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

As weather balloons rise from the earth’s surface, the pressure of the atmosphere becomes less, tending to cause the volume of the balloons to expand. However, the temperature is much lower in the upper atmosphere than at sea level. Would this temperature effect tend to make such a balloon expand or contract? Weather balloons do, in fact, expand as they rise. What does this tell you?

Short Answer

Expert verified
As weather balloons rise, atmospheric pressure decreases and temperature also decreases. According to Boyle's law, the decrease in pressure should cause the balloon to expand, while according to Charles's law, the decrease in temperature should cause the balloon to contract. However, weather balloons do expand as they rise, indicating that the pressure change has a greater effect on the balloon's volume than the temperature change.

Step by step solution

01

Relationship between Pressure and Volume

According to Boyle's law, the pressure of a gas is inversely proportional to its volume when the temperature is held constant. As the weather balloon rises, the pressure of the atmosphere decreases, which should cause the volume of the balloon to expand.
02

Relationship between Temperature and Volume

According to Charles's law, the volume of a gas is directly proportional to its temperature when the pressure is held constant. As the weather balloon rises, the atmospheric temperature decreases, which should cause the volume of the balloon to contract.
03

Combined Effect on Volume

In the case of a weather balloon, both pressure and temperature are changing simultaneously. The combined effect of these factors determines the actual change in volume.
04

What do the actual behavior of weather balloons tell us

Since weather balloons do, in fact, expand as they rise, this indicates that the reduction in atmospheric pressure has a greater effect on the volume of the balloon than the decrease in temperature.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boyle's Law
Boyle's Law is an essential principle in understanding the behavior of gases under varying pressure conditions. According to this law, the volume of a gas is inversely proportional to the pressure exerted on it, provided the temperature remains constant.
This relationship is mathematically expressed as: \[ P_1 V_1 = P_2 V_2 \]where:
  • \( P_1 \) and \( P_2 \) are the initial and final pressures, respectively,
  • \( V_1 \) and \( V_2 \) are the initial and final volumes, respectively.
In simpler terms, if you decrease the pressure on a contained gas, its volume will increase, and vice versa. For instance, with a weather balloon, as it ascends and the atmospheric pressure drops, the balloon's volume increases, demonstrating Boyle's Law in action. This increase occurs even as the surrounding temperature decreases, indicating that pressure has a more significant effect on the volume under these conditions.
Charles's Law
Charles's Law offers crucial insights into how gases behave with temperature changes, at constant pressure. According to this law, the volume of a gas is directly proportional to its absolute temperature.
The mathematical expression for Charles's Law is:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]where:
  • \( V_1 \) and \( V_2 \) are the initial and final volumes, respectively,
  • \( T_1 \) and \( T_2 \) are the initial and final temperatures, respectively, measured in Kelvin.
This law tells us that if a gas is heated, its volume will increase, assuming the pressure doesn't change. Conversely, cooling the gas will reduce its volume. For weather balloons, as they rise and encounter cooler temperatures, Charles's Law suggests that their volume should decrease. However, since balloons expand as they ascend, the effect of decreasing temperature is outweighed by the significant drop in pressure.
Weather Balloons
Weather balloons are important for gathering atmospheric data, rising to great heights to measure conditions such as temperature, pressure, and humidity. As they rise from the earth’s surface, they experience changes in both pressure and temperature.
In lower atmospheric regions, pressure decreases more significantly with altitude. This phenomenon, explained by Boyle's Law, leads to an expansion in the balloon's volume.
However, the temperature drops as altitude increases. Charles's Law tells us that this cooling would normally contract the volume. But in the case of weather balloons, the effect of decreasing pressure on volume is stronger than the cooling effect. Hence, the balloons continue to expand with altitude increase.
The actual expansion of weather balloons is a testament to the dominating influence of pressure reduction over temperature decrease. Understanding this helps meteorologists and scientists predict how balloons will behave as they ascend, ensuring the collected data's reliability.

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

The partial pressure of \(\mathrm{CH}_{4}(g)\) is 0.175 atm and that of \(\mathrm{O}_{2}(g)\) is 0.250 \(\mathrm{atm}\) in a mixture of the two gases. a. What is the mole fraction of each gas in the mixture? b. If the mixture occupies a volume of 10.5 \(\mathrm{L}\) at \(65^{\circ} \mathrm{C}\) , calculate the total number of moles of gas in the mixture. c. Calculate the number of grams of each gas in the mixture.

The oxides of Group 2A metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$\mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s)$$ A 2.85 -g sample containing only MgO and CuO is placed in a \(3.00-\mathrm{L}\) container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of \(740 .\) torr at \(20 .^{\circ} \mathrm{C}\) . After the reaction has gone to completion, the pressure inside the flask is \(390 .\) torr at \(20 .^{\circ} \mathrm{C}\) . What is the mass percent of MgO in the mixture? Assume that only the \(\mathrm{MgO}\) reacts with \(\mathrm{CO}_{2}\) .

Equal moles of sulfur dioxide gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the formation of gaseous sulfur trioxide. Assuming that the reaction goes to completion, what is the ratio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are measured at the same temperature and pressure?

A \(2.50-\mathrm{L}\) flask contains 0.60 \(\mathrm{g} \mathrm{O}_{2}\) at a temperature of \(22^{\circ} \mathrm{C}\) . What is the pressure (in atm) inside the flask?

In the "Methode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)$$ Fermentation of \(750 .\) mL grape juice (density \(=1.0 \mathrm{g} / \mathrm{cm}^{3} )\) is allowed to take place in a bottle with a total volume of 825 \(\mathrm{mL}\) until 12\(\%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) . Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C} ?\) (The density of ethanol is \(0.79 \mathrm{g} / \mathrm{cm}^{3} . )\)

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