At STP, the order of the RMS speed of molecules of \(\mathrm{H}_{2}, \mathrm{~N}_{2}, \mathrm{O}_{2}\) and HBr gases is (a) \(\mathrm{H}_{2}>\mathrm{N}_{2}>\mathrm{O}_{2}>\mathrm{HBr}\) (b) \(\mathrm{HBr}>\mathrm{O}_{2}>\mathrm{N}_{2}>\mathrm{H}_{2}\) (c) \(\mathrm{HBr}>\mathrm{H}_{2}>\mathrm{O}_{2}>\mathrm{N}_{2}\) (d) \(\mathrm{N}_{2}>\mathrm{O}_{2}>\mathrm{H}_{2}>\mathrm{HBr}\)

Short Answer

Expert verified
\(H_{2} > N_{2} > O_{2} > HBr\)

Step by step solution

01

Understand the Concept of RMS Speed

The root mean square (RMS) speed of gas molecules is given by the formula: \(v_{rms} = \sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of a gas molecule. At standard temperature and pressure (STP), \(T\) is constant, so the RMS speed of the gas molecules is inversely proportional to the square root of the molar mass of the gas.
02

Calculate Molar Mass of each Gas

Find the molar mass of each gas. For \(H_2\), the molar mass is 2 g/mol. For \(N_2\), it is 28 g/mol. For \(O_2\), it is 32 g/mol. For HBr, it is the sum of the molar masses of hydrogen and bromine, which is 1 g/mol + 80 g/mol = 81 g/mol.
03

Compare the RMS Speed Based on Molar Mass

Since the RMS speed is inversely proportional to the square root of the molar mass, we can compare the speeds of the gases by inverting the order of their molar masses. The smaller the molar mass, the higher the RMS speed.
04

Determine the Correct Order of RMS Speeds

Listing the gases in order from the smallest to the largest molar mass: \(H_2\) (2 g/mol), \(N_2\) (28 g/mol), \(O_2\) (32 g/mol), HBr (81 g/mol). Now reverse the order for RMS speeds: \(H_2\) > \(N_2\) > \(O_2\) > HBr.

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 Velocity
The concept of root mean square (RMS) velocity is fundamental in understanding the behavior of gas molecules. It represents the measure of the speed of particles within a gas, providing insight into their kinetic energy. The RMS velocity is calculated using the formula \(v_{rms} = \sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant, \(T\) is the absolute temperature, and \(m\) is the mass of an individual gas molecule.

At a given temperature and pressure, all gas molecules have the same kinetic energy. However, because the masses of different gas molecules vary, their speed also varies. Lighter molecules move faster than heavier ones, which is why RMS velocity differs between gases. This concept helps students understand why, under the same conditions, certain gases diffuse more quickly than others and has significant implications in fields like thermodynamics and statistical mechanics.
Molar Mass of Gases
Molar mass is another pivotal concept in the study of gases. It is defined as the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). The molar mass of gases can be determined by adding up the atomic masses of the elements that make up the molecule.

For instance, the molar mass of hydrogen gas \(H_2\) is twice the atomic mass of hydrogen because each molecule consists of two hydrogen atoms. Understanding the molar mass of gases enables students to make predictions about gas behavior, including RMS velocities. This property is instrumental in calculations involving the Ideal Gas Law and other gas-related formulas, which is why grasping the molar mass concept is critical for successfully navigating chemistry problems.
Boltzmann Constant
The Boltzmann constant, denoted by \(k\), is a crucial value in the realm of physics and chemistry, connecting the macroscopic and microscopic worlds. It relates the average kinetic energy of particles in a gas with the temperature of the gas. The constant has units of energy per temperature per particle, typically expressed in joules per kelvin (J/K) or in terms of \(\frac{L\cdot atm}{mol\cdot K}\).

The significance of the Boltzmann constant lies in its role within the RMS velocity formula and its broader relevance in the statistical mechanics of particle systems. It serves as a bridge between macroscopic measurements, like temperature and pressure, and microscopic properties, like the energy of particles.
Standard Temperature and Pressure
When addressing the behavior of gases, it is often necessary to reference a common baseline or standard conditions, known as standard temperature and pressure (STP). This standard is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere (atm). These conditions are significant because they allow for the comparison of data between different experiments and observations. They serve as the benchmark for calculations and tabulations of gas properties.

At STP, the volume of one mole of an ideal gas is known to be 22.414 liters, and this standard reference makes it easier to discuss and compare the behavior of gases, such as determining their RMS speed. The fact that the temperature is held constant at STP simplifies problems involving the RMS velocity, enabling students to focus on the relationship between the molar mass of the gas and its RMS velocity.

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 balloon is filled with \(\mathrm{N}_{2} \mathrm{O}\) is pricked with a sharp point and quickly plunged into a tank of \(\mathrm{CO}_{2}\) under the same pressure and temperature. The balloon will (a) be enlarged (b) shrink (c) remain unchanged in size (d) collapse completely

The pressure outside a jet plane flying at high altitude falls considerably below atmospheric pressure at sea level. The air inside the cabin must therefore be pressurized to protect the passengers. What is the pressure (in atmosphere) in the cabin if the barometer reading is \(688 \mathrm{~mm}\) of \(\mathrm{Hg}\) ? (a) \(0.905 \mathrm{~atm}\) (b) \(6.88\) atm (c) \(9.05 \mathrm{~atm}\) (d) data, insufficient

The partial pressures of \(\mathrm{N}_{2}, \mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) in a vessel are \(38 \mathrm{~cm}\) of \(\mathrm{Hg}, 190\) torr and \(0.5\) atm, respectively. The total pressure of the mixture at the same temperature is (a) \(0.96 \mathrm{~atm}\) (b) \(1.02 \mathrm{~atm}\) (c) \(1.64\) atm (d) \(1.25 \mathrm{~atm}\)

Vapour is injected at a uniform rate in a closed vessel which was initially evacuated. The pressure in the vessel (a) increases continuously (b) decreases continuously (c) first increases and then decreases (d) first increases and then becomes constant

An unknown gas ' \(\mathrm{X}\) ' has rate of diffusion measured to be \(0.88\) times that of \(\mathrm{PH}_{3}\) at the same conditions of temperature and pressure. The gas may be (a) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (b) \(\mathrm{CO}\) (c) \(\mathrm{NO}_{2}\) (d) \(\mathrm{N}_{2} \mathrm{O}\)

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