/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Calculate the temperature at whi... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the temperature at which a carbon dioxide molecule would have the same root-mean- square speed as a neon atom at \(125^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The temperature at which CO₂ has the same RMS speed as Ne at 125°C is approximately 200.57 K.

Step by step solution

01

Convert Celsius to Kelvin

To work with the kinetic theory calculations, we need the temperature in Kelvin. The conversion formula is:\[ T(K) = T(^{\circ}C) + 273.15 \]For neon at \(125^{\circ}C\), we have:\[ T = 125 + 273.15 = 398.15 \, K \]
02

Calculate RMS Speed of Neon

The root-mean-square (RMS) speed of a gas molecule is given by:\[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \]where:- \(k\) is Boltzmann's constant, \(1.38 \times 10^{-23} \text{J/K}\)- \(m\) is the mass of one moleculeFirst, find the molar mass of neon (Ne), which is approximately \(20.18 \, \text{g/mol}\), and convert it to kg:\[ m_{\text{Ne}} = \frac{20.18}{1000} \times \frac{1}{N_A} = \frac{20.18 \times 10^{-3}}{6.022 \times 10^{23}} \approx 3.35 \times 10^{-26} \, \text{kg} \]Now, substitute to find \(v_{\text{rms}}\) for neon:\[ v_{\text{rms,Ne}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 398.15}{3.35 \times 10^{-26}}} \approx 609.39 \, \text{m/s} \]
03

Find Mass of Carbon Dioxide Molecule

Use the molar mass of carbon dioxide (COâ‚‚), which is approximately \(44.01 \, \text{g/mol}\), and convert it to kg per molecule:\[ m_{\text{COâ‚‚}} = \frac{44.01}{1000} \times \frac{1}{N_A} = \frac{44.01 \times 10^{-3}}{6.022 \times 10^{23}} \approx 7.31 \times 10^{-26} \, \text{kg} \]
04

Calculate Temperature for COâ‚‚

Equating the \(v_{\text{rms}}\) of COâ‚‚ to that of Ne, using the formula \(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\), we solve for temperature \(T\) for COâ‚‚:\[ v_{\text{rms,COâ‚‚}} = v_{\text{rms,Ne}} \rightarrow \sqrt{\frac{3kT_{\text{COâ‚‚}}}{m_{\text{COâ‚‚}}}} = 609.39 \]Squaring both sides, we get:\[ \frac{3kT_{\text{COâ‚‚}}}{m_{\text{COâ‚‚}}} = (609.39)^2 \]Solving for \(T_{\text{COâ‚‚}}\):\[ T_{\text{COâ‚‚}} = \frac{(609.39)^2 \times m_{\text{COâ‚‚}}}{3k} \approx \frac{(609.39)^2 \times 7.31 \times 10^{-26}}{3 \times 1.38 \times 10^{-23}} \approx 200.57 \, K \]

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Key Concepts

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

Kinetic Theory of Gases
The kinetic theory of gases is an essential concept in understanding the behavior of gases. It provides a molecular-level explanation for pressure, temperature, and volume of gases. According to this theory, gases are composed of a large number of tiny particles that move rapidly in all directions. These particles, which are atoms or molecules, are in constant random motion, colliding with each other and the walls of the container.
One fundamental aspect of the kinetic theory is the idea of root-mean-square speed (RMS speed). The RMS speed is a measure of the average speed of gas particles, and it's influenced by the temperature and mass of the gas molecules. RMS speed is calculated using the formula:
  • \[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \] where \(v_{\text{rms}}\) is the root-mean-square speed, \(k\) is Boltzmann's constant, \(T\) is the absolute temperature in Kelvin, and \(m\) is the mass of one molecule.

    • This formula indicates that as the temperature increases, the RMS speed increases, illustrating the direct relationship between temperature and the energy of gas molecules. Understanding this relationship helps in solving problems that require comparing the speeds of different gas molecules at varied conditions.
Temperature Conversion
Temperature conversion, especially between Celsius and Kelvin, is a crucial step in chemistry and physics calculations involving gases. For any gas calculation based on the kinetic theory of gases, it is essential to work with temperature in Kelvins, as it starts from absolute zero, which is the theoretical point where particle motion stops.
To convert a temperature from Celsius (
  • \(^\circ C\) to Kelvin, the formula is: \[ T(K) = T(^{\circ}C) + 273.15 \]

    • This conversion is necessary because most scientific equations, including those in kinetic theory, require the temperature to be in Kelvin to ensure accurate and consistent results.
      For instance, in the exercise, the temperature of neon at 125°C is converted to Kelvin using this formula, giving us 398.15 K. This temperature conversion enables the calculation of the RMS speed of the neon gas molecules under the given conditions.
Molecular Mass Calculation
Molecular mass calculation is fundamental when dealing with the kinetic theory of gases, as the mass of a molecule directly affects the RMS speed. Molecular mass, also known as molar mass, is typically given in grams per mole (g/mol), but for use in kinetic theory calculations, it needs to be converted to kilograms per molecule.
Here are the steps for molecular mass calculation:
  • Find the molar mass of the substance from the periodic table or other sources. For example, neon is about 20.18 g/mol.
  • Convert the molar mass from grams per mole to kilograms by dividing by 1000.
  • Divide by Avogadro's number ( \(6.022 \times 10^{23}\) molecules/mol) to find the mass of an individual molecule.
For neon, this results in a molecular mass of approximately 3.35 \( \times 10^{-26} \) kg. Similarly, for carbon dioxide (COâ‚‚), with a molar mass of about 44.01 g/mol, the calculation gives approximately 7.31 \( \times 10^{-26} \) kg per molecule. Accurate molecular mass calculations are crucial for determining the correct RMS speed and solving kinetic gas problems effectively.

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Most popular questions from this chapter

Explain what factors not accounted for by the ideal-gas equation are corrected for by the van der Waals constants \(a\) and \(b\).

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