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It is well known that warm air in a cooler environment rises. Now consider a warm mixture of air and gasoline on top of an open gasoline can. Do you think this gas mixture will rise in a cooler environment?

Short Answer

Expert verified
Answer: Yes, the warm mixture of air and gasoline will likely rise in a cooler environment due to its less dense properties and the buoyant force acting upon it.

Step by step solution

01

Identify the properties of the gas mixture

The gas mixture consists of warm air and gasoline vapors. Warm air is less dense than cooler air, and gasoline vapors are generally lighter than air. When combining these properties, we have a warm, less dense gas mixture in comparison to the cooler, denser surrounding air.
02

Analyze the buoyancy of the gas mixture

Buoyancy is the force that determines whether a gas will rise or fall in the presence of a surrounding fluid (in this case, air). Less dense gas will experience a buoyant force that pushes it upwards, while denser gas will experience a downward force. Since our warm mixture of air and gasoline is indeed less dense than the surrounding air due to its properties, we can assume it will experience a buoyant force pushing it upwards.
03

Determine if gas mixture will rise in a cooler environment

Taking the properties and buoyancy into consideration, we can now determine if the gas mixture will rise in a cooler environment. Due to the buoyant force acting on the less dense mixture of air and gasoline, it is likely that the mixture will rise in a cooler environment. In conclusion, the warm mixture of air and gasoline on top of an open gasoline can is likely to rise in a cooler environment due to its less dense properties and the buoyant force acting upon it.

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Key Concepts

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

Density of Gases
Gases, like all matter, have density, which is the measure of mass per unit volume. The density of a gas is typically much lower than that of solids or liquids due to the large distances between the gas particles. Factors such as temperature and pressure can significantly influence the density of a gas. For instance, heating a gas at constant pressure increases its volume, thus decreasing its density. This is described by the ideal gas law, \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the amount of substance, \( R \) is the ideal gas constant, and \( T \) is the temperature. In our exercise, the mixture of warm air and gasoline is less dense than the cooler surrounding air because the warm gas particles are more spread out, or in other words, the gas expands as it warms.

This concept is essential when considering buoyancy in gases, as a gas with lower density than its environment will tend to rise, which leads us to our exercise problem where the warm mixture of air and gasoline vapors is expected to rise above a cooler environment.
Thermodynamics
The thermodynamic principles play a pivotal role in understanding buoyancy in gases. Thermodynamics deals with heat and temperature and their relation to energy and work. It is governed by the four laws of thermodynamics that dictate how energy is transferred within systems and their surroundings. In the context of gases, the first law of thermodynamics, also known as the conservation of energy, states that the internal energy of a closed system is conserved.

When we apply heat to a gas, we increase its internal energy, causing the gas particles to move more rapidly and the gas to expand. As a result, given the same pressure, the gas becomes less dense as it warms. This ties into our exercise problem where the warm mixture of air and gasoline will rise when placed in a cooler environment due to the buoyant forces.
Gas Mixtures
The behavior of gas mixtures, which are combinations of two or more different gases, is an important concept, particularly when these gases have different densities. In our problem, air is a mixture of nitrogen, oxygen, and other gases, while gasoline vapor is a hydrocarbon compound. These components each have their own unique properties including density. The mixture's overall density is determined by the proportion and densities of its components.

Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures is also relevant; it states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases. This principle helps in understanding the thermodynamic behavior of the gas mixture in relation to its surroundings. Therefore, considering the gasoline vapors are less dense than air, when mixed with warm air, the overall density of the mixture is less than the cooler surrounding air, providing a scenario where the gas mixture will rise due to buoyancy effects.

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Most popular questions from this chapter

Does the amount of heat absorbed as 1 kg of saturated liquid water boils at \(100^{\circ} \mathrm{C}\) have to be equal to the amount of heat released as \(1 \mathrm{kg}\) of saturated water vapor condenses at \(100^{\circ} \mathrm{C} ?\)

Consider a sealed can that is filled with refrigerant134a. The contents of the can are at the room temperature of \(25^{\circ} \mathrm{C} .\) Now a leak develops, and the pressure in the can drops to the local atmospheric pressure of \(90 \mathrm{kPa}\). The temperature of the refrigerant in the can is expected to drop to (rounded to the nearest integer \((a) 0^{\circ} \mathrm{C}\) (b) \(-29^{\circ} \mathrm{C}\) \((c)-16^{\circ} \mathrm{C}\) \((d) 5^{\circ} \mathrm{C}\) \((e) 25^{\circ} \mathrm{C}\)

A \(13-\mathrm{m}^{3}\) tank contains nitrogen at \(17^{\circ} \mathrm{C}\) and \(600 \mathrm{kPa}\) Some nitrogen is allowed to escape until the pressure in the tank drops to \(400 \mathrm{kPa}\). If the temperature at this point is \(15^{\circ} \mathrm{C},\) determine the amount of nitrogen that has escaped.

Is it possible to have water vapor at \(-10^{\circ} \mathrm{C} ?\)

Water initially at 200 kPa and \(300^{\circ} \mathrm{C}\) is contained in a piston-cylinder device fitted with stops. The water is allowed to cool at constant pressure until it exists as a saturated vapor and the piston rests on the stops. Then the water continues to cool until the pressure is \(100 \mathrm{kPa}\). On the \(T\) -v diagrams sketch, with respect to the saturation lines, the process curves passing through both the initial, intermediate, and final states of the water. Label the \(T, P\) and \(v\) values for end states on the process curves. Find the overall change in internal energy between the initial and final states per unit mass of water.

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