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On what law is the first law of thermodynamics based? Explain the sign conventions in the equation \(\Delta U=q+w\)

Short Answer

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The first law of thermodynamics is based on the law of conservation of energy. In the equation \(\Delta U=q+w\), the sign convention is that \(q\) and \(w\) are positive when heat is added to the system or when work is done on the system, respectively. They are negative when heat is removed from the system or when work is done by the system, respectively. This sign convention indicates the direction of energy transfer.

Step by step solution

01

Identify the Physical Law Underpinning the First Law of Thermodynamics

The law of conservation of energy serves as the foundation for the first law of thermodynamics. This law states that energy cannot be created or destroyed. It can only be transferred or transformed from one form to another. In this context, within a closed system, the total energy remains constant.
02

Understand the Equation Representing the First Law of Thermodynamics

The equation for the first law of thermodynamics is written as \(\Delta U=q+w\). Here, \(\Delta U\) is the change in internal energy of the system, \(q\) is the heat added to the system, and \(w\) is the work done on the system. This equation essentially says that the change in internal energy of a system is equal to the heat added to the system plus the work done on the system.
03

Explain the Sign Conventions

The sign convention for the heat term \(q\) is positive when heat is added to the system and negative when heat is removed from the system. Likewise, for the work term \(w\), the sign is positive when work is done on the system and negative when work is done by the system. These sign conventions indicate the direction of energy transfer. A positive \(\Delta U\) means the system has gained energy, while a negative \(\Delta U\) indicates the system has lost energy.

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Key Concepts

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

Law of Conservation of Energy
Imagine you are playing with a set of building blocks. You can stack them up in different structures, knock them down, or move them from place to place, but unless you add or remove blocks, you end up with the same number of blocks that you started with. This is the essence of the law of conservation of energy. In the realm of physics, energy behaves much like your set of blocks. It can change form—think of electrical energy lighting up a bulb or chemical energy in food turning into kinetic energy when you walk—but the total amount of energy in an isolated system remains constant. This principle is the foundation for the first law of thermodynamics.

The first law, often expressed with the formula \(\Delta U=q+w\), succinctly captures this law but in the context of thermodynamic systems. Whether energy enters a system as heat (\(q\)) or work (\(w\)), it contributes to the internal energy (\(\Delta U\)) of the system. In practical terms, this means that if you're heating a pot of water, the energy from the stove isn't lost; rather, it's transferred to the water, increasing its internal energy and potentially changing its state, say from liquid to gas.
Internal Energy Change
The term 'internal energy' refers to the total energy contained within a thermodynamic system, represented by the symbol \(U\). It encompasses all forms of energy within the system, such as kinetic energy of atoms, potential energy from molecular bonds, or even the energy due to the temperature of the system. When we describe the internal energy change (\(\Delta U\)), we talk about the difference in energy between two states of the system. A positive change implies that energy has increased (such as when a substance is heated up), whereas a negative change indicates a decrease in energy (like when a gas expands and cools down).

The subtlety here is that, while you can measure the change in internal energy, the total internal energy itself is a more elusive quantity. In the realm of education, understanding this change is crucial for predicting how a system will react to various inputs of heat and work. If students grasp this concept, they can make sense of why a refrigerator, for example, grows warm on the outside as it cools the inside: the internal energy of the refrigerated interior decreases as a result of work and heat transfer.
Thermodynamics Sign Conventions
Dealing with positive and negative signs can sometimes feel like trying to remember which way to turn a stubborn screw—'righty-tighty, lefty-loosey.' In thermodynamics, the sign conventions are crucial for understanding the direction of energy transfer. The overarching rule is that the system is the focus of our attention.

For heat (\(q\)), a positive value means that energy in the form of heat is added to the system—the opposite of getting chilly when you step out into the cold. Conversely, a negative \(q\) means the system is losing heat, much like a melting ice cube. As for work (\(w\)), if a system expands against external pressure (think of a balloon inflating), it does work on its surroundings—indicated by a negative sign. In contrast, if you compress a piston, you're doing work on the system, which is shown as a positive \(w\).

If students remember that positive signs mean 'energy in' and negative signs 'energy out' with respect to the system, they'll have an easier time determining whether energy is gained or lost. A firm grasp of these conventions helps avoid confusion and leads to better intuition around energy processes.

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Most popular questions from this chapter

A press release announcing a new fuel-cell car to the public stated that hydrogen is "relatively cheap" and "some stations in California sell hydrogen for \(\$ 5\) a kilogram. A kg has the same energy as a gallon of gasoline, so it's like paying \(\$ 5\) a gallon. But you go two to three times as far on the hydrogen." Analyze this claim.

lime is a term that includes calcium oxide (CaO, also called quicklime) and calcium hydroxide \(\left[\mathrm{Ca}(\mathrm{OH})_{2},\right.\) also called slaked lime \(] .\) It is used in the steel industry to remove acidic impurities, in airpollution control to remove acidic oxides such as \(\mathrm{SO}_{2}\), and in water treatment. Quicklime is made industrially by heating limestone \(\left(\mathrm{CaCO}_{3}\right)\) above \(2000^{\circ} \mathrm{C}:\) $$ \begin{array}{r} \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \\ \Delta H^{\circ}=177.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Slaked lime is produced by treating quicklime with water: $$ \begin{array}{r} \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s) \\ \Delta H^{\circ}=-65.2 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ The exothermic reaction of quicklime with water and the rather small specific heats of both quicklime \(\left(0.946 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) and slaked lime \(\left(1.20 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) make it hazardous to store and transport lime in vessels made of wood. Wooden sailing ships carrying lime would occasionally catch fire when water leaked into the hold. (a) If a \(500-\mathrm{g}\) sample of water reacts with an equimolar amount of \(\mathrm{CaO}\) (both at an initial temperature of \(25^{\circ} \mathrm{C}\) ), what is the final temperature of the product, \(\mathrm{Ca}(\mathrm{OH})_{2} ?\) Assume that the product absorbs all of the heat released in the reaction. (b) Given that the standard enthalpies of formation of \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O}\) are \(-635.6 \mathrm{~kJ} / \mathrm{mol}\) and \(-285.8 \mathrm{~kJ} / \mathrm{mol}\) respectively, calculate the standard enthalpy of formation of \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Predict the value of \(\Delta H_{\mathrm{f}}^{\circ}\) (greater than, less than, or equal to zero) for these elements at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Br}_{2}(g)\) \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{I}_{2}(g) ; \mathrm{I}_{2}(s)\)

An excess of zinc metal is added to \(50.0 \mathrm{~mL}\) of a \(0.100 M\) AgNO \(_{3}\) solution in a constant-pressure calorimeter like the one pictured in Figure \(6.9 .\) As a result of the reaction $$\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)$$ the temperature rises from \(19.25^{\circ} \mathrm{C}\) to \(22.17^{\circ} \mathrm{C}\). If the heat capacity of the calorimeter is \(98.6 \mathrm{~J} /{ }^{\circ} \mathrm{C},\) calculate the enthalpy change for the above reaction on a molar basis. Assume that the density and specific heat of the solution are the same as those for water, and ignore the specific heats of the metals.

Determine the amount of heat (in kJ) given off when \(1.26 \times 10^{4} \mathrm{~g}\) of \(\mathrm{NO}_{2}\) are produced according to the equation $$\begin{array}{r}2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\\\\Delta H=-114.6 \mathrm{~kJ} / \mathrm{mol}\end{array}$$

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