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Explain why a helium weather balloon expands as it rises in the air. Assume that the temperature remains constant.

Short Answer

Expert verified
The balloon expands because air pressure decreases with altitude, causing its volume to increase.

Step by step solution

01

Understanding the Ideal Gas Law

The behavior of gases can be explained through the Ideal Gas Law, which is given by the equation \( PV = nRT \). Here, \( P \) stands for pressure, \( V \) stands for volume, \( n \) is the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.
02

Analyzing the Conditions as the Balloon Rises

As the balloon rises in the atmosphere, the external air pressure \( P \) decreases. According to our assumption, the temperature \( T \) remains constant. Thus, using the Ideal Gas Law, \( nRT \) is constant, and if \( P \) decreases, \( V \) must increase to keep the equation balanced.
03

Applying the Concept to the Balloon

For the helium balloon, the decrease in external air pressure allows it to expand because the pressure inside the balloon becomes greater than the pressure outside. As a result, the balloon increases in volume to balance the pressure differences as it rises in the atmosphere.

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Key Concepts

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

Helium Weather Balloon
Helium weather balloons are used for various scientific purposes. They help in collecting data about the atmosphere by carrying sensors and instruments. These balloons are filled with helium gas, which is lighter than air. This property allows the balloon to rise as it displaces the heavier air around it. While ascending, the helium balloon travels through different layers of the atmosphere. Each layer varies in pressure and temperature. The ascent of the helium balloon is mostly due to the principles of buoyancy combined with the characteristics of helium. These balloons are essential tools in meteorology and environmental research. They allow us to gather critical data in real time and improve weather forecasting accuracy.
Gas Expansion
The phenomenon of gas expansion is central to understanding the behavior of the helium weather balloon. As the balloon ascends, it moves from an area of higher pressure to one of lower pressure. According to the Ideal Gas Law, when a gas experiences a decrease in pressure, it tends to expand.
  • The equation for the Ideal Gas Law is \( PV = nRT \).
  • When pressure \( P \) drops, volume \( V \) must increase if temperature \( T \) is constant.
This expansion occurs because the gas molecules inside the balloon have more space to move around as the outside pressure decreases. This is why a helium balloon expands as it rises into the sky. The process is an application of the gas laws, particularly emphasizing the relationship between pressure and volume.
Atmospheric Pressure
Atmospheric pressure is a measure of the force exerted by the air above the surface of the Earth. It decreases as altitude increases. This decrease is vital for the expansion of a helium weather balloon.
  • At sea level, atmospheric pressure is highest.
  • As altitude increases, atmospheric pressure decreases.
When a helium balloon rises, it encounters lower atmospheric pressure. This is because there is less air above pressing down. As the pressure outside the balloon decreases, the pressure inside the balloon, which remains initially unchanged, causes the balloon to expand. This disparity in pressure is what leads to the physical expansion of the balloon as it ascends.
Temperature Constancy
Temperature constancy plays a significant role in the expansion of gases, including those in a helium weather balloon. When analyzing situations involving gases, assuming a constant temperature simplifies calculations and helps focus on other variables such as pressure and volume.
  • Under constant temperature, volume and pressure changes are highlighted.
  • According to the Ideal Gas Law, if \( T \) is constant, then volume \( V \) and pressure \( P \) have an inverse relationship.
In a real-world scenario, the temperature might vary, but assuming it stays constant makes it easier to apply the Ideal Gas Law. By maintaining a constant temperature assumption, students can observe how changes in atmospheric pressure directly cause gas expansion, reflecting the practical dynamics of a weather balloon's ascent.

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Most popular questions from this chapter

Air entering the lungs ends up in tiny sacs called alveoli. It is from the alveoli that oxygen diffuses into the blood. The average radius of the alveoli is \(0.0050 \mathrm{~cm},\) and the air inside contains 14 percent oxygen. Assuming that the pressure in the alveoli is 1.0 atm and the temperature is \(37^{\circ} \mathrm{C},\) calculate the number of oxygen molecules in one of the alveoli. (Hint: The volume of a sphere of radius \(r\) is \(\left.\frac{4}{3} \pi r^{3} .\right)\)

Cite two pieces of evidence to show that gases do not behave ideally under all conditions. Under what set of conditions would a gas be expected to behave most ideally: (a) high temperature and low pressure, (b) high temperature and high pressure, (c) low temperature and high pressure, or (d) low temperature and low pressure?

Use the kinetic theory of gases to explain why hot air rises.

A certain hydrate has the formula \(\mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O} .\) A quantity of \(54.2 \mathrm{~g}\) of the compound is heated in an oven to drive off the water. If the steam generated exerts a pressure of 24.8 atm in a 2.00-L container at \(120^{\circ} \mathrm{C}\), calculate \(x\)

(a) What volume of air at 1.0 atm and \(22^{\circ} \mathrm{C}\) is needed to fill a \(0.98-\mathrm{L}\) bicycle tire to a pressure of \(5.0 \mathrm{~atm}\) at the same temperature? (Note that the 5.0 atm is the gauge pressure, which is the difference between the pressure in the tire and atmospheric pressure. Before filling, the pressure in the tire was \(1.0 \mathrm{~atm} .\) ) (b) What is the total pressure in the tire when the gauge pressure reads 5.0 atm? (c) The tire is pumped by filling the cylinder of a hand pump with air at 1.0 atm and then, by compressing the gas in the cylinder, adding all the air in the pump to the air in the tire. If the volume of the pump is 33 percent of the tire's volume, what is the gauge pressure in the tire after three full strokes of the pump? Assume constant temperature.

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