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A balloon contains 0.30 mol of helium. It rises, while maintaining a constant 300-K temperature, to an altitude where its volume has expanded five times. Neglecting tension forces in the balloon, how much work is done by the helium during this isothermal expansion?

Short Answer

Expert verified
The work done by the helium in the balloon during its isothermal expansion is about 9164 J.

Step by step solution

01

Understanding given values

First, we need to identify the given values from the problem. Here, the amount of Helium (\(n\)) is 0.30 mol. The temperature (\(T\)) is constant at 300 K. The balloon's volume has expanded five times, which means the final volume (\(Vf\)) is five times the initial volume (\(Vi\)). For the purpose of finding work done, we don't need to specify the actual volumes, because we only need the ratio \(Vf/Vi\) which is 5.
02

Use the formula

Now, we apply the formula for work done in an isothermal expansion: \(W = nRT \ln(\frac{Vf}{Vi})\). We substitute the given values into this equation: \(W = 0.30 \times 8.314 \times 300 \times \ln(5)\).
03

Calculating the work done

By performing the calculation, one will determine the work done by the balloon as it expands.

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Key Concepts

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

Work Done in Thermodynamics
In thermodynamics, the concept of work done is associated with a gas expanding or compressing within some boundary, such as a piston or a balloon. Specifically, when we talk about isothermal expansion, we consider a process in which the temperature remains constant while the volume of the gas changes. The isothermal process is governed by Boyle's law, which states that the pressure of a fixed amount of gas held at a constant temperature is inversely proportional to its volume.

During an isothermal expansion, the gas does work on its surroundings as it pushes against a force to increase its volume. This is a quintessential example of work conducted by a system in thermodynamics. The equation for this work can be given in its integral form, representing the work as an area under a P-V curve on a graph.

The specialized formula for calculating work done in an isothermal expansion involving an ideal gas is expressed as \(W = nRT \times \text{ln}\bigg(\frac{Vf}{Vi}\bigg)\), where \(W\) is the work done, \(n\) is the number of moles, \(R\) is the universal gas constant, \(T\) is the absolute temperature, \(Vf\) is the final volume, and \(Vi\) is the initial volume. The natural logarithm (ln) accounts for the logarithmic relationship between volume and work done in an isothermal process.

Incorporating this into an educational platform, the calculation should be made clear with step-by-step explanations, ensuring that students understand the interplay between volume changes, temperature, and work. Additionally, reinforcing the concept with visual aids, like graphs demonstrating the P-V relationship, can significantly enhance comprehension.
Ideal Gas Law
The ideal gas law is a cornerstone of chemical and physical studies, providing a simple but powerful formula that bridges the properties of ideal gases: pressure (\(P\)), volume (\(V\)), and temperature (\(T\)), with the quantity of gas in moles (\(n\)). The law is encapsulated in the equation \(PV = nRT\), where \(R\) represents the ideal gas constant. This equation is incredibly useful for making predictions about how a gas will behave under certain conditions and for relating various properties of gases to one another.

When working with exercises involving the ideal gas law, it's important to understand that all variables are state functions, meaning they describe the state of the system at a given moment. This law assumes ideal conditions where molecules do not attract or repel each other and occupy no volume in themselves, which is a fair approximation under many, but not all, conditions.

In our exercise, the ideal gas law underlies the processes we are examining. While the law itself isn't directly used to calculate work done during the isothermal expansion, it provides the theoretical framework necessary to understand the relationship between the gas's pressure, volume, and temperature during such a process. As educators, it's crucial to explain that in real-life scenarios, gases do not always behave ideally, and deviations can occur at high pressures or low temperatures.
Molar Quantity
Molar quantity, such as the amount of substance measured in moles, is a fundamental concept in chemistry and thermodynamics. The mole is a unit which provides a method for expressing the amount of a chemical substance and is directly related to the number of atoms, ions, or molecules in a given sample.

In the context of our problem involving the helium-filled balloon, the molar quantity (\(n\)) refers to the amount of helium gas, expressed in moles, contained within the balloon. This specific exercise uses the molar quantity to calculate the work done during the expansion of the balloon. Since thermodynamic equations often include the number of moles as a variable, understanding molar quantities allows students to grasp the proportional relationship between the amount of gas and the work done, or any other thermodynamic properties.

During an isothermal expansion, the work done is directly proportional to the number of moles of gas involved. Hence, more moles of gas would entail more work being done for the same change in volume and temperature. Using moles rather than mass is particularly convenient for gases, as it links the microscopic scale (molecules) to the macroscopic properties we observe and measure. It's essential to convey to students not only the significance of the molar quantity in calculations but also its role in connecting abstract concepts to tangible, real-world phenomena.

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Most popular questions from this chapter

Find the rate of heat flow into a system whose internal energy is increasing at the rate of 65 W, given that the system is doing work at the rate of 175 W.

How much of a triatomic gas with \(C_{V}=3 R\) would you have to add to \(6.0 \mathrm{~mol}\) of a monatomic gas to get a mixture whose thermodynamic behavior was like that of a diatomic gas?

One scheme for reducing greenhouse-gas emissions from coalfired power plants calls for capturing carbon dioxide and pumping it into the deep ocean, where the pressure is at least 350 atm. You're called to assess the energy cost of such a scheme for a power plant that produces electrical energy at the rate of \(1.0 \mathrm{GW}\) while at the same time emitting \(\mathrm{CO}_{2}\) at the rate of 1100 tonnes/hour. If \(\mathrm{CO}_{2}\) is extracted from the plant's smokestack at \(320 \mathrm{~K}\) and \(1 \mathrm{~atm}\) pressure and then compressed adiabatically to 350 atm, what fraction of the plant's power output would be needed for the compression? Take \(\gamma=1.3\) for \(\mathrm{CO}_{2}\). (Your answer is a rough estimate because \(\mathrm{CO}_{2}\) doesn't behave like an ideal gas at very high pressures; also, it doesn't include the energy cost of separating the \(\mathrm{CO}_{2}\) from other stack gases or of transporting it to the compression site.)

How much work does it take to compress 3.3 mol of an ideal gas to half its original volume while maintaining a constant temperature of 290 K?

Prove that the slope of an adiabat at a given point in a \(p V\) diagram is \(\gamma\) times the slope of the isotherm passing through the same point.

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