Find (a) \(\Delta E\) when a gas absorbs \(18 \mathrm{~J}\) of heat and has \(13 \mathrm{~J}\) of work done on it. (b) \(q\) when 72 J of work is done on a system and its energy is increased by \(61 \mathrm{~J}\).

Short Answer

Expert verified
In another scenario, the work done on the system is 72 J and the energy increase of the system is 61 J. How much heat is added or absorbed in this case? Answer: In the first scenario, the change in internal energy is 31 J. In the second scenario, the system loses 11 J of heat.

Step by step solution

01

Part (a) - Finding \(\Delta E\)

Given, a gas absorbs \(q = 18 \mathrm{~J}\) of heat and has \(w = 13 \mathrm{~J}\) of work done on it. We can use the equation \(\Delta E = q + w\) to find the change in internal energy. \(\Delta E = 18 \mathrm{~J} + 13 \mathrm{~J}\) \(\Delta E = 31 \mathrm{~J}\) So, the change in internal energy is \(31 \mathrm{~J}\).
02

Part (b) - Finding \(q\)

In this case, we are given that the work done on the system is \(w = 72 \mathrm{~J}\) and the energy increase of the system is \(\Delta E = 61 \mathrm{~J}\). We need to find the heat \(q\) added or absorbed by the system. We can rearrange the equation \(\Delta E = q + w\) to solve for \(q\): \(q = \Delta E - w\) \(q = 61 \mathrm{~J} - 72 \mathrm{~J}\) \(q = -11 \mathrm{~J}\) The negative sign indicates that the system is losing heat. Therefore, the system loses \(11 \mathrm{~J}\) of heat.

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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. The field studies the conversion of energy from one form to another and how it affects matter. Understanding thermodynamics is pivotal for a vast range of scientific and engineering disciplines, from designing engines and refrigerators to predicting the behavior of gases and exploring the limits of computing.

At the heart of thermodynamics lie four fundamental laws that describe different aspects of energy and its transformations. These laws apply to systems, which are typically defined as specific quantities of matter or regions in space being studied. In solving problems like our exercise here, thermodynamics provides a framework for understanding how energy transfer—whether as heat or work—affects the internal energy of a system.
First Law of Thermodynamics
The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. Instead, energy can only change from one form to another. This principle is key to understanding the internal energy change in any system.

In mathematical terms, for a given system, the change in internal energy \(\Delta E\) is equal to the heat \(q\) added to the system minus the work \(w\) done by the system on its surroundings. If we rearrange the formula, we see that \(\Delta E = q + w\), where work done on the system is considered positive and work done by the system is negative. In the exercise provided, calculating the change in internal energy or the heat exchanged becomes straightforward with this formula.

When a gas absorbs heat and has work done on it, as in part (a) of the exercise, both heat and work contribute positively to the increase in the system's internal energy. The real-world implications of this are profound, as it helps us understand the behavior of gases in engines, weather systems, and even in our own bodies.
Conservation of Energy
The conservation of energy is a universal principle that underpins all of physics, including thermodynamics. It insists that in a closed system—that is, one with no exchange of matter or energy with its surroundings—the total energy remains constant over time. This principle is reflected in countless natural phenomena and technological applications.

In the context of the first law of thermodynamics, the conservation of energy manifests in the understanding that any change in a system's internal energy must be balanced by an equal and opposite change in energy elsewhere. Whether energy leaves the system as work or enters it as heat, the total energy is conserved. Our exercise illustrates the conservation of energy through the accounting of heat absorbed and work done on the system, or vice versa. Remembering that energy takes many forms and the law applies to all energy conversion processes helps us envision the significance of energy conservation in everything from energy production to environmental policy.

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Most popular questions from this chapter

To produce silicon, used in semiconductors, from sand \(\left(\mathrm{SiO}_{2}\right)\), a reaction is used that can be broken down into three steps: $$ \begin{aligned} \mathrm{SiO}_{2}(s)+2 \mathrm{C}(s) \longrightarrow \mathrm{Si}(s)+2 \mathrm{CO}(g) & & \Delta H=689.9 \mathrm{~kJ} \\ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(g) & & \Delta H=-657.0 \mathrm{~kJ} \\ \mathrm{SiCl}_{4}(g)+2 \mathrm{Mg}(s) \longrightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s) & & \Delta H=-625.6 \mathrm{~kJ} \end{aligned} $$ (a) Write the thermochemical equation for the overall reaction for the formation of silicon from silicon dioxide; \(\mathrm{CO}\) and \(\mathrm{MgCl}_{2}\) are byproducts. (b) What is \(\Delta H\) for the formation of one mole of silicon? (c) Is the overall reaction exothermic?

Write thermochemical equations for the formation of one mole of the following compounds from the elements in their stable states at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\). (a) acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) (b) nitrogen dioxide \((g)\) (c) lead(II) bromide (s) (d) phosphorus pentachloride \((g)\)

Strontium metal is responsible for the red color in fireworks. Fireworks manufacturers use strontium carbonate, which can be produced by combining strontium metal, graphite (C), and oxygen gas. The formation of one mole of \(\mathrm{SrCO}_{3}\) releases \(1.220 \times 10^{3} \mathrm{~kJ}\) of energy. (a) Write a balanced thermochemical equation for the reaction. (b) What is \(\Delta H\) when \(10.00 \mathrm{~L}\) of oxygen at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) is used by the reaction?

A sample of sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), is contaminated by sodium chloride. When the contaminated sample is burned in a bomb calorimeter, sodium chloride does not burn. What is the percentage of sucrose in the sample if a temperature increase of \(1.67^{\circ} \mathrm{C}\) is observed when \(3.000 \mathrm{~g}\) of the sample is burned in the calorimeter? Sucrose gives off \(5.64 \times 10^{3} \mathrm{~kJ} / \mathrm{mol}\) when burned. The heat capacity of the calorimeter is \(22.51 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\).

Nitrogen oxide (NO) has been found to be a key component in many biological processes. It also can react with oxygen to give the brown gas \(\mathrm{NO}_{2}\). When one mole of NO reacts with oxygen, \(57.0 \mathrm{~kJ}\) of heat is evolved. (a) Write the thermochemical equation for the reaction between one mole of nitrogen oxide and oxygen. (b) Is the reaction exothermic or endothermic? (c) Draw an energy diagram showing the path of this reaction. (Figure \(8.4\) is an example of such an energy diagram.)

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