Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Calculate the rms speed of \(\mathrm{Br}_{2}\) molecules at \(23^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\) What is the rms speed of \(\mathrm{Br}_{2}\) at \(23^{\circ} \mathrm{C}\) and \(2.00 \mathrm{~atm} ?\)

Short Answer

Expert verified
The rms speed of Br2 is approximately 215.12 m/s at both 1.00 atm and 2.00 atm.

Step by step solution

01

Understanding the Problem

To find the root mean square (rms) speed of Br2 molecules, we need to use the formula for rms speed in terms of temperature and molar mass. The formula is: \(v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}\), where \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass.
02

Convert Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin because the gas constant requires temperature in Kelvin. Use the formula: \(T(K) = T(^{\circ}C) + 273.15\). Thus, the temperature is \(23 + 273.15 = 296.15\, K\).
03

Identify Constants and Molar Mass

The universal gas constant \(R\) is \(8.314\, \text{J/(mol}\cdot\text{K)}\). The molar mass of \(\mathrm{Br}_2\) is approximately \(159.808\, \text{g/mol}\), which converts to \(0.159808\, \text{kg/mol}\) for use in the formula.
04

Calculate RMS Speed at 1.00 ATM

Use the formula \(v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}\). Substitute \(R = 8.314\, \text{J/(mol}\cdot\text{K)}\), \(T = 296.15\, \text{K}\), and \(M = 0.159808\, \text{kg/mol}\). Calculate: \(v_{\text{rms}} = \sqrt{\dfrac{3 \times 8.314 \times 296.15}{0.159808}}\).
05

Compute RMS Speed

Perform the calculation: \(v_{\text{rms}} \approx \sqrt{\dfrac{7395.9672}{0.159808}} \approx \sqrt{46256.0714} \approx 215.12\, \text{m/s}\). Thus, the rms speed at 1.00 atm is approximately \(215.12\, \text{m/s}\).
06

Verify the Dependence on Pressure

The rms speed formula \(v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}\) does not explicitly depend on pressure. Therefore, for the same temperature, the rms speed remains the same regardless of changes in pressure. Thus, the rms speed at 2.00 atm will also be \(215.12\, \text{m/s}\).

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 Vaia!

Key Concepts

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

Root Mean Square Speed
Root mean square speed (rms speed) is a measure of the average speed of particles in a gas, calculated from the velocities of the molecules. If you're wondering why it's not just called average speed, think of it as a way to account for both the directional motion and the number of molecules. The root mean square speed is determined by using the formula:\[ v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}} \]This formula includes the gas constant \( R \), the temperature \( T \) in Kelvin, and the molar mass \( M \) in kilograms per mole.
Remember, rms speed is crucial for understanding how fast molecules in a sample move at a given temperature. This speed affects how often molecules collide, which is vital for reactions and behavior of gases.
Molecular Speed
Molecular speed refers to the speed at which individual gas molecules move. While rms speed gives us a particular average, molecular speeds can vary greatly. In gases, molecules are constantly in motion, colliding with each other, and bouncing off the walls of their container. These speeds depend on:
  • Temperature: Higher temperatures increase molecular speeds.
  • Molar Mass: Heavier molecules move more slowly than lighter ones at the same temperature.
It's fascinating to note that at any given temperature, lighter molecules like hydrogen will have faster molecular speeds compared to heavier ones like bromine.
Gas Laws
The behavior of gases is described by the gas laws, which link together pressure, volume, temperature, and the number of particles. Although the exercise deals with rms speed, which doesn't explicitly depend on pressure, understanding gas laws help in knowing how gases behave under different conditions. The key gas laws include:
  • Boyle's Law: Pressure inversely relates to volume for a constant temperature and gas amount.
  • Charles's Law: Volume directly relates to temperature for a constant pressure.
  • Avogadro's Law: Volume directly relates to the number of gas molecules at constant temperature and pressure.
These provide a foundation for understanding more complex gas behaviors and calculations.
Molecular Mass
Molecular mass, which is crucial when calculating rms speed, is the mass of a molecule. It affects how quickly a molecule moves within a gas at a given temperature. The larger the molecular mass, the slower the molecule's speed will be.For \( \text{Br}_2 \), the molecular mass is approximately 159.808 grams per mole. However, for calculations involving rms speed, it's essential to convert this to kilograms per mole, giving us 0.159808 kg/mol. This conversion is crucial because the other constants used in the calculation, like the gas constant \( R \), are in terms of energy, which requires mass to be in kilograms.
Temperature Conversion
Temperature conversion, particularly from Celsius to Kelvin, is a staple in chemistry calculations. Many equations in thermodynamics require the temperature to be input in Kelvin.The conversion formula is straightforward:\[ T(K) = T(^{\circ}C) + 273.15 \]
For example, to convert 23°C to Kelvin, we add 273.15, resulting in 296.15 K. This step ensures your calculations align with the units required by the formulas.
Understanding and mastering temperature conversion is vital for correctly calculating molecular velocities and other temperature-dependent phenomena.

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

A mixture contained calcium carbonate, \(\mathrm{CaCO}_{3}\), and magnesium carbonate, \(\mathrm{MgCO}_{3}\). A sample of this mixture weighing \(7.85 \mathrm{~g}\) was reacted with excess hydrochloric acid. The reactions are $$ \mathrm{CaCO}_{3}(g)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) $$ \(\mathrm{MgCO}_{3}(s)+2 \mathrm{HCl}(a q) \longrightarrow\) $$ \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(t)+\mathrm{CO}_{2}(g) $$ If the sample reacted completely and produced \(1.94 \mathrm{~L}\) of carbon dioxide, \(\mathrm{CO}_{2}\), at \(25^{\circ} \mathrm{C}\) and \(785 \mathrm{mmHg}\), what were the percentages of \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgCO}_{3}\) in the mixture?

A \(41.41-\mathrm{mL}\) sample of a \(0.1250 \mathrm{M}\) acid reacts with an excess of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to form \(150.0 \mathrm{~mL} \mathrm{CO}_{2}\) at \(646 \mathrm{mmHg}\) and \(27^{\circ} \mathrm{C}\). If the acid is either \(\mathrm{HCl}\) or \(\mathrm{H}_{2} \mathrm{SO}_{4}\), which is it?

An experiment calls for \(3.70 \mathrm{~mol}\) of chlorine, \(\mathrm{Cl}_{2}\) What volume will this be if the gas volume is measured at \(36^{\circ} \mathrm{C}\) and \(3.30 \mathrm{~atm} ?\)

Ethanol, the alcohol used in automobile fuels, is produced by the fermentation of sugars present in plants. Corn is often used as the sugar source. The following equation represents the fermentation of glucose, the sugar in corn, by yeast to produce ethanol and carbon dioxide. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ Ethanol is combusted in an automobile engine according to the equation $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) $$ What would be the total volume of \(\mathrm{CO}_{2}\) gas formed at STP when \(3.00 \mathrm{~kg}\) of sugar is fermented and the ethanol is then combusted in an automobile engine?

A vessel containing \(39.5 \mathrm{~cm}^{3}\) of helium gas at \(25^{\circ} \mathrm{C}\) and \(106 \mathrm{kPa}\) was inverted and placed in cold ethanol. As the gas contracted, ethanol was forced into the vessel to maintain the same pressure of helium. If this required \(7.5 \mathrm{~cm}^{3}\) of ethanol, what was the final temperature of the helium?

See all solutions

Recommended explanations on Chemistry 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.

Sign-up for free