Question 62: Thermodynamics & KTG

Question

The mean energy per molecule for a diatomic gas is

Accepted Answer

For an ideal gas, mean kinetic energy per molecule
$=\frac{f}{2} k_B T$
For diatomic gas, degree of freedom,
$\begin{gathered}
f=5 \
	herefore 	ext { K.E per molecule }=\frac{5 k_B T}{2}
\end{gathered}$

Answer

$\frac{5 k_B T}{N}$

Answer

$\frac{5 k_B T}{2 N}$

Answer

$\frac{5 k_B T}{2}$

Answer

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

Thermodynamics & KTG

The mean energy per molecule for a diatomic gas is

Choose the correct answer: