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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 decreases by 457 Joules.

Step by step solution

01

Understanding the Relationship

The change in internal energy for an ideal gas can be determined using the first law of thermodynamics, which states that \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system. For a perfectly insulated container, \( Q = 0 \). Therefore, \( \Delta U = -W \).
02

Setting Up the Equation

Since \( Q = 0 \) (as the container is perfectly insulated) and the work \( W \) done on the gas is given as \( 457 \mathrm{~J} \), substitute \( W \) into the equation: \( \Delta U = 0 - 457 \mathrm{~J} \), simplifying to \( \Delta U = -457 \mathrm{~J} \).
03

Interpreting the Result

The negative sign in \( \Delta U = -457 \mathrm{~J} \) indicates a decrease in internal energy because the work was done on the gas (i.e., the gas did not do the work).
04

Finalizing the Solution

Thus, the internal energy of the ideal Helium gas decreases by 457 Joules.

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Key Concepts

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

Ideal Gas
An ideal gas is a theoretical model that simplifies the study of gas behaviors under various conditions. In this model, gas particles are assumed to have no volume and do not exert forces on one another, except when they collide. This assumption allows us to predict the behavior of gases using mathematical equations. One crucial equation in thermodynamics for ideal gases is the Ideal Gas Law, represented as \( PV = nRT \), where:
  • \( P \) is the pressure
  • \( V \) is the volume
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature
For our exercise, it is important to understand that helium gas behaves as an ideal gas, simplifying our calculations on changes in internal energy.
Work and Energy
Work and energy are foundational concepts in physics that describe how forces and energy interact. When you perform work on a system, you transfer energy to or from that system. This exercise considers the work done by applying a force to compress or expand a gas. Specifically, when work is done on the gas, energy is transferred to the gas. However, in this context, it leads to changes in the internal energy of the gas, without involving changes in heat due to the insulation. Understanding that work involves changes in energy helps us analyze how energy transformations affect gas properties in systems, such as our perfectly insulated helium gas case.
First Law of Thermodynamics
The first law of thermodynamics is a fundamental principle that governs the behavior of energy in physical systems. It states that energy cannot be created or destroyed, only transformed from one form to another. In mathematical terms, it's expressed as:\[\Delta U = Q - W\]Where:
  • \( \Delta U \) represents the change in internal energy of the system
  • \( Q \) is the heat added to the system
  • \( W \) is the work done by the system
For our problem with a perfectly insulated container, no heat transfer occurs (\( Q = 0 \)). Therefore, the change in internal energy \( \Delta U \) is essentially \(-W\), the negative of the work done on the system. This principle highlights how work directly influences the internal energy of the gas in an insulated system.
Internal Energy Change
The internal energy change of a gas is a crucial concept in thermodynamics, representing the total energy stored within the gas molecules. This energy comprises kinetic energy, due to the motion of the particles, and potential energy resulting from interactions between particles.In our exercise, because the helium gas is in a perfectly insulated container, no heat is exchanged with the surroundings. Therefore, the entire focus is on the work done on the gas, which changes the internal energy. Here, the work performed (\(457 \text{ J}\)) is the only factor altering the internal energy, leading to a decrease by that amount. This is reflected in the equation \( \Delta U = -457 \text{ J} \), explaining a reduction in energy, as the gas has absorbed work performed on it without a heat exchange.

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Most popular questions from this chapter

