/*! 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 49 Suppose that the typical speed \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that the typical speed \(\left(v_{\mathrm{rms}}\right)\) of carbon dioxide molecules (molar mass is \(44.0 \mathrm{g} / \mathrm{mol}\) ) in a flame is found to be \(1350 \mathrm{m} / \mathrm{s} .\) What temperature does this indicate?

Short Answer

Expert verified
The temperature indicated by the given rms speed of carbon dioxide molecules in the flame is approximately 3221 Kelvin.

Step by step solution

01

Write the formula for rms speed

The formula relating the root-mean-square (rms) speed, temperature, and molar mass is: \(v_{rms} = \sqrt{\dfrac{3RT}{M}}\) Where: - \(v_{rms}\) is the rms speed of the molecules - R is the universal gas constant (R = 8.314 J/mol K) - T is the temperature (in Kelvin) - M is the molar mass of the gas (in kg/mol)
02

Convert the molar mass of carbon dioxide to kg/mol

Given the molar mass of carbon dioxide is \(44.0 \frac{g}{mol}\), we need to convert it to kg/mol for accurate calculation. Divide the given value by 1000 to convert grams to kilograms: \(44.0 \frac{g}{mol} = \dfrac{44}{1000} \frac{kg}{mol} = 0.044 \frac{kg}{mol}\)
03

Rearrange the formula to find the temperature

We need to find the temperature (T) from the rms speed formula. Rearrange the formula to solve for T: \(T = \dfrac{Mv^2_{rms}}{3R}\)
04

Substitute the given values and calculate the temperature

Now, we will plug in the given values of \(v_{rms} (1350 \frac{m}{s})\) and the calculated molar mass (M = 0.044 kg/mol) into the rearranged formula and solve for T: \(T = \dfrac{(0.044 \frac{kg}{mol})\times (1350 \frac{m}{s})^2}{3 \times (8.314 \frac{J}{mol \cdot K})}\) \(T = \dfrac{(0.044 \frac{kg}{mol})\times (1822500 \frac{m^2}{s^2})}{24.942 \frac{J}{mol \cdot K}}\) \(T = \dfrac{80295 \frac{kg \cdot m^2}{mol \cdot s^2}}{24.942 \frac{J}{mol \cdot K}}\) \(T \approx 3221 K\) The temperature indicated by the given rms speed of carbon dioxide molecules in the flame is approximately 3221 Kelvin.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Carbon Dioxide Molecules
Carbon dioxide (COâ‚‚) is a common gas found in Earth's atmosphere. It consists of one carbon atom covalently bonded to two oxygen atoms. This simple molecule plays a critical role in numerous biological and environmental processes, such as respiration and photosynthesis. Understanding the behavior of carbon dioxide molecules at different temperatures is important in various scientific fields.
When we consider the root-mean-square (rms) speed of carbon dioxide molecules, we are looking at a statistical measure of the speed of particles in a gas. It is relevant in thermodynamics as it relates to the kinetic energy of molecules:
  • The rms speed is affected by the molecular weight of the gas as well as temperature.
  • Gases with lighter molecules generally have higher rms speeds at a given temperature compared to gases with heavier molecules.
  • For COâ‚‚, with a molar mass of 44.0 g/mol, the rms speed will be lower than that of lighter gases like helium.
Universal Gas Constant
The universal gas constant, often denoted as R, is a fundamental constant used in various equations of state in thermodynamics. It represents the proportionality factor that relates the energy scale to the temperature scale. R is essential in the ideal gas law equation and other calculations involving gases:
- R has a fixed value of 8.314 J/mol·K, which stands for joules per mole per Kelvin. - It allows for the connection between the macroscopic and microscopic properties of a gas. - In equations like the rms speed formula, it helps link temperature, molar mass, and speed. - R plays a critical role in energy-related calculations, such as determining the amount of heat involved in gas processes. Understanding R provides insights into the behavior of gases under different conditions and is central to thermodynamic problems, like calculating the temperature of carbon dioxide in a flame.
Molar Mass Conversion
Molar mass conversion is a process of transforming the units of molar mass from grams per mole (\(g/mol\)) to kilograms per mole (\(kg/mol\)). This conversion is important when using formulas that require consistent units, such as the rms speed equation.
To correctly convert the molar mass of carbon dioxide, follow these steps:
  • Identify the given molar mass, which is usually in grams per mole, like 44.0 \(g/mol\).
  • Convert it to kilograms per mole by dividing by 1000, since there are 1000 grams in a kilogram.
  • For COâ‚‚, this conversion results in \(0.044\) \(kg/mol\).
This conversion ensures that the units used in calculations align, allowing accurate results when determining the temperature or other properties.
Temperature Calculation
Temperature calculation involves rearranging the rms speed formula to solve for temperature, T. By understanding the relationship between speed, molar mass, and the universal gas constant, we can derive the temperature of a gas.
Here's how you can calculate temperature:
  • Start with the rms speed formula: \(v_{rms} = \sqrt{\dfrac{3RT}{M}}\)
  • Rearrange it to solve for temperature: \(T = \dfrac{Mv_{rms}^2}{3R}\)
  • Substitute in the known values: the rms speed of \(1350 \mathrm{m/s}\), the molar mass \(0.044 \mathrm{kg/mol}\), and the universal gas constant \(8.314 \mathrm{J/mol \cdot K}\).
Perform the calculation step by step:
- Calculate \(v_{rms}^2 = (1350)^2\)
- Plug these numbers into the rearranged formula to find the temperature.
- Simplifying will yield the temperature as approximately \(3221 \mathrm{K}\).This process helps us understand the kinetic energy and thermal state of the gas particles in a given environment, such as a flame.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You mix 5 moles of \(\mathrm{H}_{2}\) at \(300 \mathrm{K}\) with 5 moles of He at \(360 \mathrm{K}\) in a perfectly insulated calorimeter. Is the final temperature higher or lower than \(330 \mathrm{K}\) ?

