Question

Equal volumes of two monatomic gases, A and B at same temperature and pressure are mixed. The ratio of specific heats $\left( {{C}_{p}}/{{C}_{v}} \right)$ of the mixture will be:
[A] 1.50
[B] 3.3
[C] 1.67
[D] 0.83

Answer 1

Hint: To solve this firstly identify the degree of freedoms for a monatomic gas. For every degree of freedom, the contribution of molar internal energy to kinetic energy is $\dfrac{1}{2}RT$. Using the equipartition theorem you can find out ${{C}_{v}}$ and then use the relation ${{C}_{p}}-{{C}_{v}}=R$ to find out the unknown term.

Complete answer:
We know that ${{C}_{p}}$ is the specific heat capacity of a gas at constant pressure and it is basically the amount of heat capacity that is required to raise the temperature of a substance by one degree at a constant pressure.
 On the other hand, ${{C}_{v}}$ is the specific heat capacity of a gas at constant volume and it is the amount of heat capacity that is required to raise the temperature of a substance by one degree at a constant volume.
Now let us discuss the heat capacity values at constant temperature and pressure for a monatomic gas to solve the given question.
 We know that according to the equipartition theorem, each degree of freedom will contribute $\dfrac{1}{2}RT$ to the molar internal energy. Gases can move around in 3 directions and this makes their degree of freedom as 3.
Therefore, U for a monatomic gas is $\dfrac{3}{2}RT$.
So, we can write that ${{C}_{v}}$ is $\dfrac{3}{2}R$ (as it is the energy with respect to temperature)
And now from here we can find out the ${{C}_{p}}$ using the relation,
$\begin{align}
& {{C}_{p}}-{{C}_{v}}=R \\
& Or,{{C}_{p}}=R+{{C}_{v}}=R+\dfrac{3R}{2}=\dfrac{5R}{2} \\
\end{align}$
Therefore, from here we can write down that $\dfrac{{{C}_{p}}}{{{C}_{v}}}=\dfrac{\dfrac{5R}{2}}{\dfrac{3R}{2}}=\dfrac{5}{3}=1.67$
We can understand from the above calculation that the ratio of specific heat capacities is 1.67.

Therefore the correct answer is option [C] 1.67.

Note:
We have used the relation ${{C}_{p}}-{{C}_{v}}=R$ to calculate the molar heat capacity at constant pressure but we should remember that this is true for ideal gases. Here we have monatomic gases therefore we considered them as ideal gases. Rotation in a gas molecule affects its degree of freedom and thus affects these values too.