(a) Calculate the work for a system that releases \(346 \mathrm{~kJ}\) of heat in a process for which the decrease in internal energy is \(125 \mathrm{~kJ}\). (b) Is work done on or by the system during this process?

Short Answer

Expert verified
Work done by the system is 221 kJ.

Step by step solution

01

Understand the first law of thermodynamics

The first law of thermodynamics can be stated as \(\Delta U = Q + W\), where \(\Delta U\) is the change in internal energy, \(Q\) is the heat exchanged with the surroundings, and \(W\) is the work done by or on the system. A positive value for \(Q\) means heat is added to the system, while a positive value for \(W\) means work is done by the system. In this context, we are given \(Q = -346 \, \mathrm{kJ}\) (since it is released) and \(\Delta U = -125 \, \mathrm{kJ}\).
02

Plug in given values into the first law equation

Using the equation \(\Delta U = Q + W\), we substitute the given values to find \(W\): \(\text{\[\Delta U = Q + W\]\text{\[-125 \, \mathrm{kJ} = -346 \, \mathrm{kJ} + W\]\text{\[W = -125 \, \mathrm{kJ} - (-346 \, \mathrm{kJ})\]\text{\[W = -125 \, \mathrm{kJ} + 346 \, \mathrm{kJ}\]\text{\[W = 221 \, \mathrm{kJ}\]\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy within physical systems. At the core of thermodynamics are the laws that describe how energy moves and changes form. One of the fundamental principles of this discipline is the conservation of energy, meaning that within an isolated system, energy can be transformed but not created or destroyed. Understanding thermodynamics is essential not only in physics but also in engineering, chemistry, and biology because it helps explain how systems react to changes in their environment, such as temperature and pressure.

In many real-world applications, thermodynamics explains how engines work, how refrigerators cool, and even the principles that govern our weather. For students tackling homework in thermodynamics, the key lies in grasping how these physical quantities—heat, work, and internal energy—interact to lead to the observed behavior of systems.
Internal Energy
Internal energy, often represented with the symbol 'U,' is the total energy contained within a system. This energy is the sum of the kinetic and potential energies of all the particles that make up the system. In a more tangible sense, internal energy could be seen as the energy required to create a system without adding any extra heat or work.

Changes in a system's internal energy can occur due to heat transfer or work done by or on the system. An increase in internal energy can result from either adding heat to the system or performing work on it, while a decrease can happen when the system releases heat or does work on the surroundings. It's crucial to recognize that while energy can be transferred in and out of a system, the internal energy itself is not something that can be directly measured—it can only be inferred by observing the system's response to changes in heat and work.
Work and Heat in Thermodynamics
The concepts of work and heat are intimately related in thermodynamics. They are both forms of energy transfer between a system and its surroundings.

Heat

Heat, or 'Q', refers to energy transfer due to a temperature difference between the system and its environment. If the system gains heat, 'Q' is positive; if it loses heat, 'Q' is negative. This directional convention helps us understand the flow of heat energy.

Work

Work, or 'W', is the energy transfer when a force moves something over a distance. For example, when a gas expands in a cylinder, it does work on the piston. In thermodynamics, if the system does work on its surroundings, 'W' is considered positive. If work is done on the system, it's negative.
  • These two forms of energy transfer are interconnected through the first law of thermodynamics, 'ΔU = Q + W'.
  • Understanding the directional flow of heat and work is crucial in solving problems.
  • The exercise provided exemplifies the principles of how heat release and work are accounted for when calculating the change in a system’s internal energy.

Grasping the interplay of work and heat allows students to not just solve mathematical problems, but also to comprehend the physical processes driving the changes in energy they calculate. In the referenced exercise, recognizing that work is positive when it is done by the system (heat released, in this case) is central to correctly applying the first law of thermodynamics and reaching a correct solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the lattice enthalpy of solid potassium bromide, \(\mathrm{KBr}(\mathrm{s}) \rightarrow \mathrm{K}^{+}(\mathrm{g})+\mathrm{Br}^{-}(\mathrm{g})\), from the following information: $$ \begin{aligned} &\Delta H_{i}{ }^{\circ}\left(\mathrm{KBr}, \text { s) }=-394 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\right. \\ &\Delta H_{i}{ }^{\circ}(\mathrm{K}, \mathrm{g})=+89.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \end{aligned} $$ First ionization energy of \(\mathrm{K}(\mathrm{g})=+425.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) \(\Delta H_{\text {vap }}{ }^{\circ}\left(\mathrm{Br}_{2}, \mathrm{l}\right)=+30.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) Br-Br bond dissociation cuthalpy \(=+192.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) Electron attachment to \(\mathrm{Br}(\mathrm{g})\) : $$ \mathrm{Br}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{Be}^{-}(\mathrm{g}), \quad \Delta H^{*}=-331.0 \mathrm{~kJ} $$

The heat capacity of a certain empty calorimeter is \(488.1 \mathrm{~J} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1}\). When \(25.0 \mathrm{~mL}\) of \(0.700 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq})\) was mixed in that calorimeter with \(25.0 \mathrm{~mL}\) of \(0.700 \mathrm{M} \mathrm{HCl}\) (aq), both initially at \(20.00^{\circ} \mathrm{C}\), the temperature increased to \(21.34^{\circ} \mathrm{C}\). Calculate the enthalpy of neutralization in kilojoules per mole of HCI.

Use the enthalpies of formation in Appendix \(2 \mathrm{~A}\) to calculate the standard enthalpy of the following reactions: (a) the replacement of deuterium by ordinary hydrogen in heavy water: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{D}_{2}(\mathrm{~g})\) (b) the removal of sulfur from the hydrogen sulfide and sulfur dioxide in natural gas: \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{~S}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (c) the oxidation of ammonia: \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

A calorimeter has a measured heat capacity of \(6.27 \mathrm{~kJ} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1}\). The combustion of \(1.84 \mathrm{~g}\) of magnesium led to a temperature change from \(21.30^{\circ} \mathrm{C}\) to \(28.56^{\circ} \mathrm{C}\). Calculare the enthalpy change of the reaction \(2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{MgO}(\mathrm{s})\).

In the manufacture of nitric acid by the oxidation of ammonia, the first product is nitric oxide, which is then cxidized to nitrogen dioxide. From the standard reaction enthalpies $$ \begin{gathered} \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) \\ \Delta H^{\circ}=+180.5 \mathrm{~kJ} \\ \mathrm{~N}_{2}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) \\ \Delta H^{\circ}=+66.4 \mathrm{~kJ} \end{gathered} $$ calculate the standard reaction enthalpy for the oxidation of nitric oxide to nitrogen dioxide: $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) $$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free