The energy per mole per degree of freedom of an ideal gas is

Question

The energy per mole per degree of freedom of an ideal gas is (A)  (3/2) kBT  (B)  (1/2)kBT (C)  (3/2) RT  (D)  (1/2) RT

Tardigrade Admin · Accepted Answer

The energy per mole per degree of freedom of an ideal gas is E=((1/2) KB T)(NA) Where, T = absolute temperature NA= Avogadro's number KB= Boltzmann's constant ∵ R=NA KB Therefore, E=(1/2) R T

Q. The energy per mole per degree of freedom of an ideal gas is

$\frac{3}{2} k_{B} T$

$\frac{1}{2} k_{B} T$

$\frac{3}{2} RT$

$\frac{1}{2} RT$

The energy per mole per degree of freedom of an ideal gas is
$E = (\frac{1}{2} K_{B} T) (N_{A})$
Where, $T$ = absolute temperature
$N_{A} =$ Avogadro's number
$K_{B} =$ Boltzmann's constant
$∵ R = N_{A} K_{B}$
Therefore, $E = \frac{1}{2} RT$

Q. The energy per mole per degree of freedom of an ideal gas is

23​kB​T

21​kB​T

23​RT

21​RT

The energy per mole per degree of freedom of an ideal gas is E=(21​KB​T)(NA​) Where, T = absolute temperature NA​= Avogadro's number KB​= Boltzmann's constant ∵R=NA​KB​ Therefore, E=21​RT

$\frac{3}{2} k_{B} T$

$\frac{1}{2} k_{B} T$

$\frac{3}{2} RT$

$\frac{1}{2} RT$

The energy per mole per degree of freedom of an ideal gas is
$E = (\frac{1}{2} K_{B} T) (N_{A})$
Where, $T$ = absolute temperature
$N_{A} =$ Avogadro's number
$K_{B} =$ Boltzmann's constant
$∵ R = N_{A} K_{B}$
Therefore, $E = \frac{1}{2} RT$