Question

Consider two cylinders of gas identical in all respects except that one contains oxygen o2 and the other helium he. both cylinders initially contain the same volume of gas at 0°c and 1 atm of pressure and are closed by a movable piston at one end. both gases are now compressed adiabatically to one-third their original volume. (a) which gas will show the greater temperature increase?

Answer 1

1) For a reversible adiabatic process, the ideal gas law can be written also as 
$TV^{\gamma-1}=cost.$
or equivalently
$T_iV_i^{\gamma-1}=T_fV_f^{\gamma-1}$
where $T_i$ and $T_f$ are the initial and final temperature of the gas, and $V_i$ and $V_f$ are its initial and final volume. $\gamma$ is the adiabatic index, given by
$\gamma= \frac{C_P}{C_V}= \frac{f+2}{f}$
where f is the number of degrees of freedom of the molecule. For helium, which is monoatomic gas, we have f=3, therefore
$\gamma_{He}= \frac{5}{3}$
Instead, for oxigen ($O_2$) which is a diatomic gas, we have f=5, therefore
$\gamma_{O_2}= \frac{7}{5}$

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:
$T_f=( \frac{V_i}{V_f})^{\gamma-1} T_i$
And  since we know that both gases are compressed to one-third of their original volume, i.e.
$V_f= \frac{1}{3}V_i$
this means
$T_f=3^{\gamma-1} T_i$

So now we can calculate the final temperature for each gas, since we know the initial temperature: $T_i=0^{\circ}C=273~K$ 

- For helium: 
$T_f=3^{ \frac{5}{3}-1} T_i = 3^{ \frac{2}{3} } \cdot 273~K = 568~K$

- For oxigen:
$T_f=3^{ \frac{7}{5}-1 } T_i = 3^{ \frac{2}{5} }\cdot 273~K = 424~K$

So, helium shows the greater temperature increase.