Chapter 20: Q. 65 (page 569)

An experiment you're designing needs a gas with $γ = 1.50$ . You recall from your physics class that no individual gas has this value, but it occurs to you that you could produce a gas with $γ = 1.50$ by mixing together a monatomic gas and a diatomic gas. What fraction of the molecules need to be monatomic?

Short Answer

Expert verified

The Fraction of molecules needs at monatomic $= \frac{1}{2}$ .

Step by step solution

Step: 1 Change in temperature:

In the micro-system, the potential energy between the bonds is zero, and the kinetic energy of a monatomic gas is translational. If the temperature varies by a certain amount, $∆ T$ , then the thermal energy for system is given by equation in the form

$Δ E_{th} = \frac{i}{2} n R Δ T$

The monatomic gas equation is,

$Δ E_{th} = \frac{3}{2} n_{1} R Δ T$

The diatomic gas equation is,

$Δ E_{th} = \frac{5}{2} n_{2} R Δ T$

The sum of energy is

$Δ E_{th} = \frac{3}{2} n_{1} R Δ T + \frac{5}{2} n_{2} R Δ T$

The molar specific heat is calculated as follows:

$Δ E_{th} = (n_{1} + n_{2}) C_{V} Δ T .$

Step: 2 Value of CV:

We are give $γ = 1.5, so C_{V} is$

$\begin{matrix} C_{p} = γ C_{V} \\ C_{p} = 1.5 C_{V} \\ C_{V} + R = 1.5 C_{v} \\ C_{V} = 2 R . \end{matrix}$

Step: 3 Finding fraction of molecules:

The molar specific heat of the the mixture is

$\begin{aligned} \frac{3}{2} n_{1} R Δ T + \frac{5}{2} n_{2} R Δ T = (n_{1} + n_{2}) C_{V} Δ T \\ (\frac{3}{2} n_{1} + \frac{5}{2} n_{2}) R = (n_{1} + n_{2}) 2 R \\ \frac{3}{2} n_{1} - 2 n_{1} = - \frac{5}{2} n_{2} + 2 n_{2} \\ - 0.5 n_{1} = - 0.5 n_{2} \\ n_{1} = n_{2} \end{aligned}$

The molarity in each group is the same. The total mass is ${\frac{N}{N}}_{A}$ where $N_{A}$ is the number calculated by Avogadro. The monatomic molecule's number is multiplied. $2$ to exceed the diatomic's number of molecules As a result, the diatomic gas's fraction of monatomic gas is $\frac{1}{2}$ .