartition theorem,the rotational kinetic energy of a rigid diatomic molecule is given by $\frac{1}{2} I \omega^2 = k_B T$,where $I$ is the moment of inertia,$\omega$ is the angular speed,and $k_B$ is the Boltzmann constant.Thus,the rms angular speed $\omega_{rms}$ is given by $\omega_{rms} = \sqrt{\frac{2 k_B T}{I}}$.Given: $T = 87^{\circ} C = 87 + 273 = 360 	ext{ K}$,$I = 2.76 	imes 10^{-46} 	ext{ kg} \cdot m^2$,and $k_B = 1.38 	imes 10^{-23} 	ext{ J} \cdot K^{-1}$.Substituting the values:$\omega_{rms} = \sqrt{\frac{2 	imes 1.38 	imes 10^{-23} 	imes 360}{2.76 	imes 10^{-46}}}$$\omega_{rms} = \sqrt{\frac{2.76 	imes 10^{-23} 	imes 360}{2.76 	imes 10^{-46}}}$$\omega_{rms} = \sqrt{360 	imes 10^{23}} = \sqrt{36 	imes 10^{24}} = 6 	imes 10^{12} 	ext{ rad} \cdot s^{-1}$.

$A$ diatomic gas consisting of rigid molecules is at a temperature of $87^{\circ} C$. If the moment of inertia of the rotating diatomic rigid molecule is $2.76 \times 10^{-46} kg \cdot m^2$,then the rms angular speed of the molecule is (Boltzmann constant $= 1.38 \times 10^{-23} J \cdot K^{-1}$)

Similar Questions

The number of translational degrees of freedom for a diatomic gas is

The temperature of argon,kept in a vessel,is raised by $1^\circ C$ at a constant volume. The total heat supplied to the gas is a combination of translational and rotational energies. Their respective shares are

For a gas,$\gamma = 7/5$. The gas may probably be

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