On the basis of the kinetic theory of gases,t | vedclass

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

Question

(A) According to the equipartition theorem,the average kinetic energy associated with each degree of freedom for a single molecule is $\frac{1}{2}kT$,

Vedclass · Accepted Answer

$\frac{1}{2}RT$

Vedclass · Answer

(A) According to the equipartition theorem,the average kinetic energy associated with each degree of freedom for a single molecule is $\frac{1}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.For $1\, mole$ of gas,the number of molecules is equal to the Avogadro number $(N_A)$.Therefore,the mean kinetic energy for $1\, mole$ per degree of freedom is $\frac{1}{2}kT 	imes N_A$.Since $k 	imes N_A = R$ (the universal gas constant),the expression becomes $\frac{1}{2}RT$.

On the basis of the kinetic theory of gases,the mean kinetic energy $(K.E.)$ of $1\, mole$ per degree of freedom is:

Similar Questions

Obtain an expression for the momentum of a gas transferred to the walls of a container in time $\Delta t$.

Consider a gas with $n$ molecules. The average number of molecules moving in a fixed direction is ..........

The value of universal gas constant is $8.3\, J/mol/K$. The mean kinetic energy of $32\, g$ of oxygen at $-73^{\circ}C$ will be ...... $J$.

In the case of hydrogen and oxygen at $N.T.P.$,which of the following quantities is/are the same?

The mean energy of a molecule of an ideal gas is

Vedclass Test Series

Exam Paper Generator

Online Exam Module