Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 5

According to the ideal gas law, what happens to the pressure of a gas in a container when: (a) You double the absolute temperature of the gas? (b) You double the number of liters the container can hold? (c) You double the number of moles of the gas? (d) You double both the absolute temperature of the gas and the number of liters the container can hold?

Short Answer

Expert verified
(a) When the absolute temperature of the gas is doubled, the pressure of the gas also doubles. (b) When the volume of the container is doubled, the pressure of the gas is halved. (c) When the number of moles of the gas is doubled, the pressure of the gas also doubles. (d) When both the absolute temperature of the gas and the number of liters the container can hold are doubled, the pressure of the gas remains constant.

Step by step solution

01

Rewrite the Ideal Gas Law equation

Let's rewrite the Ideal Gas Law equation with the given change: \(P_1V_1 = n_1R(T_1)\) and \(P_2V_2 = n_2R(T_2)\), where T2 is twice the initial temperature (T1).
02

Compare the two situations

Since volume (V), the number of moles (n), and the ideal gas constant (R) remain constant in this scenario, we can write: \(P_1V = nRT_1\) and \(P_2V = nR(2T_1)\).
03

Find the new pressure

Divide the second equation by the first one: \(\dfrac{P_2V}{P_1V} = \dfrac{nR(2T_1)}{nRT_1}\). Simplify the equation to find P2 in terms of P1: \(\dfrac{P_2}{P_1} = 2\). So, when the absolute temperature of the gas is doubled, the pressure of the gas is also doubled. (b) Doubling the number of liters the container can hold:
04

Rewrite the Ideal Gas Law equation

Let's rewrite the Ideal Gas Law equation with the given change: \(P_1V_1 = n_1RT_1\) and \(P_2V_2 = n_2RT_2\), where V2 is twice the initial volume V1.
05

Compare the two situations

Since the temperature (T), the number of moles (n), and the ideal gas constant (R) remain constant in this scenario, we can write: \(P_1V_1 = nRT\) and \(P_2(2V_1) = nRT\).
06

Find the new pressure

Divide the second equation by the first one: \(\dfrac{P_2(2V_1)}{P_1V_1} = \dfrac{nRT}{nRT}\). Simplify the equation to find P2 in terms of P1: \(\dfrac{P_2}{P_1} = \dfrac{1}{2}\). So, when the volume of the container is doubled, the pressure of the gas is halved. (c) Doubling the number of moles of the gas:
07

Rewrite the Ideal Gas Law equation

Let's rewrite the Ideal Gas Law equation with the given change: \(P_1V_1 = n_1RT_1\) and \(P_2V_2 = n_2RT_2\), where n2 is twice the initial number of moles n1.
08

Compare the two situations

Since the temperature (T), the volume (V), and the ideal gas constant (R) remain constant in this scenario, we can write: \(P_1V = n_1RT\) and \(P_2V = (2n_1)RT\).
09

Find the new pressure

Divide the second equation by the first one: \(\dfrac{P_2V}{P_1V} = \dfrac{(2n_1)RT}{n_1RT}\). Simplify the equation to find P2 in terms of P1: \(\dfrac{P_2}{P_1} = 2\). So, when the number of moles of the gas is doubled, the pressure of the gas is also doubled. (d) Doubling both the absolute temperature of the gas and the number of liters the container can hold:
10

Rewrite the Ideal Gas Law equation

Let's rewrite the Ideal Gas Law equation with both changes: \(P_1V_1 = n_1RT_1\) and \(P_2V_2 = n_2RT_2\), where T2 is twice the initial temperature T1, and V2 is twice the initial volume V1.
11

Compare the two situations

Since the number of moles (n) and the ideal gas constant (R) remain constant, we can write: \(P_1V_1 = nRT_1\) and \(P_2(2V_1) = nR(2T_1)\).
12

Find the new pressure

Divide the second equation by the first one: \(\dfrac{P_2(2V_1)}{P_1V_1} = \dfrac{nR(2T_1)}{nRT_1}\). Simplify the equation to find P2 in terms of P1: \(\dfrac{P_2}{P_1} = 1\). So, when both the absolute temperature of the gas and the number of liters the container can hold are doubled, the pressure of the gas remains constant.

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.

Start your free trial

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

Rewrite the ideal gas law solving for \(n\). Also show how all units cancel to leave you with just units of moles.

In the laboratory preparation of oxygen gas, the following reaction is carried out: \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) (a) How many moles of oxygen gas can be produced from the decomposition of \(500.0 \mathrm{~g}\) of \(\mathrm{KClO}_{3} ?\) (b) What volume in liters would the oxygen gas produced in part (a) occupy at \(24.0{ }^{\circ} \mathrm{C}\) and \(750.0 \mathrm{~mm} \mathrm{Hg} ?\) (c) What volume in liters would this gas occupy at STP?

When a balloon is placed in a freezer, the size of the balloon decreases. Explain why.

In an ideal gas, the molecules are considered to have no forces.

(a) If the temperature of a gas is doubled while the pressure is kept constant, the volume of the gas (b) If the pressure of a gas is halved while the temperature is kept constant, the volume of the gas
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

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.

Sign-up for free