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(a) Write an equation that expresses the first law of thermodynamics in terms of heat and work. (b) Under what conditions will the quantities \(q\) and \(w\) be negative numbers?

Short Answer

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(a) The First Law of Thermodynamics equation is: ΔU = q + w (b) The conditions for negative values of \(q\) and \(w\) are: - \(q\) is negative when heat is removed from the system. - \(w\) is negative when work is done on the system.

Step by step solution

01

(a) Formulate the First Law of Thermodynamics equation

The First Law of Thermodynamics equation is given by: ΔU = q + w where ΔU represents the change in internal energy, q represents the heat added to the system, and w represents the work done by the system.
02

(b) Identify conditions for negative q and w values

Let's analyze the conditions that make the heat (\(q\)) and work (\(w\)) quantities negative. 1. Negative heat (q < 0): Heat is considered negative when it's being removed from the system. This is because heat leaving the system causes the internal energy of the system to decrease. 2. Negative work (w < 0): Work is considered negative when the work is done on the system. This is because when work is performed on the system, the internal energy of the system increases as a consequence of an external force acting on it. So, - \(q\) will be negative when heat is removed from the system. - \(w\) will be negative when work is done on the system.

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Key Concepts

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

Understanding Internal Energy
When we talk about internal energy in thermodynamics, we're focusing on the energy contained within a system. This energy is due to the microscopic motions of atoms and molecules. In simpler terms, internal energy is like the internal bank account of energy that a system has.
It includes kinetic and potential energy at the molecular level.
Internal energy can change due to two main factors: heat and work. When heat is added to a system, the internal energy increases. Likewise, when heat is removed, it decreases.
Similarly, when work is done by the system, the internal energy decreases, and when work is done on the system, the internal energy increases. These changes are captured by the First Law of Thermodynamics, which shows how a system exchanges energy with its surroundings through heat and work.
This is crucial because it helps us understand energy transformations in scientific and engineering problems involving gases, engines, and even biological systems.
Heat Transfer Basics
Heat transfer is a key concept in thermodynamics, and it describes the movement of heat energy from one place to another. Heat always flows from a hotter object to a cooler one, as per the Second Law of Thermodynamics.
There are three main types of heat transfer:
  • Conduction: Transfer of heat through direct contact of particles.
  • Convection: Transfer of heat by the movement of fluids or gases.
  • Radiation: Transfer of heat through electromagnetic waves, like the warmth you feel from the sun.
In the context of the First Law of Thermodynamics, the symbol \(q\) represents heat. If \(q\) is positive, it indicates heat added to the system. Conversely, if \(q\) is negative, heat is leaving or being removed from the system.
This movement of heat affects the internal energy and, therefore, the state of the system.
Exploring Work Done
Work done in thermodynamics is a measure of energy transfer which occurs when a force moves an object. It's like the energy you expend to push a heavy box across the floor. In terms of a system, it's about how energy is transferred through moving boundaries or changes in volume. Work can be either positive or negative. According to the First Law of Thermodynamics, when the system does work, we often consider it positive \((w > 0)\), as energy is leaving the system.
However, when work is done on the system, it's negative \((w < 0)\), meaning the system is gaining energy due to external influence.
Understanding these basics of work helps explain phenomena in engines where gases expand and do work, or in refrigeration cycles, where work is done on a system to remove heat. Work done alongside heat transfer ties back into changing the internal energy of a system, completing our picture of energy conservation and transfer as described by the First Law of Thermodynamics.

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Most popular questions from this chapter

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) \(1.36 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) and \((\mathbf{b}) 10.4 \mathrm{~g}\) of \(\mathrm{KCl} .(\mathbf{c})\) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2},\) is likely to be feasible under ordinary conditions? Explain your answer.

Consider the following reaction: $$ 2 \mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow 2 \mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \quad \Delta H=+252.8 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(24.0 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is decomposed by this reaction at constant pressure. (c) For a given sample of \(\mathrm{CH}_{3} \mathrm{OH},\) the enthalpy change during the reaction is \(82.1 \mathrm{~kJ}\). How many grams of methane gas are produced? (d) How many kilojoules of heat are released when \(38.5 \mathrm{~g}\) of \(\mathrm{CH}_{4}(g)\) reacts completely with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{CH}_{3} \mathrm{OH}(g)\) at constant pressure?

Consider the following reaction: $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) \quad \Delta H=-1204 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(3.55 \mathrm{~g}\) of \(\mathrm{Mg}(s)\) reacts at constant pressure. (c) How many grams of \(\mathrm{MgO}\) are produced during an enthalpy change of \(-234 \mathrm{~kJ}\) ? (d) How many kilojoules of heat are absorbed when \(40.3 \mathrm{~g}\) of \(\mathrm{MgO}(s)\) is decomposed into \(\mathrm{Mg}(s)\) and \(\mathrm{O}_{2}(g)\) at constant pressure?

Under constant-volume conditions, the heat of combustion of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is \(16.49 \mathrm{~kJ} / \mathrm{g}\). A \(3.00-\mathrm{g}\) sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.94 to \(24.62^{\circ} \mathrm{C} .(\mathbf{a})\) What is the total heat capacity of the calorimeter? (b) If the size of the sucrose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

Can you use an approach similar to Hess's law to calculate the change in internal energy, \(\Delta E,\) for an overall reaction by summing the \(\Delta E\) values of individual reactions that add up to give the desired overall reaction?

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