Question

The specific heat at constant volume of mixture of $${\text{N}}_2$$ and $$\text{He}$$ ( $${\text{N}}_2:\text{He}=3:2$$ ) will be:
A. $$1.7R$$
B. $$1.5R$$
C. $$1.9R$$
D. $$2.1R$$

Answer 1

Hint: First of all, we will need the values of specific heat at constant volume for nitrogen and helium. After that, we will directly use the formula, which gives the specific heat of the mixture. We will substitute the required values and manipulate to obtain the answer.

Formula used:
The formula that gives the specific heat at constant volume, for a mixture of two gases is:
$$C_{{\text{v}}_{\text{mix}}}={\displaystyle \frac{n_1{C}_{{\text{v}}_{\text{He}}}+n_2{C}_{{\text{v}}_{{\text{N}}_{\text{2}}}}}{n_1+n_2}}$$ …… (1)
Where,
$$C_{{\text{v}}_{\text{mix}}}$$ indicates the specific heat of the mixture.
$$n_1$$ indicates the number of moles of helium.
$$C_{{\text{v}}_{\text{He}}}$$ indicates the specific heat of nitrogen.
$$n_2$$ indicates the number of moles of nitrogen.
$$C_{{\text{v}}_{{\text{N}}_{\text{2}}}}$$ indicates the specific heat of nitrogen.

Complete step by step answer:
In the given question, we are supplied with the following data:
There are two gases given nitrogen and helium.
The ratio of the two gases, nitrogen to helium is $$3:2$$.
Both the gases are mixed.
We are asked to find the specific heat at a constant volume of mixture of $${\text{N}}_2$$ and $$\text{He}$$.

Let us proceed to solve the numerical.
We know that helium is a monatomic gas. So, its specific heat can be written as:
$$C_{{\text{v}}_{\text{He}}}={\displaystyle \frac{3}{2}}R$$

Again, nitrogen is a di-atomic gas, so its specific heat can be written as:
$$C_{{\text{v}}_{{\text{N}}_{\text{2}}}}={\displaystyle \frac{5}{2}}R$$

Now, we substitute the required values in the equation (1) and we get:
$$\begin{gathered} C_{{\text{v}}_{\text{mix}}}={\displaystyle \frac{n_1{C}_{{\text{v}}_{\text{He}}}+n_2{C}_{{\text{v}}_{{\text{N}}_{\text{2}}}}}{n_1+n_2}} \\ \Rightarrow C_{{\text{v}}_{\text{mix}}}={\displaystyle \frac{2\times {\displaystyle \frac{3}{2}}R+3\times {\displaystyle \frac{5}{2}}R}{2+3}} \\ \Rightarrow C_{{\text{v}}_{\text{mix}}}={\displaystyle \frac{3R+{\displaystyle \frac{15}{2}}R}{5}} \\ \Rightarrow C_{{\text{v}}_{\text{mix}}}=2.1R \end{gathered}$$
Hence, the specific heat at constant volume of mixture of $${\text{N}}_2$$ and $$\text{He}$$ is $$2.1R$$.

So, the correct answer is “Option D”.

Note:
It is important to remember that the specific heat at constant volume and the specific heat at constant pressure changes depending on the type of gases taken. The values directly depend on the degrees of freedom. The minimum degrees of freedom are three whether it is monoatomic, diatomic or polyatomic. These three degrees of freedom are associated with the translational motion of the molecule.