/*! 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 20 Why does warm air rise?... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Why does warm air rise?

Short Answer

Expert verified
Warm air rises because heating it causes its molecules to move faster and spread apart, making the air less dense than the cooler air around it. Less dense air floats upwards due to the principle of buoyancy.

Step by step solution

01

Explanation of Molecular Activity with Heat

When air is heated, the molecules in the air start to move faster and spread apart due to kinetic energy. This phenomenon is based on the theory that heat is a form of energy that triggers motion in molecules.
02

The Density Difference

As a result of the molecules spreading apart, the air becomes less dense compared to the cooler air around it. Density is defined as mass divided by volume. When heated, the same mass of air occupies more volume due to molecular movement, thus reducing its density.
03

The Concept of Buoyancy

Buoyancy is the tendency of an object to float in a fluid. In our atmosphere, fluid is the surrounding air. Now, less dense objects (or substances) tend to rise in more dense ones. Hence, because warm air is less dense than the surrounding cooler air, it rises.

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.

Molecular Activity with Heat
Understanding how heat affects air begins at the molecular level. Imagine millions of tiny air molecules fluttering around us. When air is exposed to heat, these molecules absorb energy and start moving faster. This is due to an increase in kinetic energy, which is the energy of motion. As the molecules move quicker, they bounce off each other and spread out more.

Think of it like a crowd of people. In cooler temperatures, they stand closer together because they are moving slowly. But once the temperature rises, it's like everyone starts dancing, needing more space and hence spreading out. This spreading out of air molecules when heated is a crucial concept because it leads to changes in air density, which plays a key role in why warm air rises.
Density Difference
Density is the ratio of mass to volume, and it changes with the molecular activity we discussed earlier. When air heats up and molecules spread apart, the volume of air increases without an increase in mass; thus, its density decreases. It's as if you have the same amount of substance, but it's taking up more room.

To visualize this, imagine a balloon filled with air. As the air inside the balloon heats up, the balloon expands because the air molecules are taking up more space. However, the overall weight of the air inside the balloon remains the same. This decrease in density is critical to understand because, in nature and our atmosphere, regions of differing densities interact strongly, shaping weather patterns and influencing environmental conditions.
Concept of Buoyancy
Buoyancy can be thought of as the ability to float. It's not just for boats and swimming pool toys but also applies to air in our atmosphere. In fluids, including gases like the air around us, buoyant forces come into play. These forces depend on density. Since warm air is less dense than cool air, as we've seen, it is subject to the buoyant force that causes it to rise.

The concept of buoyancy can be understood using the principle of Archimedes, which states that the upward buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. Therefore, when a parcel of warm air with lower density than its cooler surroundings 'displaces' some of the surrounding air, the difference in weight between the warm air and the displaced cool air leads to a net upward force, causing the warm air to rise.

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

Nitric oxide is produced in the reaction between copper metal and nitric acid: $$ 3 \mathrm{Cu}(s)+8 \mathrm{HNO}_{3}(a q) \longrightarrow{ }_{3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{NO}(g)} $$ What mass of copper is required to produce \(15.0 \mathrm{~L}\) of \(\mathrm{NO}\) at 725 torr and \(20.0^{\circ} \mathrm{C}\) ?

Solve the following equations for \(x\). (a) \(14 x+16=44\) (b) \(\frac{1}{3} x+3=5\) (c) \(12 x=4(12-x)\) (d) \(3 x+15=4 x+12\)

Given the following amounts of gases, calculate the number of moles of each gas. Calculate the volume each amount of gas would occupy at STP. (a) \(5.8 \mathrm{~g} \mathrm{NH}_{3}\) (b) \(48 \mathrm{~g} \mathrm{O}_{2}\) (c) \(10.8 \mathrm{~g}\) He

Boyle used a U-tube to investigate gas properties. As shown in the figure, a gas was trapped in the closed arm of the U-tube at \(29.9 \mathrm{in} \mathrm{Hg}\), and the pressure was varied by adding mercury to the open arm. The total pressure exerted on the gas is the sum of the atmospheric pressure ( \(29.9\) in \(\mathrm{Hg}\) ) and the pressure due to the addition of mercury as measured by the difference in mercury height. Boyle recorded the following data: $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { Length of Gas Column } \\ \text { (in) } \end{array} & \begin{array}{c} \text { Difference Between } \\ \text { Mercury Levels (in), } \Delta h \end{array} \\ \hline 48 & 0.0 \\ \hline 44 & 2.8 \\ \hline 40 & 6.2 \\ \hline 36 & 10.1 \\ \hline 32 & 15.1 \\ \hline 28 & 21.2 \\ \hline 24 & 29.7 \\ \hline 22 & 35.0 \\ \hline 20 & 41.6 \\ \hline 18 & 48.8 \\ \hline 16 & 58.2 \\ \hline \end{array} $$ Graph these data. What does the graph show about the relationship between volume and pressure?

Solve the following equations for \(x\). (a) \(5 x+1=3\) (b) \(0.412 / x=2.00\) (c) \(x^{2}=32-x^{2}\) (d) \(2 x=6-x\)

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.