The total internal energy of one mole of rigid diatomic gas is

Question

The total internal energy of one mole of rigid diatomic gas is (A) (3/2) R T (B) (7/2) R T (C) (5/2) R T. (D) (9/2) R T

Tardigrade Admin · Accepted Answer

A rigid diatomic molecule has 5 degrees of freedom: 3 translational and 2 rotational. According to the law of equipartition of energy, the total internal energy of one mole of rigid diatomic gas is U=(5/2) kB T × NA=(5/2) R T (∵ R=kB NA)

Q. The total internal energy of one mole of rigid diatomic gas is

$\frac{3}{2} RT$

$\frac{7}{2} RT$

$\frac{5}{2} RT$ .

$\frac{9}{2} RT$

A rigid diatomic molecule has $5$ degrees of freedom:
$3$ translational and $2$ rotational.
According to the law of equipartition of energy, the total internal energy of one mole of rigid diatomic gas is
$U = \frac{5}{2} k_{B} T \times N_{A} = \frac{5}{2} RT$
$(∵ R = k_{B} N_{A})$

Q. The total internal energy of one mole of rigid diatomic gas is

23​RT

27​RT

25​RT.

29​RT

A rigid diatomic molecule has 5 degrees of freedom: 3 translational and 2 rotational. According to the law of equipartition of energy, the total internal energy of one mole of rigid diatomic gas is U=25​kB​T×NA​=25​RT (∵R=kB​NA​)

$\frac{3}{2} RT$

$\frac{7}{2} RT$

$\frac{5}{2} RT$ .

$\frac{9}{2} RT$

A rigid diatomic molecule has $5$ degrees of freedom:
$3$ translational and $2$ rotational.
According to the law of equipartition of energy, the total internal energy of one mole of rigid diatomic gas is
$U = \frac{5}{2} k_{B} T \times N_{A} = \frac{5}{2} RT$
$(∵ R = k_{B} N_{A})$