degrees of freedomthermodynamicsvibrations
For ideal diatomic molecules such as $\text{H}_2$, $\text{N}_2$ and $\text{O}_2$, at what temperature does the vibrational degree of freedom significantly contributes to the calculations such as that of the specific molar heat capacity? Please explain without using statistical mechanics.
P.S. What I meant is how do I find $T_{vib}$ in the graph below.
"At high temperatures, the specific heat at constant volume $C_v$ has three degrees of freedom from rotation, two from translation, and two from vibration." I can't understand this line. $C_v$ is a physical quantity not a dynamical system. So how can it have a degrees of freedom?? You can say the degrees of freedom of an atom or molecule is something but it is wrong if you say the degrees of freedom of some physical quantity(like temperature, specific heat etc.) is something. Degrees of freedom is the number of independent coordinates necessary for specifying the position and configuration in space of a dynamical system.
Now to answer your question, we know that the energy per mole of the system is $\frac{1}{2} fRT$. where $f$= degrees of freedom the gas.
$\therefore$ molar heat capacity, $C_v=(\frac{dE}{dT})_v=\frac{d}{dT}(\frac{1}{2}fRT)_v=\frac{1}{2}fR$
Now, $C_p=C_v+R=\frac{1}{2}fR+R=R(1+ \frac{f}{2})$
$\therefore$ $\gamma=1+ \frac{2}{f}$
Now for a diaatomic gas:- 
A diaatomic gas has three translation(along x,y,z asis) and two rotational(about y and z axis) degrees of freedom. i.e. total degrees of freedom is $5$.
Hence $C_v=\frac{1}{2}fR=\frac{5}{2}R$ and $C_p=R(1+ \frac{f}{2})=R(1+ \frac{5}{2})=\frac{7}{2}R$
Your formula for Cv is wrong. The formula u used is for that condition when Vibrational mode is absent. Actual in general formula is Cv= ( ft/2 + fr/2 + fv) R Now using these u will get the value as 4R.
Best Answer
The kinetic energy of the gas molecules is of order $kT$, so when two gas molecules collide there is up to around $2kT$ of energy available. If the spacing of the vibrational energy levels is less than or equal to $2kT$ then there is a non-zero probability that the kinetic enegry could be used to excite a vibrational transition. This would cool the gas molecules, because it reduces their kinetic energy, so extra heat is required to return the gas to its original temperature and the end result is to increase the specific heat.
So vibrational transitions start to contribute to the specific heat when the temperature rises to the point where $kT$ is roughly equal to the spacing of the vibrational energy levels.
There isn't a precise point where the vibrational modes cut in because the gas molecules have a spread of energies. Vibrational transitions will start to be excited at below $kT$ and won't fully contribute to the specific heat until above $kT$. However it's a good guideline as to when they become important.