/*! 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 12 If you have any two gases in dif... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 12

If you have any two gases in different containers that are the same size at the same pressure and same temperature, what is true about the moles of each gas? Why is this true?

Short Answer

Expert verified
The moles of each gas are equal (n_1 = n_2) under the same pressure, volume, and temperature conditions. This is true because of the Ideal Gas Law, which dictates the relationship between the pressure, volume, temperature, and moles of a gas. In this case, since pressure, volume, and temperature are kept the same for both gases, their moles must also be equivalent.

Step by step solution

01

Recall the Ideal Gas Law

The Ideal Gas Law is given by the equation \(PV = nRT\), where P is pressure, V is volume, n is the moles of gas, R is the ideal gas constant, and T is temperature.
02

Apply the given conditions to the Ideal Gas Law

Since the pressures, volumes, and temperatures of both gases are the same, we can rewrite the Ideal Gas Law for both gases as follows: For Gas 1: \(P_1V_1 = n_1RT_1\) For Gas 2: \(P_2V_2 = n_2RT_2\) Given that \(P_1 = P_2\), \(V_1 = V_2\), and \(T_1 = T_2\), we can now compare these two equations.
03

Compare the equations for both gases

Since the pressure, volume, and temperature are the same for both gases, the equations become: \(PV = n_1R\cancel{T}\) \(PV = n_2R\cancel{T}\) Since both equations are equal to the same value, we can equate them: \(n_1RT = n_2RT\)
04

Simplify the equation and find the relationship between moles

We can now divide both sides of the equation by RT to find the relationship between the moles of each gas: \(n_1\cancel{R}\cancel{T} = n_2\cancel{R}\cancel{T}\) \(n_1 = n_2\)
05

Conclusion

From the analysis, we find that the moles of each gas are equal (n_1 = n_2) under the same pressure, volume, and temperature conditions. This is true because of the Ideal Gas Law, which dictates the relationship between the pressure, volume, temperature, and moles of a gas. In this case, since pressure, volume, and temperature are kept the same for both gases, their moles must also be equivalent.

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 91影视!

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

A gas sample containing \(1.50 \mathrm{~mol}\) at \(25^{\circ} \mathrm{C}\) exerts a pressure of 400 . torr. Some gas is added to the same container and the temperature is increased to \(50 .{ }^{\circ} \mathrm{C}\). If the pressure increases to 800 . torr, how many moles of gas were added to the container? Assume a constant-volume container.

At room temperature, water is a liquid with a molar volume of \(18 \mathrm{~mL}\). At \(105^{\circ} \mathrm{C}\) and 1 atm pressure, water is a gas and has a molar volume of over \(30 \mathrm{~L}\). Explain the large difference in molar volumes.

A glass vessel contains \(28 \mathrm{~g}\) nitrogen gas. Assuming ideal behavior, which of the processes listed below would double the pressure exerted on the walls of the vessel? a. Adding enough mercury to fill one-half the container. b. Raising the temperature of the container from \(30 .{ }^{\circ} \mathrm{C}\) to \(60 .{ }^{\circ} \mathrm{C}\). c. Raising the temperature of the container from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). d. Adding 28 g nitrogen gas.

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be produced by the following reaction: $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ Hydrogen at STP flows into a reactor at a rate of \(16.0 \mathrm{~L} / \mathrm{min}\) Carbon monoxide at STP flows into the reactor at a rate of \(25.0\) L/min. If \(5.30 \mathrm{~g}\) methanol is produced per minute, what is the percent yield of the reaction?

Without looking at a table of values, which of the following gases would you expect to have the largest value of the van der Waals constant \(b: \mathrm{H}_{2}, \mathrm{~N}_{2}, \mathrm{CH}_{4}, \mathrm{C}_{2} \mathrm{H}_{6}\), or \(\mathrm{C}_{3} \mathrm{H}_{8}\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation
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.