Question

You are designing an experiment that requires a gas with \(\gamma=1.60 .\) However, from your physics lectures, you remember that no gas has such a \(\gamma\) value. However, you also remember that mixing monatomic and diatomic gases can yield a gas with such a \(\gamma\) value. Determine the fraction of diatomic molecules in a mixture of gases that has this value.

Short answer

Answer: The percentage of diatomic molecules in a mixture with an adiabatic index of 1.6 is approximately 81.25%.

Step-by-step solution

Review properties of diatomic and monatomic gases

Monatomic gases have a constant volume specific heat capacity of \(C_V^{mono} = \frac{3}{2} R\), whereas diatomic gases have a constant volume specific heat capacity of \(C_V^{di} = \frac{5}{2} R\), where R is the specific gas constant. In the same manner, the constant pressure specific heat capacities of monatomic and diatomic gases are \(C_P^{mono} = \frac{5}{2} R\) and \(C_P^{di} = \frac{7}{2} R\), respectively.

Find the specific heat capacities of the mixture

Let the fraction of diatomic molecules in the mixture be denoted as x. Therefore, the fraction of monatomic molecules is (1-x). The specific heat capacity of the mixture at constant volume is given by: \(C_V^{mix} = x C_V^{di} + (1 - x) C_V^{mono} = x\left(\frac{5}{2}R\right) + (1 - x)\left(\frac{3}{2}R\right)\) Similarly, the specific heat capacity of the mixture at constant pressure is given by: \(C_P^{mix} = x C_P^{di} + (1 - x) C_P^{mono} = x\left(\frac{7}{2}R\right) + (1 - x)\left(\frac{5}{2}R\right)\)

Establish the relationship between specific heat capacities and adiabatic index

Recall that the adiabatic index \(\gamma\) can be expressed as the ratio of specific heat capacities: \(\gamma = \frac{C_P}{C_V}\) We can write the desired adiabatic index for the mixture in terms of its specific heat capacities: \(1.6 = \frac{C_P^{mix}}{C_V^{mix}}\) We now need to solve for x in terms of the specific heat capacities.

Solve for the fraction of diatomic molecules (x)

Firstly, substitute the expressions of \(C_P^{mix}\) and \(C_V^{mix}\) in terms of x: \(1.6 = \frac{x\left(\frac{7}{2}R\right) + (1 - x)\left(\frac{5}{2}R\right)}{x\left(\frac{5}{2}R\right) + (1 - x)\left(\frac{3}{2}R\right)}\) We can simplify this expression and solve for x: \(1.6 = \frac{7x + 5 - 5x}{5x + 3 - 3x}\) Now cross-multiply and further simplify: \(1.6(5x + 3 - 3x) = 7x + 5 - 5x\) \(8x - 4.8x = 2.6\) \(x = \frac{2.6}{3.2}\) \(x \approx 0.8125\)

Interpret the result

The fraction of diatomic molecules in a mixture of gases with an adiabatic index of 1.6 is approximately 0.8125, or 81.25%. This means that the mixture should contain 81.25% diatomic molecules and 18.75% monatomic molecules to achieve the desired adiabatic index of 1.6.