1) For a reversible adiabatic process, the ideal gas law can be written also as
[tex]TV^{\gamma-1}=cost.[/tex]
or equivalently
[tex]T_iV_i^{\gamma-1}=T_fV_f^{\gamma-1}[/tex]
where [tex]T_i[/tex] and [tex]T_f[/tex] are the initial and final temperature of the gas, and [tex]V_i[/tex] and [tex]V_f[/tex] are its initial and final volume. [tex]\gamma[/tex] is the adiabatic index, given by
[tex]\gamma= \frac{C_P}{C_V}= \frac{f+2}{f} [/tex]
where f is the number of degrees of freedom of the molecule. For helium, which is monoatomic gas, we have f=3, therefore
[tex]\gamma_{He}= \frac{5}{3} [/tex]
Instead, for oxigen ([tex]O_2[/tex]) which is a diatomic gas, we have f=5, therefore
[tex]\gamma_{O_2}= \frac{7}{5} [/tex]
2) Using the initial relationship written at point 1), we can now calculate the increase in temperature for both gases. First of all, let's rewrite the initial equation as:
[tex]T_f=( \frac{V_i}{V_f})^{\gamma-1} T_i [/tex]
And since we know that both gases are compressed to one-third of their original volume, i.e.
[tex]V_f= \frac{1}{3}V_i [/tex]
this means
[tex]T_f=3^{\gamma-1} T_i[/tex]
So now we can calculate the final temperature for each gas, since we know the initial temperature: [tex]T_i=0^{\circ}C=273~K[/tex]
- For helium:
[tex]T_f=3^{ \frac{5}{3}-1} T_i = 3^{ \frac{2}{3} } \cdot 273~K = 568~K [/tex]
- For oxigen:
[tex]T_f=3^{ \frac{7}{5}-1 } T_i = 3^{ \frac{2}{5} }\cdot 273~K = 424~K[/tex]
So, helium shows the greater temperature increase.
