Question

Equal volumes of monoatomic and diatomic gases at the same temperature are given equal quantities of heat. Then,
A. the temperature of diatomic gas will be more.
B. the temperature of monoatomic gas will be more.
C. The temperature of both will be zero.
D. nothing can be said.

Answer 1

Hint: Use the law of Equipartition of energy, which states that The energy of a gas is equally distributed among the possible degrees of freedom of the molecule such that each degree of freedom gets an energy of $e = \dfrac{1}{2}{K_B}T$.

Complete step-by-step answer:

A degree of freedom is an independent way of a motion that the molecules can exhibit. A mono-atomic gas molecule is just like a point mass. It can move freely along any of the three directions - $X$, $Y$, and $Z$ independent of the other two. Thus, we can say a mono-atomic gas has 3 degrees of freedom.

1. Now, If we consider a diatomic gas, as in fig(2), it can have the following independent and dissimilar motions:

2. It can move freely along three axes: $X$, $Y$, and $Z$. Thus we get three degrees of freedom.

3. It can rotate with $X$, $Y$, and $Z$ axis as the axis of rotation. Among these, Rotation about $X$ axis and $Z$ axis are similar and hence could be merged as one degree of freedom. Also, in case of rotation about the $Y$ axis, the masses are very close to the axis and hence, they have a negligible contribution to energy. So this degree of freedom is neglected. Apart from rotation and translation, the molecules can vibrate along the $Y$ axis, By stretching and squashing the bond between them. This adds another degree of freedom.

So a diatomic gas has a total of 5 degrees of freedom, whereas a mono-atomic gas has only 3. So when an equal amount of energy - say $E$ is given to both of them, Since energy gets distributed among all degrees of freedom, A mono-atomic gas would have more energy in each degree of freedom compared to a diatomic gas. Now we know from the law of Equipartition of energy that the energy of each degree of freedom is directly related to the temperature of the gas.
Energy per degree of freedom - $e = \dfrac{1}{2}{K_B}T$
So since a mono-atomic gas has more energy per degree of freedom, its temperature would be higher than that of the diatomic gas.
Mathematically - for mono-atomic gas - degrees of freedom = 3
Energy per degree of freedom $\dfrac{E}{3}$
$\dfrac{E}{3} = \dfrac{1}{2}{K_B}{T_m}$
for di-atomic gas -
degrees of freedom = 5

Energy per degree of freedom $\dfrac{E}{5}$
$\dfrac{E}{5} = \dfrac{1}{2}{K_B}{T_d}$
Comparing (1) and (2) -
${T_m} = \dfrac{5}{3}{T_d}$

Note: We have assumed that all the energy supplied to the system has gone into increasing the temperature. This is not true in general and some amount of energy also goes into doing work. But still, a molecule with higher degrees of freedom has less energy in each d.o.f and hence a lower temperature.