The mean kinetic energy of one mole of gas per degree of freedom (on the basis of kinetic theory of gases) is

Question

The mean kinetic energy of one mole of gas per degree of freedom (on the basis of kinetic theory of gases) is (A) (1/2) k T (B) (3/2) k T (C) (3/2) R T (D) (1/2) R T

Tardigrade Admin · Accepted Answer

The mean kinetic energy of one mole of gas n degree of freedom. E=(n/2)RT The mean kinetic energy of one mole of gas per degree of freedom. Eâ²=(E/n)=((n/2) R T/n) Eâ²=(1/2)RT

Q. The mean kinetic energy of one mole of gas per degree of freedom (on the basis of kinetic theory of gases) is

$\frac{1}{2} k T$

$\frac{3}{2} k T$

$\frac{3}{2} R T$

$\frac{1}{2} R T$

The mean kinetic energy of one mole of gas $n$ degree of freedom.
$E = \frac{n}{2} RT$
The mean kinetic energy of one mole of gas per degree of freedom.
$E^{'} = \frac{E}{n} = \frac{\frac{n}{2} RT}{n}$
$E^{'} = \frac{1}{2} RT$

Q. The mean kinetic energy of one mole of gas per degree of freedom (on the basis of kinetic theory of gases) is

21​kT

23​kT

23​RT

21​RT

The mean kinetic energy of one mole of gas n degree of freedom. E=2n​RT The mean kinetic energy of one mole of gas per degree of freedom. E′=nE​=n2n​RT​ E′=21​RT

$\frac{1}{2} k T$

$\frac{3}{2} k T$

$\frac{3}{2} R T$

$\frac{1}{2} R T$

The mean kinetic energy of one mole of gas $n$ degree of freedom.
$E = \frac{n}{2} RT$
The mean kinetic energy of one mole of gas per degree of freedom.
$E^{'} = \frac{E}{n} = \frac{\frac{n}{2} RT}{n}$
$E^{'} = \frac{1}{2} RT$