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Chapter 19: Problem 52

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 a mixture has to have to obtain this value.

Short Answer

Expert verified
Answer: To achieve an adiabatic index value of 1.60, the mixture should contain 25% diatomic molecules and 75% monatomic molecules.

Step by step solution

01

Recall the adiabatic index (γ) values for monatomic and diatomic gases

Monatomic gases, which consist of single atoms, have an adiabatic index (γ) value of 5/3 (approximately 1.67), while diatomic gases, made up of two atoms bonded together, have a γ value of 7/5 (approximately 1.4). Let the mass fraction of diatomic molecules be denoted by x and that of monatomic molecules be denoted by (1-x).
02

Write the equation relating the adiabatic indexes of the mixed gases

To find the γ value for a mixture of monatomic and diatomic gases, we can use the formula: \[ \gamma_{mixture} = \frac{x\gamma_{diatomic} + (1-x)\gamma_{monatomic}}{x+(1-x)}\] Where; \(x\) = mass fraction of diatomic molecules, \(\gamma_{diatomic}\) = γ value for diatomic gas (7/5), \(\gamma_{monatomic}\) = γ value for monatomic gas (5/3), and \(\gamma_{mixture}\) = desired γ value of the mixture (1.60)
03

Solve the equation for the mass fraction of diatomic molecules

Substitute the given values into the equation and solve for x: \[ 1.60 = \frac{x \times \frac{7}{5} + (1-x) \times \frac{5}{3}}{x+(1-x)} \] Now solve for x: \[ 1.60 = \frac{7x}{5} + \frac{5(1-x)}{3} \] Clear the denominators by multiplying both sides by 15: \[ 24 = 21x + 25 - 25x \] Combine the terms containing x: \[ 24 - 25 = -4x \] Now, divide by -4 to get the value of x: \[ x = \frac{-1}{-4} \] \[ x = \frac{1}{4} \]
04

Convert the fraction to percentage for better understanding

To express the mass fraction of diatomic molecules in percentage, multiply by 100: \[ x \% = \frac{1}{4} \times 100\% = 25\%\] So, to obtain a gas mixture with a γ value of 1.60, the mixture should contain 25% diatomic molecules and 75% monatomic molecules.

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Most popular questions from this chapter

In a period of \(6.00 \mathrm{~s}, 9.00 \cdot 10^{23}\) nitrogen molecules strike a section of a wall with an area of \(2.00 \mathrm{~cm}^{2}\). If the molecules move with a speed of \(400.0 \mathrm{~m} / \mathrm{s}\) and strike the wall head on in elastic collisions, what is the pressure exerted on the wall? (The mass of one \(\mathrm{N}_{2}\) molecule is \(4.68 \cdot 10^{-26} \mathrm{~kg}\).)

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