Question

Consider the following statements:
For diatomic gasses, the ratio $\dfrac{{{C}_{p}}}{{{C}_{v}}}$is equal to
$(1)\,\,1.40$( Lower temperature)
$(2)\,\,1.66$(moderate temperature)
$(3)\,\,1.29$( Higher temperature)
Which of the above statements is correct
A.$1,\,2$and $3$
B.$1$and $2$
C.$2$and $3$
D.$1$and $3$

Answer 1

Hint: Generally heat capacity means the amount of heat energy per unit mass that is required to raise the temperature ${{1}^{o}}C$. Now the ratio of heat capacity at constant pressure$({{C}_{p}})$and heat capacity at constant volume$({{C}_{v}})$ varies for different monoatomic and diatomic gasses at a particular temperature.

Formula Used:Heat capacity at constant volume,${{C}_{v}}=\dfrac{3}{2}R$
And ${{C}_{p}}-{{C}_{v}}=R$
Here ${{C}_{p}}$denotes heat capacity at constant pressure and $R$denotes Universal gas constant.

Complete step-by-step solution:According to the law of Equipartition principle, for any dynamical system in thermal equilibrium, the energy system is equally distributed amongst the various degrees of freedom, and energy associated with each degree of freedom per molecule is $\dfrac{1}{2}KT$, Where $K$Boltzmann constant.
For a system of $n$molecules, each with $f$degrees of freedom,
Average internal energy,$U=n\times f\times \dfrac{1}{2}KT$
Or, $U=f\times \dfrac{1}{2}RT$
For monoatomic gas, there is only one translational degree of freedom in the direction of three dimensions. Hence $f=3$
$U=3\times \dfrac{1}{2}RT=\dfrac{3}{2}RT$
For diatomic gasses, there is $3$translational, $2$rotational degrees of freedom at low temperatures. But at high temperatures, $2$vibrational degrees of freedom come into play.
$U=5\times \dfrac{1}{2}RT=\dfrac{5}{2}RT$ [At low temperature,$f=(3+2)=5$]
$U=7\times \dfrac{1}{2}RT=\dfrac{7}{2}RT$ [At high temperature,$f=(3+2+2)=7$]
Differentiating both sides $dT$at constant volume$v$,
${{\left[ \dfrac{\partial U}{\partial T} \right]}_{v}}=\dfrac{\partial }{\partial T}{{\left[ \dfrac{5}{2}RT \right]}_{v}}$
Or,${{C}_{v}}=\dfrac{5}{2}R$
And ${{C}_{p}}=R+\dfrac{5}{2}R$ [since ${{C}_{p}}=R+{{C}_{v}}$]
Or,${{C}_{p}}=\dfrac{7}{2}R$
$\therefore \dfrac{{{C}_{p}}}{{{C}_{v}}}=\dfrac{\dfrac{7}{2}R}{\dfrac{5}{2}R}=\dfrac{7}{5}=1.4$
Similarly at high temperatures, ${{C}_{v}}=\dfrac{7}{2}R$
And ${{C}_{p}}=R+\dfrac{7}{2}R=\dfrac{9}{2}R$
$\therefore \dfrac{{{C}_{p}}}{{{C}_{v}}}=\dfrac{\dfrac{9}{2}R}{\dfrac{7}{2}R}=\dfrac{9}{7}=1.29$
Therefore for diatomic gasses, the ratio $\dfrac{{{C}_{p}}}{{{C}_{v}}}$ equals to $1.40$at low temperature and $1.29$at high temperature.

Thus, option (D) is correct.

Note: For monoatomic gas the heat capacity ratio $\dfrac{{{C}_{p}}}{{{C}_{v}}}$ equals to $1.66$as they do not possess rotational and vibrational degrees of freedom, but have only one translational degree of freedom along the three coordinate systems.