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

(a) List two experimental conditions under which gases deviate from ideal behavior. (b) List two reasons why the gases deviate from ideal behavior. (c) Explain how the function \(P V / R T\) can be used to show how gases behave nonideally.

Short Answer

Expert verified
(a) Two experimental conditions under which gases deviate from ideal behavior are high pressure and low temperature. (b) Gases deviate from ideal behavior due to intermolecular forces and the finite size of molecules. (c) The function \(PV/RT\) can be used to show how gases behave nonideally by comparing its calculated value to the ideal value of 1. A value different from 1 indicates nonideal behavior, influenced by experimental conditions and the type of gas being observed.

Step by step solution

01

(a) Experimental Conditions of Deviation from Ideal Behavior

Two experimental conditions under which gases deviate from ideal behavior are: 1. High pressure: The deviation from ideal behavior becomes more significant when gases are subjected to high pressure, as the gas molecules are forced closer together and can no longer be treated as points. 2. Low temperature: At low temperatures, the speed of gas molecules decreases, increasing the chance of intermolecular attractions, which in turn cause more significant deviations from ideal behavior.
02

(b) Reasons for Deviation from Ideal Behavior

There are two primary reasons why gases deviate from ideal behavior: 1. Intermolecular forces: In an ideal gas, it is assumed that there are no intermolecular forces between gas molecules. However, in reality, gas molecules are influenced by weak attractive and repulsive forces. These interactions lead to deviation from ideal behavior. 2. Finite size of molecules: Ideal gas assumptions state that gas molecules have no volume or size. However, in real gases, molecules have a finite size, and when the volume or pressure becomes significant, the volume occupied by the gas molecules themselves will affect the behavior of the gas.
03

(c) Using the Function \(PV/RT\) to Show Nonideal Behavior

The function \(PV/RT\) is used to describe the behavior of a gas: - \(P\) represents pressure - \(V\) represents volume - \(R\) is the gas constant - \(T\) is the temperature For an ideal gas, the value of \(PV/RT\) should be equal to 1 at all conditions. However, in real gases, deviations from ideal behavior result in the value of \(PV/RT\) being either greater or lesser than 1, depending on the experimental conditions and type of gas. When \(PV/RT\) is greater than 1, it indicates that the gas is experiencing stronger intermolecular forces and/or larger molecular volume than what the ideal gas model assumes. This is typically observed when gases are under high pressure or at low temperatures. On the other hand, when \(PV/RT\) is less than 1, it means that the gas is behaving more ideally than under normal circumstances, which can be attributed to reduced intermolecular forces or negligibly small molecular volume. In conclusion, the function \(PV/RT\) can be used to demonstrate how gases deviate from ideal behavior by comparing its calculated value to the ideal value of 1. A value different from 1 is indicative of nonideal behavior, with deviations influenced by experimental conditions and the type of gas being observed.

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!

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 and hydrogen gases react to form ammonia gas as follows: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ At a certain temperature and pressure, \(1.2 \mathrm{~L}\) of \(\mathrm{N}_{2}\) reacts with \(3.6 \mathrm{~L}\) of \(\mathrm{H}_{2}\). If all the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are consumed, what volume of \(\mathrm{NH}_{3}\), at the same temperature and pressure, will be produced?

An aerosol spray can with a volume of \(250 \mathrm{~mL}\) contains \(2.30 \mathrm{~g}\) of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) as a propellant. (a) If the can is at \(23^{\circ} \mathrm{C}\), what is the pressure in the can? (b) What volume would the propane occupy at STP? (c) The can says that exposure to temperatures above \(130^{\circ} \mathrm{F}\) may cause the can to burst. What is the pressure in the can at this temperature?

As discussed in the "Chemistry Put to Work" box in Section \(10.8\), enriched uranium can be produced by gaseous diffusion of \(\mathrm{UF}_{6}\). Suppose a process were developed to allow diffusion of gaseous uranium atoms, \(\mathrm{U}(g)\). Calculate the ratio of diffusion rates for ${ }^{235} \mathrm{U}\( and \){ }^{238} \mathrm{U}$, and compare it to the ratio for UF \(_{6}\) given in the essay.

A piece of solid carbon dioxide with a mass of \(5.50 \mathrm{~g}\) is placed in a \(10.0\) -L vessel that already contains air at 705 torr and \(24^{\circ} \mathrm{C}\). After the carbon dioxide has totally vaporized, what is the partial pressure of carbon dioxide and the total pressure in the container at \(24^{\circ} \mathrm{C}\) ?

The typical atmospheric pressure on top of \(\mathrm{Mt}\). Everest \((29,028 \mathrm{ft})\) is about 265 torr. Convert this pressure to (a) \(\mathrm{atm}\), (b) \(\mathrm{mm} \mathrm{Hg}\), (c) pascals, (d) bars.

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