One type of gas mixture used in anesthesiology is a \(50 \% / 50 \%\) mixture (by volume) of nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) and oxygen \(\left(\mathrm{O}_{2}\right),\) which can be premixed and kept in a cylinder for later use. Because these two gases don't react chemically at or below 2000 psi, at typical room temperatures they form a homogeneous single- gas phase, which can be considered an ideal gas. If the temperature drops below \(-6^{\circ} \mathrm{C},\) however, \(\mathrm{N}_{2} \mathrm{O}\) may begin to condense out of the gas phase. Then any gas removed from the cylinder will initially be nearly pure \(\mathrm{O}_{2}\). As the cylinder empties, the proportion of \(\mathrm{O}_{2}\) will decrease until the gas coming from the cylinder is nearly pure \(\mathrm{N}_{2} \mathrm{O}\). In a hospital, pure oxygen may be delivered at 50 psi (gauge pressure) and then mixed with \(\mathrm{N}_{2} \mathrm{O}\). What volume of oxygen at \(20^{\circ} \mathrm{C}\) and 50 psi (gauge pressure) should be mixed with \(1.7 \mathrm{~kg}\) of \(\mathrm{N}_{2} \mathrm{O}\) to get a \(50 \% / 50 \%\) mixture by volume at \(20^{\circ} \mathrm{C} ?\) A. \(0.21 \mathrm{~m}^{3}\) B. \(0.27 \mathrm{~m}^{3}\) C. \(1.9 \mathrm{~m}^{3}\) D. \(100 \mathrm{~m}^{3}\)

A \(20.0 \mathrm{~L}\) tank contains \(0.225 \mathrm{~kg}\) of helium at \(18.0^{\circ} \mathrm{C}\). The molar mass of helium is \(4.00 \mathrm{~g} / \mathrm{mol}\). (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?

The atmosphere of the planet Mars is \(95.3 \%\) carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) and about \(0.03 \%\) water vapor. The atmospheric pressure is only about \(600 \mathrm{~Pa}\), and the surface temperature varies from \(-30^{\circ} \mathrm{C}\) to \(-100^{\circ} \mathrm{C}\). The polar ice caps contain both \(\mathrm{CO}_{2}\) ice and water ice. Could there be liquid \(\mathrm{CO}_{2}\) on the surface of Mars? Could there be liquid water? Why or why not?

One type of gas mixture used in anesthesiology is a \(50 \% / 50 \%\) mixture (by volume) of nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) and oxygen \(\left(\mathrm{O}_{2}\right),\) which can be premixed and kept in a cylinder for later use. Because these two gases don't react chemically at or below 2000 psi, at typical room temperatures they form a homogeneous single- gas phase, which can be considered an ideal gas. If the temperature drops below \(-6^{\circ} \mathrm{C},\) however, \(\mathrm{N}_{2} \mathrm{O}\) may begin to condense out of the gas phase. Then any gas removed from the cylinder will initially be nearly pure \(\mathrm{O}_{2}\). As the cylinder empties, the proportion of \(\mathrm{O}_{2}\) will decrease until the gas coming from the cylinder is nearly pure \(\mathrm{N}_{2} \mathrm{O}\). You have a cylinder that contains \(500 \mathrm{~L}\) of the gas mixture pressurized to 2000 psi (gauge pressure). A regulator sets the gas flow to deliver \(8.2 \mathrm{~L} / \mathrm{min}\) at atmospheric pressure. Assume that this flow is slow enough that the expansion is isothermal and the gases remain mixed. How much time will it take to empty the cylinder? A. 1 h B. \(33 \mathrm{~h}\) C. \(57 \mathrm{~h}\) D. \(140 \mathrm{~h}\)

If you have ever hiked or climbed to high altitudes in the mountains, you surely have noticed how short of breath you get. This occurs because the air is thinner, so each breath contains fewer \(\mathrm{O}_{2}\) molecules than at sea level. At the top of Mount Everest, the pressure is only \(\frac{1}{3}\) atm. Air contains \(21 \%\) O \(_{2}\) and \(78 \%\) N \(_{2}\), and an average human breath is \(0.50 \mathrm{~L}\) of air. At the top of Mount Everest, (a) how many \(\mathrm{O}_{2}\) molecules does each breath contain when the temperature is \(-15^{\circ} \mathrm{F},\) and \((\mathrm{b})\) what percent is this of the number of \(\mathrm{O}_{2}\) molecules you would get from a breath at sea level at \(-15^{\circ} \mathrm{F}\) ?

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