A gas sample in a piston assembly expands, doing \(536 \mathrm{~kJ}\) of work on its surroundings at the same time that \(214 \mathrm{~kJ}\) of heat is added to the gas. (a) What is the change in internal energy of the gas during this process? (b) Will the pressure of the gas be higher or lower when these changes are completed?

Short Answer

Expert verified
The change in internal energy of the gas is \( -322 \mathrm{~kJ} \). The pressure of the gas will be lower when these changes are completed.

Step by step solution

01

Recall the first law of thermodynamics

The first law of thermodynamics states that the change in internal energy, \(\Delta U\), of a system is equal to the heat added to the system, \(Q\), minus the work done by the system on the surroundings, \(W\). The relationship is given by \(\Delta U = Q - W\).
02

Calculate the change in internal energy

Using the values provided, compute the change in internal energy by substituting \(Q = 214 \mathrm{~kJ}\) and \(W = 536 \mathrm{~kJ}\) into the first law of thermodynamics equation: \(\Delta U = Q - W = 214 \mathrm{~kJ} - 536 \mathrm{~kJ} = -322 \mathrm{~kJ}\).
03

Determine the pressure change

If the gas is expanding and doing work on its surroundings, its internal energy decreases if the amount of heat added is less than the work done by the gas. As internal energy is related to temperature, a decrease in internal energy suggests a decrease in temperature, given that the gas follows an ideal behavior. A lower temperature generally means lower pressure for a fixed volume, implying that the pressure of the gas would be lower at the end of this process.

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Key Concepts

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

Understanding Internal Energy
Internal energy is a fundamental concept in thermodynamics, representing the total energy contained within a system, which may consist of the kinetic energy due to the motion of molecules, and potential energy arising from intermolecular forces. When studying the behavior of gases, internal energy is crucial because it influences temperature, pressure, and volume of the gas.

Let’s consider the piston assembly example. The change in internal energy, signified as \(\Delta U\), includes all forms of energy transfers, such as heat and work. When heat is added to the gas, the internal energy increases, and when the gas does work on its surroundings by expanding, the internal energy decreases. A negative \(\Delta U\) indicates that the energy of the gas is reduced after the process. This reduction could lead to a decrease in temperature, as heat is a form of energy transfer that increases molecular motion and thus the internal energy.

Understanding internal energy is key for interpreting more than just temperature changes. This concept is integral to predicting the behavior of substances in various thermodynamic processes and scenarios.
Fundamentals of Thermodynamics
Thermodynamics is the science that deals with heat, work, and the forms of energy involved in chemical or physical processes. It is based on several fundamental laws, with the first law of thermodynamics addressing energy conservation in systems. The principle works on the premise that energy can be transferred or transformed but not created or destroyed.

In any thermodynamic process, especially in cases like our piston assembly, multiple forms of energy interact. The first law helps us assess the system's energy change by accounting for both heat transfer and work done. The role of heat in this context is particularly significant because it represents energy in transit—often as a result of a temperature difference—and can change the internal energy and thus the state of the system. Work, on the other hand, represents a transfer of energy when an external force is applied over a distance, such as the gas pushing against the piston.

Unraveling the interplay between heat, work, and internal energy allows for a comprehensive understanding of the processes governing the behavior and equilibrium of systems, which is what thermodynamics seeks to achieve.
Work and Heat in Thermodynamics
When discussing work and heat in thermodynamics, we often refer to them as two distinct paths for energy transfer into or out of a system. In our piston example, the gas does work on the surroundings by expanding against a force, and heat is simultaneously added to the system.

  • Work (\(W\)): The process by which energy is exchanged due to a force causing displacement. If a gas expands, it does positive work on its surroundings, and if compressed, the work done on the gas is positive, increasing its internal energy.
  • Heat (\(Q\)): It is the form of energy transfer driven by a temperature difference. Adding heat to a system increases its internal energy unless it is utilized for performing work.

Interrelation of Work and Heat

Heat and work are related through the first law of thermodynamics: \(\Delta U = Q - W\). This equation is a statement of energy conservation. If more work is done by the system than the heat added to it, the internal energy decreases, resulting in potential changes in temperature and pressure, as seen with the gas in the piston.

Understanding how these two modes of energy transfer interplay provides a clearer picture regarding the outcome of thermodynamic processes. It’s no wonder that students are often advised to treat these concepts with diligence to fully appreciate the broad implications they hold in the field of thermodynamics.

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Most popular questions from this chapter

Each of the four cylinders of a new type of combustion engine has a displacement of \(3.60 \mathrm{~L}\). (The volume of the cylinder expands \(3.60 \mathrm{~L}\) each time the fuel is ignited.) (a) If each piston in the four cylinders is displaced against a pressure of \(1.80 \mathrm{kbar}\) and each cylinder is ignited once per second, how much work can the engine do in \(1.00\) minute? (b) Is the work positive or negative with respect to the engine and its contents?

In a combustion cylinder, the total internal energy change produced from the burning of a fuel is \(-2.573 \mathrm{~kJ}\). The cooling system that surrounds the cylinder absorbs \(947 \mathrm{~kJ}\) as heat. How much work can be done by the fuel in the cylinder?

Using standard enthalpies of formation from Appendix \(2 \mathrm{~A}\), calculate the standard reaction enthalpy for each of the following reactions: (a) the final stage in the production of nitric acid: $$ 3 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) $$ (b) the industrial synthesis of boron trifluoride: $$ \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{CaF}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{BF}_{3}(\mathrm{~g})+3 \mathrm{CaO}(\mathrm{s}) $$ (c) the formation of a sulfide by the action of hydrogen sulfide on an aqueous solution of a base: $$ \mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})+2 \mathrm{KOH}(\mathrm{aq}) \longrightarrow \mathrm{K}_{2} \mathrm{~S}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

For a certain reaction at constant pressure, \(\Delta H=-15 \mathrm{~kJ}\), and \(22 \mathrm{~kJ}\) of expansion work is done on the system. What is \(\Delta U\) for this process?

The standard enthalpies of combustion of graphite and diamond are \(-393.51 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) and \(-395.41 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\), respectively. Calculate the enthalpy of the graphite \(\rightarrow\) diamond transition.

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