/*! 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 31 What is the temperature of a gas... [FREE SOLUTION] | 91Ó°ÊÓ

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What is the temperature of a gas of \(\mathrm{CO}_{2}\) molecules whose rms speed is \(329 \mathrm{m} / \mathrm{s}\) ?

Short Answer

Expert verified
The temperature of the gas is approximately 300 K.

Step by step solution

01

Understand the Root Mean Square Speed Formula

The root mean square (rms) speed formula for gas molecules is given by \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( v_{rms} \) is the root mean square speed, \( k \) is Boltzmann’s constant \( (1.38 \times 10^{-23} \; \mathrm{J/K}) \), \( T \) is the temperature in Kelvin, and \( m \) is the mass of a single molecule.
02

Calculate the Mass of a Single \( \mathrm{CO}_2 \) Molecule

First, find the molecular mass of \( \mathrm{CO}_2 \). The molar mass of \( \mathrm{CO}_2 \) is approximately \( 44.01 \; \mathrm{g/mol} \), which needs to be converted to kilograms per molecule using Avogadro's number: \( 44.01 \; \mathrm{g/mol} = 44.01 \times 10^{-3} \; \mathrm{kg/mol} \), and dividing by \( 6.022 \times 10^{23} \; \mathrm{mol^{-1}} \). So, \( m = \frac{44.01 \times 10^{-3}}{6.022 \times 10^{23}} \; \mathrm{kg} \).
03

Rearrange the Formula to Solve for Temperature

Rearrange the rms speed equation to solve for \( T \):\[ T = \frac{m v_{rms}^2}{3k} \]
04

Substitute the Known Values and Calculate Temperature

Substitute the values:- \( v_{rms} = 329 \; \mathrm{m/s} \)- \( k = 1.38 \times 10^{-23} \; \mathrm{J/K} \)- \( m = \frac{44.01 \times 10^{-3}}{6.022 \times 10^{23}} \; \mathrm{kg} \)Calculate:\[ T = \frac{\left(\frac{44.01 \times 10^{-3}}{6.022 \times 10^{23}}\right) (329)^2}{3 \times 1.38 \times 10^{-23}} \approx 300 \; \mathrm{K} \].

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Key Concepts

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

Root Mean Square Speed
The root mean square (rms) speed is an important concept in understanding the motion of gas molecules. It provides a measure of the average speed of these tiny particles in a container. Unlike the average speed, the rms speed considers the direction and velocity of each molecule, ensuring precise motion calculations.

The formula for calculating the rms speed is \[ v_{rms} = \sqrt{\frac{3kT}{m}} \]Here,
  • \( v_{rms} \) represents the root mean square speed.
  • \( k \) is Boltzmann's constant, a fundamental constant in physics with the value approximately \( 1.38 \times 10^{-23} \; \mathrm{J/K} \).
  • \( T \) is the temperature of the gas in Kelvin.
  • \( m \) is the mass of a single molecule of the gas.
The rms speed gives essential insights into how temperature affects the movement of particles. As the temperature increases, the rms speed of the molecules increases, indicating that they move faster.
Boltzmann's Constant
Let's explore Boltzmann's constant, a crucial factor in the equation for root mean square speed. Boltzmann's constant \( (k) \) bridges the gap between macroscopic and microscopic physical properties. It links the temperature of a gas to the kinetic energy of its molecules.

With a value of approximately \( 1.38 \times 10^{-23} \; \mathrm{J/K} \), Boltzmann's constant is vital in statistical mechanics, helping to quantify how temperature relates to energy at the molecular level. This connection is a cornerstone of the kinetic theory of gases, which explains the macroscopic properties of gases in terms of the movements of their molecules.

Additionally, Boltzmann's constant appears in many other important equations in physics and chemistry, such as in the ideal gas law and the distribution of molecular speeds, making it fundamental for calculations involving gases.
Molecular Mass Calculation
Calculating the molecular mass of gases is a critical step when solving problems involving their motion and temperature. In our example, the gas was \( \mathrm{CO}_2 \), which has a molar mass of about \( 44.01 \; \mathrm{g/mol} \). To use this in the context of kinetic energy or rms speed, this molar mass needs to be converted to the mass of an individual molecule.

The steps are straightforward:
  • First, convert the molar mass from grams per mole to kilograms per mole: \( 44.01 \; \mathrm{g/mol} = 44.01 \times 10^{-3} \; \mathrm{kg/mol} \).
  • Next, use Avogadro's number, \( 6.022 \times 10^{23} \; \mathrm{mol^{-1}} \), to find the mass of a single molecule: \( m = \frac{44.01 \times 10^{-3}}{6.022 \times 10^{23}} \; \mathrm{kg} \).
This individual molecular mass can then be used in the rms speed formula to relate molecular movement to temperature. Understanding how to convert and calculate molecular mass is essential for applying key concepts of gas dynamics.

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Most popular questions from this chapter

Standard temperature and pressure (STP) is defined as a temperature of \(0^{\circ} \mathrm{C}\) and a pressure of \(101.3 \mathrm{kPa}\). What is the volume occupied by one mole of an ideal gas at STP?

A balloon is filled with helium at a pressure of \(2.4 \times 10^{5} \mathrm{Pa}\). The balloon is at a temperature of \(18^{\circ} \mathrm{C}\) and has a radius of \(0.25 \mathrm{m} .\) (a) How many helium atoms are contained in the balloon? (b) Suppose we double the number of helium atoms in the balloon, keeping the pressure and the temperature fixed. By what factor does the radius of the balloon increase? Explain.

If you put a helium-filled balloon in the refrigerator, (a) will its volume increase, decrease, or stay the same? (b) Choose the best explanation from among the following: I. Lowering the temperature of an ideal gas at constant pressure results in a reduced volume. II. The same amount of gas is in the balloon; therefore, its volume remains the same. III. The balloon can expand more in the cool air of the refrigerator, giving an increased volume.

Amount of Helium in a Blimp The Goodyear blimp Spirit of Alcron is \(62.6 \mathrm{m}\) long and contains \(7023 \mathrm{m}^{3}\) of helium. When the temperature of the helium is \(285 \mathrm{K}\), its absolute pressure is \(112 \mathrm{kPa} .\) Find the mass of the helium in the blimp.

A gas has a temperature of \(310 \mathrm{K}\) and a pressure of \(101 \mathrm{kPa}\). (a) Find the volume occupied by 1.25 mol of this gas, assuming it is ideal. (b) Assuming the gas molecules can be approximated as small spheres of diameter \(2.5 \times 10^{-10} \mathrm{m},\) determine the fraction of the volume found in part (a) that is occupied by the molecules. (c) In determining the properties of an ideal gas, we assume that molecules are points of zero volume. Discuss the validity of this assumption for the case considered here.

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