# Query about working out the Specific heat ratio of a gas

The specific heat ratio for $$\rm CO_2$$ at room temperature is $$1.28$$ according to my tables. Since $$C_V= \left.\frac{\partial U}{\partial T}\right|_V$$ and $$C_P=\left.\frac{\partial U}{\partial T}\right|_V+\left(P+\left.\frac{\partial U}{\partial V}\right|_T\right) \left.\frac{\partial V}{\partial T}\right|_P$$ and $$γ=\frac{C_P}{C_V}$$, then $$γ=1+\frac{\left(P+\left.\frac{\partial U}{\partial V}\right|_T\right) \left.\frac{\partial V}{\partial T}\right|_P}{\left.\frac{\partial U}{\partial T}\right|_V}.$$

Given the equipartition theorem, we can treat the internal energy as a linear function of temperature $$U(t)=\frac{f}{2} RT$$ where $$f$$ is the number of degrees of freedom of the molecule.

Basically, my question is, with the internal energy defined as above, the value for $$\gamma$$ seems to only depend on the number of degrees of freedom. For monoatomic Helium which only has 3 degrees of freedom, $$\gamma=\frac{C_P}{C_V} = \frac{5}{3}$$, my table agrees with this value. For N$$_2$$ a diatomic molecule that has 3 translational degrees of freedom and 2 rotational, $$\gamma=\frac{C_P}{C_V} = \frac{7}{5}$$ which again is the accepted value. But using this reasoning, I can't get the right value of $$\gamma$$ for CO$$_2$$. Why does it work for some gases but not for others?

• How many degrees of freedom does CO2 have? – Chet Miller Mar 14 at 1:11
• It should have the same as n2. All the atoms in the molecule lie on a straight line so it works out to have 3 translational and 2 rotational. – Vishal Jain Mar 14 at 5:34
• I think CO2 can't be considered as an ideal gas, since interaction between molecules is note negligible (as it is in Helium and N2 gases). The interactions modify $U(T)$ and consequently the specific heat ratio $\gamma$. – Matteo Mar 14 at 8:34
• Carbon dioxide sp. heat ratio is accounted for the fact that the vibrations are not available at low temperature. It is cross posting through it seems you assert opposite things. In chemistry you said vibrations must be accessible to get 1.3 – Alchimista Mar 14 at 9:35
• @Alchimista If vibrations are not available at low temperatures then that implies a C02 molecule only has 5 degrees of freedom which gives the incorrect value of gamma. Can you please clarify what you mean by "Carbon dioxide sp. heat ratio is accounted for the fact that the vibrations are not available at low temperature." – Vishal Jain Mar 14 at 9:57