/*! 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 112 Interstellar space contains most... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Interstellar space contains mostly hydrogen atoms at a concentration of about 1 atom/cm \(^{3}\). (a) Calculate the pressure of the \(\mathrm{H}\) atoms. (b) Calculate the volume (in liters) that contains \(1.0 \mathrm{~g}\) of \(\mathrm{H}\) atoms. The temperature is \(3 \mathrm{~K}\)

Short Answer

Expert verified
We are given the concentration of hydrogen atoms as 1 atom/cm\(^{3}\), and the temperature as 3 K. To calculate the pressure, we will use the ideal gas law \(PV = nRT\), where \(n\) is the number of moles, \(R\) is the ideal gas constant, \(P\) is the pressure, and \(V\) is the volume. Since \(n = \frac{N}{N_A}\), where \(N\) is the number of hydrogen atoms and \(N_A\) is Avogadro's number, we must convert the concentration from atoms/cm\(^{3}\) to moles/cm\(^{3}\).

Step by step solution

01

Identify Given Values for Pressure Calculation

We are given the concentration of hydrogen atoms as 1 atom/cm\(^{3}\), and the temperature as 3 K. To calculate the pressure, we will use the ideal gas law \(PV = nRT\), where \(n\) is the number of moles, \(R\) is the ideal gas constant, \(P\) is the pressure, and \(V\) is the volume. Since \(n = \frac{N}{N_A}\), where \(N\) is the number of hydrogen atoms and \(N_A\) is Avogadro's number, we must convert the concentration from atoms/cm\(^{3}\) to moles/cm\(^{3}\).

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 91Ó°ÊÓ!

Key Concepts

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

Hydrogen Atoms
Hydrogen atoms are the simplest atoms in the universe, consisting of just one proton and one electron. These atoms are extremely light and are the most abundant element in the cosmos. In interstellar space, such as the region between stars, hydrogen atoms usually exist as neutral hydrogen atoms rather than molecules due to the low temperatures and low densities found there.

Given that outer space contains about 1 hydrogen atom per cubic centimeter, this might seem sparse compared to Earth's atmosphere. However, this is typical of interstellar space where such low densities are normal. This sparse density of hydrogen plays a significant role in understanding various astrophysical phenomena, such as the formation of stars and galaxies.
Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry and physics, used to solve problems involving gases. The law is expressed as:\[ PV = nRT \]

Here, \(P\) is the pressure of the gas, \(V\) is the volume, \(n\) is the number of moles of the gas, \(R\) is the ideal gas constant (8.314 J/mol·K), and \(T\) is the temperature in Kelvin.

This law assumes that gases are composed of tiny particles in rapid, random motion, which perfectly translate kinetic energy into the pressure exerted by the gas within a container. While it provides a very good approximation for many gases under a wide range of conditions, it performs best under conditions of low pressure and high temperature. Known deviations occur in gases that are either highly compressed or at very low temperatures, but in the vast, low-pressure environment of space, the ideal gas law holds quite well.
Pressure Calculation
To calculate the pressure of hydrogen atoms in interstellar space, we need to switch from thinking about individual atoms to thinking about moles—the standard chemical unit. Here’s a simplified process to follow:

The concentration given is 1 atom/cm\(^3\). First, convert this concentration into moles using Avogadro's number, which is approximately 6.022 x 10\(^{23}\) atoms/mol. This conversion helps because the ideal gas law is naturally defined with moles, not atoms.

Once converted, we can apply the ideal gas law \(PV = nRT\), rearranging it to solve for pressure \(P\):\[ P = \frac{nRT}{V} \]

Plugging in the given temperature (3 K) and the converted concentration, we obtain the pressure of this diffuse interstellar gas. The result will indicate the influence of these sparse hydrogen atoms in the vast emptiness of space.
Moles Conversion
Moles are a central concept in chemistry, serving as the link between particle counts and mass. One mole represents Avogadro's number—approximately 6.022 x 10\(^{23}\) particles, whether they be atoms, molecules, or ions.

To convert between these units, especially when dealing with gases like hydrogen, you need to connect the atomic scale to the macroscopic scale. For instance, if you are given a mass of hydrogen, you can find the number of moles using:\[ n = \frac{m}{M} \] where \(m\) is the mass of the gas, and \(M\) is the molar mass of hydrogen, about 2 g/mol for diatomic hydrogen (H\(_2\)), but 1 g/mol for atomic hydrogen.

Such conversions are vital for using the ideal gas law, as this law depends on moles rather than the mass or a simple count of atoms. Being comfortable with these conversions helps better understand, predict, and manipulate chemical and physical systems.

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

Nitrogen forms several gaseous oxides. One of them has a density of \(1.33 \mathrm{~g} / \mathrm{L}\) measured at \(764 \mathrm{mmHg}\) and \(150^{\circ} \mathrm{C}\). Write the formula of the compound.

A mixture of methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) is stored in a container at \(294 \mathrm{mmHg}\). The gases are burned in air to form \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). If the pressure of \(\mathrm{CO}_{2}\) is 356 \(\mathrm{mmHg}\), measured at the same temperature and volume as the original mixture, calculate the mole fractions of the gases.

The apparatus shown here can be used to measure atomic and molecular speeds. Suppose that a beam of metal atoms is directed at a rotating cylinder in a vacuum. A small opening in the cylinder allows the atoms to strike a target area. Because the cylinder is rotating, atoms traveling at different speeds will strike the target at different positions. In time, a layer of the metal will deposit on the target area, and the variation in its thickness is found to correspond to Maxwell's speed distribution. In one experiment it is found that at \(850^{\circ} \mathrm{C}\) some bismuth (Bi) atoms struck the target at a point \(2.80 \mathrm{~cm}\) from the spot directly opposite the slit. The diameter of the cylinder is \(15.0 \mathrm{~cm},\) and it is rotating at 130 revolutions per second. (a) Calculate the speed (in \(\mathrm{m} / \mathrm{s}\) ) at which the target is moving. (Hint: The circumference of a circle is given by \(2 \pi r\), where \(r\) is the radius.) (b) Calculate the time (in seconds) it takes for the target to travel \(2.80 \mathrm{~cm} .\) (c) Determine the speed of the \(\mathrm{Bi}\) atoms. Compare your result in part (c) with the \(u_{\mathrm{rms}}\) of \(\mathrm{Bi}\) at \(850^{\circ} \mathrm{C}\). Comment on the difference.

What is the mass of the solid \(\mathrm{NH}_{4} \mathrm{Cl}\) formed when \(73.0 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) is mixed with an equal mass of \(\mathrm{HCl} ?\) What is the volume of the gas remaining, measured at \(14.0^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) ? What gas is it?

Estimate the distance (in \(\mathrm{nm}\) ) between molecules of water vapor at \(100^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm} .\) Assume ideal behavior. Repeat the calculation for liquid water at \(100^{\circ} \mathrm{C},\) given that the density of water is \(0.96 \mathrm{~g} / \mathrm{cm}^{3}\) at that temperature. Comment on your results. (Assume each water molecule to be a sphere with a diameter of \(0.3 \mathrm{nm} .\) ) (Hint: First calculate the number density of water molecules. Next, convert the number density to linear density, that is, the number of molecules in one direction.)

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.