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\(\bullet\) Suppose you do 457 \(\mathrm{J}\) of work on 1.18 moles of ideal He gas in a perfectly insulated container. By how much does the internal energy of this gas change? Does it increase or decrease?

Short Answer

Expert verified
The internal energy increases by 457 J.

Step by step solution

01

Understanding the First Law of Thermodynamics

The First Law of Thermodynamics states that the change in the internal energy (\(\Delta U\)) of a system is equal to the heat added to the system (\(Q\)) minus the work done by the system (\(W\)): \[\Delta U = Q - W\]In this problem, since the container is perfectly insulated, no heat is exchanged with the surroundings (\(Q = 0\)). Therefore, the change in internal energy is entirely due to the work done on the gas.
02

Applying the Information Given

We are given that 457 J of work is done on the gas. When work is done on the system, it is considered negative work in the thermodynamics convention used here. Therefore, \(W = -457 \ \mathrm{J}\).
03

Calculating the Change in Internal Energy

Substitute the work value into the First Law of Thermodynamics equation:\[\Delta U = Q - W = 0 - (-457) = 457 \ \mathrm{J}\]This means the internal energy increases by 457 J, showing that the internal energy has increased due to the work added.

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Key Concepts

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

Internal Energy Change
When the internal energy of a system changes, it is often due to external influences such as heat transfer or work being done on the system. In our context, internal energy (\(\Delta U\)) is the energy contained within the system itself. This exercise uses the First Law of Thermodynamics, which relates internal energy change to heat and work. For any system:\[\Delta U = Q - W\]where \(Q\) is the heat added to the system, and \(W\) is the work done by the system.In this problem, since no heat is exchanged (\(Q = 0\)), the change in internal energy is solely from the work. Thus, when work is done on the system, internal energy increases. The exercise clarifies this with a calculation showing the energy increase by 457 J.
Ideal Gas
An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. This concept is crucial because it simplifies many calculations in thermodynamics. Ideal gases follow the ideal gas law, expressed as:\[PV = nRT\]where \(P\) is pressure, \(V\) is volume, \(n\) is the amount of moles, \(R\) is the universal gas constant, and \(T\) is temperature.In this exercise, helium gas (He) is typically treated as an ideal gas. This assumption allows us to focus on key work and energy changes without complex molecular interactions affecting our calculations. It's important to remember that while ideal gas behavior provides a good approximation for many gases, real gases may deviate from this ideal behavior under certain conditions.
Work on Gas
The concept of work in thermodynamics can be thought of as energy transfer to or from a system through force applied over distance. Work done on a gas is especially important in changing its internal energy.
  • When work is done on the gas, energy is added to the system, and the internal energy increases.
  • If the system does work on the surroundings, energy is lost, and internal energy decreases.
In this exercise, 457 J of work is done on the helium gas. Following the convention where work done on the system is negative (as energy is added to the system), we relate this to the increase in internal energy. Thus, knowing the work done provides a direct measure of how much the internal energy changes.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, and temperature, and their relation to energy, radiation, and physical properties of matter. Its principles are fundamental in explaining how energy transfers and transformations occur. The First Law of Thermodynamics applies here as it equates changes in internal energy to heat and work. It helps us understand that:
  • Energy within a closed system is conserved. Any energy change in the system is due to transfers of energy by heat or work.
  • The law can predict how energy will transfer even in more complex interactions.
Applying the First Law allows us to determine the change in internal energy by considering work done on the gas. Recognizing that no heat is exchanged due to insulation further simplifies this problem. The First Law forms the backbone for many calculations across various applications in science and engineering.

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Most popular questions from this chapter

\(\bullet\) A gas in a cylinder is held at a constant pressure of \(2.30 \times\) \(10^{5}\) Pa and is cooled and compressed from 1.70 \(\mathrm{m}^{3}\) to 1.20 \(\mathrm{m}^{3} .\) The internal energy of the gas decreases by \(1.40 \times 10^{5} \mathrm{J}.\) (a) Find the work done by the gas. (b) Find the amount of the heat that flowed into or out of the gas, and state the direction (inward or outward) of the flow.

\(\bullet\) In a certain chemical process, a lab technician supplies 254 J of heat to a system. At the same time, 73 J of work are done on the system by its surroundings. What is the increase in the internal energy of the system?

\(\bullet\) A gas in a cylinder expands from a volume of 0.110 \(\mathrm{m}^{3}\) to 0.320 \(\mathrm{m}^{3} .\) Heat flows into the gas just rapidly enough to keep the pressure constant at \(1.80 \times 10^{5} \mathrm{Pa}\) during the expansion. The total heat added is \(1.15 \times 10^{5} \mathrm{J}\) . (a) Find the work done by the gas. (b) Find the change in internal energy of the gas.

\(\bullet$$\bullet\) A flask with a volume of 1.50 \(\mathrm{L}\) , provided with a stopcock, contains ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) at 300 \(\mathrm{K}\) and atmospheric pressure \(\left(1.013 \times 10^{5} \mathrm{Pa}\right) .\) The molar mass of ethane is 30.1 \(\mathrm{g} / \mathrm{mol}\) . The system is warmed to a temperature of \(380 \mathrm{K},\) with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

\(\bullet$$\bullet\) The surface of the sun. The surface of the sun has a temperature of about 5800 \(\mathrm{K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is \(1.67 \times\) \(10^{-27} \mathrm{kg} .\) (b) What would be the mass of an atom that had half the rms speed of hydrogen?

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