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

Ordinary hydrogen is a mixture of the isotopes II and D. In which of the following propcrtics would the two gases \(\mathrm{II}_{2}\) and \(\mathrm{D}_{2}\) differ? (1) Number of molecules present in a given volume at STP (2) Rate of diffusion under the same conditions of temperature and pressure (3) Colour (4) Number of orbital clectrons

Short Answer

Expert verified
\(\text{H}_2\) and \(\text{D}_2\) differ in their rate of diffusion under the same conditions of temperature and pressure.

Step by step solution

01

Understanding Isotopes

Isotopes are atoms of the same element that have different numbers of neutrons. In this exercise, the isotopes II (Hydrogen-1) and D (Deuterium, Hydrogen-2) differ by one neutron.
02

Evaluating Property 1: Number of Molecules Present in a Given Volume at STP

Under standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. This means that both \(\text{H}_2\) and \(\text{D}_2\) will have the same number of molecules in a given volume at STP.
03

Evaluating Property 2: Rate of Diffusion

Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Since Deuterium (\text{D}_2) has a higher molar mass than Hydrogen (\text{H}_2), \(\text{H}_2\) will diffuse faster than \(\text{D}_2\).
04

Evaluating Property 3: Colour

Both \(\text{H}_2\) and \(\text{D}_2\) are colourless gases. Hence, they do not differ in colour.
05

Evaluating Property 4: Number of Orbital Electrons

Both isotopes, Hydrogen-1 and Deuterium, have the same number of protons and electrons. Therefore, they do not differ in the number of orbital electrons.
06

Conclusion

Based on the analysis of the properties, \(\text{H}_2\) and \(\text{D}_2\) differ only in their rate of diffusion under the same conditions of temperature and pressure.

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.

Graham's Law of Diffusion
Graham's law of diffusion helps us understand how gases spread or mix with each other. It tells us that lighter gases diffuse faster than heavier ones. This is because the rate at which a gas diffuses is inversely proportional to the square root of its molar mass. So, if you have two gases, the one with the lower molar mass will move quicker.
For example, hydrogen gas \(\text{H}_2\) and deuterium gas \(\text{D}_2\) differ in mass. \(\text{D}_2\) is heavier due to the presence of an extra neutron in each molecule. Thus, \(\text{H}_2\) will diffuse faster than \(\text{D}_2\). This idea can be represented using the equation for Graham's law: \[ \frac{\text{Rate of diffusion of gas 1}}{\text{Rate of diffusion of gas 2}} = \sqrt{\frac{\text{Molar mass of gas 2}}{\text{Molar mass of gas 1}}} \]
In this way, we can predict how quickly a gas will diffuse compared to another.
Standard Temperature and Pressure (STP)
Understanding STP is essential when studying the properties of gases. STP stands for Standard Temperature and Pressure, and it is used as a reference point in gas calculations.
Standard temperature is 0 degrees Celsius (273.15 Kelvin), and standard pressure is 1 atmosphere (atm). At these conditions, one mole of any gas occupies a volume of 22.4 liters.
This means that hydrogen \(\text{H}_2\) and deuterium \(\text{D}_2\) at STP will have the same number of molecules if they have the same volume. The formula \[ \text{PV} = \text{nRT} \] denotes the ideal gas law, where:
  • \(\text{P}\) = pressure
  • \(\text{V}\) = volume
  • \(\text{n}\) = number of moles
  • \(\text{R}\) = universal gas constant
  • \(\text{T}\) = temperature
This relationship helps us see that under the same conditions of STP, the volume occupied by a gas is directly proportional to the number of moles of gas, regardless of its type.
Molecular Properties of Gases
Gases have unique molecular properties that define their behavior. Some of these properties include:
  • Diffusion: The process in which gas molecules spread out to fill the available space. Graham's law, as discussed earlier, is a fundamental principle of diffusion.
  • Compressibility: Gases can be compressed into smaller volumes when pressure is applied due to the significant amounts of empty space between molecules.
  • Expansion: Gases expand to fill their containers, so if you increase the volume of a container, the gas will spread to fill the entire space.
  • Low Density: Gases generally have much lower densities compared to liquids and solids because their molecules are far apart.
  • Elastic Collisions: Gas molecules collide with each other and the walls of their container without losing energy. This is why pressure is exerted by gases.
  • Temperature and Pressure Relationship: According to the ideal gas law, at a constant volume, the pressure of a gas is directly proportional to its temperature, and at a constant pressure, the volume is directly proportional to its temperature.
Each of these properties is significant in understanding how gases behave under various conditions, and they help explain why gases like \(\text{H}_2\) and \(\text{D}_2\) act differently under the same conditions.

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

Which of the following is correct for hydrides of lanthanides and actinidcs? (1) They are less denser than the parent metal. (2) They have fairly high heat of formation. (3) Their electronegativity values are fairly close to those of Group I and II elements. (4) All are correct.

\Lambdatomic hydrogen is obtained by passing ordinary hydrogen through (1) a suitable catalyst maintained at high temperature under high pressure (2) a solution containing zinc and sulphuric acid (3) an clectric arc (4) a silent clectric discharge at ordinary tempcrature

Which of the following statement is false? (1) Hydrogen can be placed in the halogen group because it forms hydrides like chlorides. (2) The best explanation for not placing hydrogen with the group of alkali metals or halogens is that the IP of \(\mathrm{H}_{2}\) is too higher than alkali metal but too low than halogens. (3) The tendency of hydrogen to gain one electron to acquire He configuration resembles halogens. (4) Hydrogen molecules are diatomic and form \(\mathrm{H}\) ions similar to halogens.

Which of the following statements regarding saline hydridcs is falsc? (1) Saline hydrides are formed only by thosc clements which have very low electronegativity values and can transfer clectrons to the hydrogen atom. (2) Salt-like hydrides are solids with ionic lattices. (3) In fused state, saline hydrides conduct electricity. (4) The density of salt-like hydrides is lesser than the parent metal.

The melting points of most of the solid substances, increases with increase in pressure. However, ice melts at a lower temperature lower than its usual melting point when the pressure is increased. This is because (1) ice is less denser than water (2) pressure generates heat (3) the chemical bonds break under pressure (4) ice is not a solid

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