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

Two gases \(\mathrm{X}\) and \(\mathrm{Y}\) have their molecular speed in ratio of \(3: 1\) at certain temperature. The ratio of their molecular masses \(M_{x}: M_{y}\) is (a) \(1: 3\) (b) \(3: 1\) (c) \(1: 9\) (d) \(9: 1\)

Short Answer

Expert verified
The ratio of their molecular masses is \(1:9\).

Step by step solution

01

Understand the relationship between molecular speed and mass

The relationship between the molecular speeds and masses of gases can be described by the equation derived from the kinetic theory of gases: \(v_{x}/v_{y} = \sqrt{M_{y}/M_{x}}\), where \(v_{x}\) and \(v_{y}\) are the speeds of gases X and Y, respectively, and \(M_{x}\) and \(M_{y}\) are their molecular masses.
02

Substitute the given speed ratio

The given speed ratio of gas X to gas Y is 3:1, which means \(v_{x}/v_{y} = 3/1\). Substituting this into the equation from Step 1, we get \(3/1 = \sqrt{M_{y}/M_{x}}\).
03

Square both sides of the equation

To get rid of the square root, square both sides of the equation: \((3/1)^2 = (\sqrt{M_{y}/M_{x}})^2\) which simplifies to \(9/1 = M_{y}/M_{x}\).
04

Express the ratio of molecular masses

The ratio of molecular masses can be written as \(M_{x} : M_{y}\), which from the previous step is found to be \(1:9\). Therefore, the correct answer is (c) \(1:9\).

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.

Kinetic Theory of Gases
The kinetic theory of gases is a fundamental principle that describes the behavior of gases in terms of the movement of their particles. According to this theory, gases consist of tiny particles in constant, random motion. The energy of these particles is directly related to the temperature of the gas; higher temperatures indicate faster moving particles.

A crucial aspect of the kinetic theory is the understanding that the pressure exerted by a gas in a container results from collisions of the gas particles with the container's walls. This pressure is also affected by the speed of the molecules and their mass. A formula that ties these variables together is derived from the kinetic theory: \( v_x / v_y = \sqrt{M_y / M_x} \), where \(v_x \) and \(v_y \) are the speeds of two different gases, and \(M_x \) and \(M_y \) are their respective molecular masses.
Molecular Mass
Molecular mass, also known as molecular weight, is the sum of the atomic masses of all the atoms in a molecule. It's an essential factor when comparing different gases because it directly influences a molecule's speed at a given temperature. According to the kinetic theory of gases, lighter molecules move faster than heavier ones at the same temperature.

In the case of comparing gases, it's often useful to consider the relative molecular masses. For instance, in our original exercise, the ratio between two gases \(M_x : M_y\) is sought. This ratio can provide insight into the relative speeds of the gases, as seen in the step-by-step solution. Understanding how molecular mass impacts the behavior and properties of gases is crucial in fields like chemistry and physics.
Speed of Gases
The speed of gas molecules is an important concept within the kinetic theory of gases. It explains how molecular speed can be related to the kinetic energy and temperature of the gas. At a given temperature, lighter gas molecules (with lower molecular mass) tend to move faster than heavier molecules.

When we analyze two different gases at the same temperature and apply the kinetic theory, we can compare the speed of one gas to another using the relationship \(v_x / v_y = \sqrt{M_y / M_x}\), as in the textbook solution. The ratio of the speeds given in the problem (\(3:1\)) can be used to find the ratio of their molecular masses. This concept explains why the molecular speed ratio provided in the example leads us to the molecular mass ratio between the two gases being \(1:9\), as seen in step 4 of the given solution.

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

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) \(1.25 \mathrm{~atm}\)

When the pressure of a sample of gas is increased from \(0.50\) to \(100 \mathrm{~atm}\) at constant temperature, its volume decreases from \(2.01\) to \(13 \mathrm{ml}\). What could cause the deviation from Boyle's law? (a) Volume of the gas molecules is a significant fraction of the volume of container at higher pressure. (b) The force of attraction between the gas molecules is greater when the pressure is higher. (c) The molecules are dimerized at the higher pressure. (d) The collision of the molecules on the walls of the container are no longer elastic at higher pressure.

The \(P V-P\) isotherms of 1 mole of different gases at \(273 \mathrm{~K}\), if the limit of pressure tending to zero, converge to a value of \(P V=\) (a) \(11.21\) -atm (b) \(22.41\) -atm (c) zero (d) \(22.41\)

If Avogadro's number were to tend to infinity, the phenomenon of Brownian motion would (a) remain completely unaffected (b) become more vigorous than that observed with the present finite value of Avogadro's number, for all sizes of the Brownian particles (c) become more vigorous than that observed with the present finite value of Avogadro's number, only for relatively large Brownian particles (d) become practically unobservable, as the molecular impact would tend to balance one another for practically all sizes of Brownian particles

One mole of each gases \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) with van der Waal's constant \(\left(\mathrm{atm} 1^{2} \mathrm{~mol}^{-2}\right)\) \(1.348,6.823,4.390\) and \(2.438\), respec- tively, are kept separately in four different vessels of equal volumes at identical temperature. Their pressures are observed to be \(P_{A}, P_{\mathrm{B}}, P_{\mathrm{C}}\) and \(P_{\mathrm{D}}\), respectively. On the basis of this data alone, the order of pressure of gases is (assume other van der Waal's constant to be nearly same for all gases) (a) \(P_{\mathrm{A}}

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