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A gas absorbs \(45 \mathrm{kJ}\) of heat and does \(29 \mathrm{kJ}\) of work. Calculate \(\Delta E\).

Short Answer

Expert verified
The change in internal energy of the gas (ΔE) is \(16 \mathrm{kJ}\).

Step by step solution

01

Write down the given values.

We are given the following information: - Heat absorbed by the gas (q) = 45 kJ - Work done by the gas (W) = 29 kJ
02

Apply the first law of thermodynamics.

According to the first law of thermodynamics, the change in internal energy (ΔE) can be calculated as: ΔE = q - W
03

Substitute the given values and calculate ΔE.

Now let us substitute the given values of q and W into the equation: ΔE = 45 kJ - 29 kJ
04

Find the difference to obtain ΔE.

Calculating the difference, we get: ΔE = 16 kJ Thus, the change in internal energy of the gas (ΔE) is 16 kJ.

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Key Concepts

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

Internal Energy
Internal energy is a key concept in the study of thermodynamics, representing the total energy stored within a system. This includes the potential and kinetic energy of the particles that make up the material. Internal energy can be affected by changes in temperature, phase, or even chemical composition.

In terms of the kinetic theory, the internal energy of a gas is the sum of the kinetic energies of its molecules. Since the internal energy is a function of the state of the system, it is expressed as a state variable. This means that the change in internal energy, denoted as \(\Delta E\), is the difference in energy between two states of the system, regardless of how the change occurred.

It's crucial to grasp that internal energy is not directly measurable; we can only determine changes in it. As seen in our exercise, when a gas absorbs heat or performs work, its internal energy changes. The computation of \(\Delta E\) gives us insight into how energy is redistributed within a system but does not account for energy's origin or destination, which might involve interactions with the surroundings.
Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. Among the core principles of thermodynamics is the first law, which asserts the conservation of energy within a closed system.

Thermodynamics revolves around the concepts of energy transfer, energy transformation, and, ultimately, energy conservation. The laws of thermodynamics establish the framework for understanding how systems respond to changes in their environment, especially with respect to energy exchanges. This field is vast, influencing engineering, chemistry, biology, and environmental sciences.

The first law, often stated as \(\Delta E = q - W\), provides a foundation for understanding how the internal energy of a system changes through heat transfer and work done. Our exercise embodies this law, as we assess the change in a system's internal energy due to specific processes.
Heat Transfer
Heat transfer is the process by which thermal energy moves from one place to another. It's fundamental to thermodynamics, as it is one of the main ways that systems exchange energy.

There are three methods of heat transfer: conduction, convection, and radiation. In the case of our exercise, we are focused on the heat absorbed by a gas, without specifying the method. What's essential is that the thermodynamic quantity of heat, denoted by \(q\), represents energy in transit.

When a substance absorbs heat, its internal energy increases, typically leading to a rise in temperature or a change in state. The measurement of heat transfer, as illustrated in the problem, is crucial for calculating changes in internal energy and helps in understanding how energy flows from one system to another, which is pivotal in various scientific and engineering applications.
Work Done
Work done in the context of thermodynamics is the energy transferred when a force moves an object over a distance. It is another way that energy can be transferred between a system and its surroundings.

In our given problem, the gas does work on its surroundings by expanding and pushing against an external force, like a piston in an engine. This kind of work is called 'expansion work' and can be described by the formula \(W = P\Delta V\), where \(P\) is pressure and \(\Delta V\) is the change in volume. However, the exercise simplifies the scenario by providing the work term \(W\) directly.

The sign convention for work is important: work done by the system on its surroundings is considered positive, and work done on the system by the surroundings is negative. This aligns with our exercise, as the positive work value indicates that energy is leaving the system due to the work done by the gas. Understanding work in thermodynamics helps in predicting how systems will behave when they interact physically with their environment.

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Most popular questions from this chapter

A system undergoes a process consisting of the following two steps: Step 1: The system absorbs \(72 \mathrm{J}\) of heat while \(35 \mathrm{J}\) of work is done on it. Step 2: The system absorbs \(35 \mathrm{J}\) of heat while performing \(72 \mathrm{J}\) of work. Calculate \(\Delta E\) for the overall process.

Consider the following changes: a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\) d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\) At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Consider the dissolution of \(\mathrm{CaCl}_{2}\) : $$\mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{kJ}$$ An \(11.0-\mathrm{g}\) sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water, with both substances at \(25.0^{\circ} \mathrm{C}\). Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of \(4.18 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\).

The specific heat capacity of silver is \(0.24 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\). a. Calculate the energy required to raise the temperature of \(150.0 \mathrm{g}\) Ag from \(273 \mathrm{K}\) to \(298 \mathrm{K}\). b. Calculate the energy required to raise the temperature of 1.0 mole of Ag by \(1.0^{\circ} \mathrm{C}\) (called the molar heat capacity of silver). c. It takes \(1.25 \mathrm{kJ}\) of energy to heat a sample of pure silver from \(12.0^{\circ} \mathrm{C}\) to \(15.2^{\circ} \mathrm{C}\). Calculate the mass of the sample silver.

When 1.00 L of \(2.00 M \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{L}\) of \(0.750 M \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid (BaSO\(_{4}\)) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{g} / \mathrm{mL},\) calculate the enthalpy change per mole of BaSO\(_{4}\) formed.

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