/*! 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 37 Consider two 5.0 - \(\mathrm{L}\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider two 5.0 - \(\mathrm{L}\) containers, each filled with gas at \(25^{\circ} \mathrm{C}\). One container is filled with helium and the other with \(\mathrm{N}_{2}\). The density of gas in the two containers is the same. What is the relationship between the pressures in the two containers?

Short Answer

Expert verified
Pressure of helium is 7 times that of nitrogen.

Step by step solution

01

Understand the Problem

We are given two containers with the same volume and temperature. One container has helium gas, and the other has nitrogen gas. The density of the gas inside each container is the same. We need to find how the pressures in each container relate to each other.
02

Use the Ideal Gas Law

Start with the ideal gas law equation, which is \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is moles of gas, \(R\) is the gas constant, and \(T\) is temperature. Since \(V\) and \(T\) are the same for both gases, we can compare the pressures by examining \(n\) for each gas.
03

Translate Density into Moles

Density is mass per unit volume. We can express the density \(d\) as \(d = \frac{m}{V}\). Given the density is the same for both gases, we have \(\frac{m_{\text{He}}}{M_{\text{He}}} = \frac{m_{N_2}}{M_{N_2}}\) with \(M\) representing the molar mass. Thus, \(m_{\text{He}} = d \cdot V \cdot M_{\text{He}}\) and \(m_{N_2} = d \cdot V \cdot M_{N_2}\). We convert masses to moles using \(n = \frac{m}{M}\).
04

Express Moles in Terms of Density

For helium: \(n_{\text{He}} = \frac{d \cdot V}{M_{\text{He}}}\). For nitrogen: \(n_{N_2} = \frac{d \cdot V}{M_{N_2}}\). Since the densities are equal, we can relate these moles to the pressures by \(P_{\text{He}} = \frac{n_{\text{He}}RT}{V}\) and \(P_{N_2} = \frac{n_{N_2}RT}{V}\).
05

Combine and Compare Equations

Now, substitute the expressions for \(n\) from Step 4 into the ideal gas law formulas for pressure:\[P_{\text{He}} = \frac{d \cdot RT}{M_{\text{He}}}\]\[P_{N_2} = \frac{d \cdot RT}{M_{N_2}}\]Divide the entire equation for nitrogen by that for helium:\[\frac{P_{N_2}}{P_{\text{He}}} = \frac{M_{\text{He}}}{M_{N_2}}\]
06

Calculate the Molar Mass Ratio

Helium has a molar mass \(M_{\text{He}} = 4 \text{ g/mol}\) and nitrogen has a molar mass \(M_{N_2} = 28 \text{ g/mol}\).Thus, \[\frac{M_{\text{He}}}{M_{N_2}} = \frac{4}{28} = \frac{1}{7}\]
07

Finalize the Pressure Relationship

Substituting the molar mass ratio back:\[\frac{P_{N_2}}{P_{\text{He}}} = \frac{1}{7}\]Therefore, \(P_{N_2} = \frac{1}{7}P_{\text{He}}\).

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.

Gas Density
Gas density is a measure of how much mass of a gas is present in a given volume. It is mathematically expressed as the mass of the gas divided by its volume: \( d = \frac{m}{V} \). In simpler terms, density tells you how 'heavy' or 'light' a gas is in a particular space.
Understanding gas density is crucial in this exercise because we are dealing with equal densities of two different gases—helium (\( \text{He} \)) and nitrogen (\( \text{N}_2 \)).
This is important to grasp because even though these gases have drastically different molar masses, the problem sets them equal by density.
- Helium is much lighter than nitrogen due to its smaller molar mass.- But in our containers, they have the same density, meaning the mass per unit volume is equal.
This equal density implies a specific relationship between quantities of each gas present, which will later help us understand their pressure differences.
Molar Mass
Molar mass is the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). It plays a key role when using the ideal gas law to determine other properties of the gas.
In this exercise, helium has a molar mass of 4 g/mol while nitrogen's molar mass is 28 g/mol, a significant difference. This difference in molar mass is crucial because:
  • It allows us to calculate the number of moles from a given mass, using the formula \( n = \frac{m}{M} \).
  • It determines how much gas is present for a fixed amount of mass and thus can greatly impact the pressure of that gas.
  • With equal densities given, a lighter molar mass gas (helium) will result in more moles than a heavier molar mass gas (nitrogen).
This disparity highlights how, even with the same amount of mass per volume (density) between two containers, their chemical identity (molar mass) will influence pressures differently.
Gas Pressure Relationship
Pressure in the context of gases is the force exerted by the gas particles when they collide with the walls of their container. The ideal gas law establishes the relationship between pressure and other related variables: \( PV = nRT \).
In this particular exercise, we compared the pressure in two containers of gas that have the same density but different chemical natures. Under the conditions set by the problem, the pressures are affected mainly by the number of moles and not the physical space or temperature, since these are constant.
The step-by-step solution led us to compare the pressures by observing their molar mass differences, resulting in a pressure relationship of \( P_{N_2} = \frac{1}{7} P_{He} \).
This relationship tells us that despite having the same gas density and volume, the gas particles' different identities cause nitrogen's pressure to be significantly less than that of helium:
  • Because helium is a lighter gas, its molecules are more numerous in the same volume, explaining why \( P_{He} \) is greater.
  • The difference encapsulated in the 1/7 ratio reflects pure chemical nature decisions about how gas atoms and molecules behave under uniform physical conditions.
Thus, understanding gas pressure relationships requires balancing density, molar mass, and how these factors influence the molecular kinetic energy observed as pressure.

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

State Avogadro's law. Explain why two volumes of hydrogen react with one volume of oxygen to form two volumes of steam.

What is the definition of pressure?

Use the postulates of the kinetic-molecular theory to explain each phenomenon. (a) \(\mathrm{Br}_{2}(\mathrm{~g})\) is reddish brown and transparent; \(\mathrm{Br}_{2}(\ell)\) is very dark brown and very little light passes through it. (b) When equal volumes of \(\mathrm{Br}_{2}(\mathrm{~g})\) and \(\mathrm{N}_{2}(\mathrm{~g})\) at the same \(T\) and \(P\) are brought into contact, they mix rapidly and the color is only half as dark as the initial \(\mathrm{Br}_{2}\) color.

Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O},\) is a volatile organic compound that is sometimes released from insulation used in home construction, and it can be trapped and build up inside the home. When this happens, people exposed to the formaldehyde can suffer adverse health effects. The U. S. National Institute of Occupational Health and Safety (NIOSH) guideline for the maximum allowable concentration of formaldehyde in air in the workplace is \(16 \mathrm{ppb}\) (parts per billion) for an eight-hour average exposure. (a) Determine the partial pressure of formaldehyde at the maximum allowable level of \(16 \mathrm{ppb}\). (b) Calculate how many molecules of formaldehyde are present in each cubic centimeter of air when formaldehyde is present at \(16 \mathrm{ppb}\). (c) Calculate how many total molecules of formaldehyde are present in a room: \(15.0 \mathrm{ft}\) long \(\times 10.0 \mathrm{ft}\) wide \(X\) \(8.00 \mathrm{ft}\) high (at \(16 \mathrm{ppb}\) ).

At \(25^{\circ} \mathrm{C}\), the measured pressure of acetic acid vapor, \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{g})\), is significantly lower than that predicted by the ideal gas law. Explain this difference.

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.