In a sample of hydrogen sulfide \((M=34.1 \mathrm{g} / \mathrm{mol})\) at a temperature of \(3.00 \times 10^{2} \mathrm{K},\) estimate the ratio of the number of molecules that have speeds very close to \(v_{\mathrm{rms}}\) to the number that have speeds very close to \(2 v_{\mathrm{rms}}\)

How is momentum related to the pressure exerted by a gas? Explain on the molecular level, considering the behavior of molecules.

(a) People often think of humid air as "heavy." Compare the densities of air with \(0 \%\) relative humidity and \(100 \%\) relative humidity when both are at 1 atm and \(30^{\circ} \mathrm{C} .\) Assume that the dry air is an ideal gas composed of molecules with a molar mass of \(29.0 \mathrm{g} / \mathrm{mol}\) and the moist air is the same gas mixed with water vapor. (b) As discussed in the chapter on the applications of Newton's laws, the air resistance felt by projectiles such as baseballs and golf balls is approximately \(F_{\mathrm{D}}=C \rho A v^{2} / 2,\) where \(\rho\) is the mass density of the air, \(A\) is the cross-sectional area of the projectile, and \(C\) is the projectile's drag coefficient. For a fixed air pressure, describe qualitatively how the range of a projectile changes with the relative humidity. (c) When a thunderstorm is coming, usually the humidity is high and the air pressure is low. Do those conditions give an advantage or disadvantage to home-run hitters?

The gauge pressure in your car tires is 18. The gase \(2.50 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) at a temperature of \(35.0^{\circ} \mathrm{C}\) when you drive it onto a ship in Los Angeles to be sent to Alaska. What is their gauge pressure on a night in Alaska when their temperature has dropped to \(-40.0^{\circ} \mathrm{C}\) ? Assume the tires have not gained or lost any air.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.