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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Monatomic Gases
Monatomic gases are gases that consist of individual atoms not bonded to any other atom. These gases are usually noble gases like helium (He), neon (Ne), and argon (Ar). Because these atoms move independently, they have simple kinetic properties. Monatomic gases are characterized by having an adiabatic index, (\( \gamma \)), of \( 5/3 \), which is approximately \( 1.67 \). This value arises from the ratio of heat capacities at constant pressure (\( C_p \)) to the heat capacities at constant volume (\( C_v \)).

The understanding of monatomic gases is crucial as they represent the simplest form of gaseous substance. Their straightforward molecular structure allows them to have high mobility and low viscosity, which affects how they conduct heat and sound. This simplicity also translates to their thermodynamic properties, making them easy to reference in thermodynamic calculations.
  • Monatomic gases have simple, predictable behaviors due to their single-atom structure.
  • They are important in the study of thermodynamics due to their straightforward properties.
Diatomic Gases
Unlike monatomic gases, diatomic gases consist of molecules that are made from two atoms bonded together. Common examples include nitrogen (\( N_2 \)), oxygen (\( O_2 \)), and hydrogen (\( H_2 \)). Diatomic gases have a more complicated internal structure than monatomic gases, leading to different thermochemical behaviors.

The adiabatic index (\( \gamma \)) for diatomic gases is \( 7/5 \), which is approximately \( 1.4 \). The presence of two atoms allows for rotational and vibrational energy modes, in addition to translational motion. These modes contribute to a lower \( \gamma \) value compared to monatomic gases, as more heat energy is absorbed internally before contributing to temperature change or pressure work.

The complexity of diatomic gases makes them instrumental in many chemical reactions and physical processes, particularly in the Earth's atmosphere.
  • Diatomic gases have additional energy modes due to their bonded atoms.
  • They play significant roles in natural processes and the composition of Earth's atmosphere.
Molecule Mixtures
The concept of molecule mixtures comes into play when different types of gases are mixed to achieve desired properties or conditions. In the context of this problem, mixing monatomic and diatomic gases provides a way to achieve an adiabatic index that is not standard in pure gases.

By carefully selecting the proportions of each gas, a mixture's properties can be tailored to meet specific needs, like attaining a particular \( \gamma \) value. For a mixture of monatomic and diatomic gases, the adiabatic index of the mixture can be calculated by taking a weighted average of each component's \( \gamma \) value. This results in the equation: \[ \gamma_{mixture} = x \gamma_{diatomic} + (1-x) \gamma_{monatomic} \]where \( x \) represents the mass fraction of the diatomic gas.

The ability to manipulate the properties of gas mixtures is highly valuable in chemical engineering and physics, enabling the design of environments that meet specific thermal and mechanical requirements. Such mixtures are common in various industrial processes and scientific applications.
  • Gas mixtures allow tailored thermal and mechanical properties.
  • Weighted averages of components' properties determine the mixture's characteristics.

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Most popular questions from this chapter

A tank of compressed helium for inflating balloons is advertised as containing helium at a pressure of 2400 psi, which, when allowed to expand at atmospheric pressure, will occupy a volume of \(244 \mathrm{ft}^{3}\). Assuming no temperature change takes place during the expansion, what is the volume of the tank in cubic feet?

What is the total mass of all the oxygen molecules in a cubic meter of air at normal temperature \(\left(25^{\circ} \mathrm{C}\right)\) and pressure \(\left(1.01 \cdot 10^{5} \mathrm{~Pa}\right) ?\) Note that air is about \(21 \%\) (by volume) oxygen (molecular \(\mathrm{O}_{2}\) ), with the remainder being primarily nitrogen (molecular \(\mathrm{N}_{2}\) ).

6.00 liters of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergo the following processes: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the pressure, volume, and temperature of the gas in states 2 and \(3 .\) How many moles of the gas are there?

1 .00 mol of molecular nitrogen gas expands in volume very quickly, so no heat is exchanged with the environment during the process. If the volume increases from \(1.00 \mathrm{~L}\) to \(1.50 \mathrm{~L},\) determine the work done on the environment if the gas's temperature dropped from \(22.0^{\circ} \mathrm{C}\) to \(18.0^{\circ} \mathrm{C}\). Assume ideal gas behavior.

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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