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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.

Short Answer

Expert verified
The change in internal energy for the overall process, ΔE, can be calculated by adding the change in internal energy for each step: ΔE1 = 107 J and ΔE2 = -37 J. Therefore, ΔE_overall = ΔE1 + ΔE2 = 107 J + (-37 J) = \(70 \mathrm{~J}\).

Step by step solution

01

Calculate heat and work for Step 1 of the process

In Step 1, we are given that the system absorbs 72 J of heat and 35 J of work is done on it. Since work is being done on the system, it has a negative value. So, we have: Q1 = 72 J W1 = -35 J
02

Calculate heat and work for Step 2 of the process

In Step 2, the system absorbs 35 J of heat and performs 72 J of work. Since work is being done by the system, it has a positive value. So, we have: Q2 = 35 J W2 = 72 J
03

Calculate ΔE for each step using the first law of thermodynamics

Using the first law of thermodynamics, ΔE = Q - W, calculate the change in internal energy for each step: ΔE1 = Q1 - W1 = 72 J - (-35 J) = 72 J + 35 J = 107 J ΔE2 = Q2 - W2 = 35 J - 72 J = -37 J
04

Calculate ΔE for the overall process by adding the results of each step

Now, add the change in internal energy for each step to find the change in internal energy for the overall process: ΔE_overall = ΔE1 + ΔE2 = 107 J + (-37 J) = 70 J The change in internal energy for the overall process is 70 J.

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Key Concepts

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

Understanding Internal Energy
Internal energy is a key concept in thermodynamics that describes the total energy contained within a system. This energy results from the movement and interactions of the molecules within the system. There are three main components to internal energy:
  • Kinetic energy of the molecules moving around.
  • Potential energy due to the forces acting between the molecules.
  • Other forms of energy, such as chemical energy in bonds.

In our exercise, internal energy is involved when we calculate \( \Delta E \), which refers to the change in internal energy. The first law of thermodynamics helps us determine how internal energy changes during a process, by considering the heat added to the system and the work done by or on it. Understanding how internal energy shifts in a thermodynamic process is crucial for solving related physics problems.
Decoding Heat in Thermodynamic Processes
Heat is the transfer of thermal energy from one system to another, resulting from a difference in temperature. In our exercise, heat is either absorbed or released by the system, impacting its internal energy.
The sign convention for heat is important in this context:
  • Heat absorbed by the system is considered positive.
  • Heat released by the system is considered negative.
For example, in Step 1 of the exercise, the system absorbs 72 J of heat, resulting in a positive heat value. Conversely, in Step 2, it absorbs 35 J of heat. Understanding how to account for heat transfer helps us use the first law of thermodynamics effectively in calculating changes in the system's internal energy.
Clarifying Work in Thermodynamics
Work, in the context of thermodynamics, refers to the energy transferred when a force moves an object. Work can be done on the system or by the system, and this distinction affects how we calculate changes in internal energy.
The sign convention for work is as follows:
  • Work done on the system is considered negative, as it increases the system's energy.
  • Work done by the system is considered positive, as it decreases the system's energy.
In Step 1 of our exercise, 35 J of work is done on the system, and is therefore negative. In Step 2, 72 J of work is done by the system, which makes it positive. By understanding how work interacts with heat in a thermodynamic process, we can accurately apply the first law of thermodynamics to calculate internal energy changes.
Thermodynamic Processes and the First Law
A thermodynamic process is a change in the state of a thermodynamic system that can involve changes in pressure, volume, temperature, or internal energy. These processes can be isothermal, adiabatic, isobaric, or isochoric, depending on the constraints involving heat transfer and work done.
The first law of thermodynamics, also known as the law of energy conservation, provides a framework for understanding these processes: \[ \Delta E = Q - W \]- \( \Delta E \) is the change in internal energy- \( Q \) is the heat added to the system- \( W \) is the work done by the system
In our exercise, this law is applied to two distinct steps to determine the overall change in internal energy. By understanding this relationships, we can predict how different actions impact the system, reinforcing the importance of the first law in analyzing thermodynamic processes.

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Most popular questions from this chapter

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

A coffee-cup calorimeter initially contains \(125 \mathrm{~g}\) water at \(24.2^{\circ} \mathrm{C}\). Potassium bromide \((10.5 \mathrm{~g})\), also at \(24.2^{\circ} \mathrm{C}\), is added to the water, and after the KBr dissolves, the final temperature is \(21.1^{\circ} \mathrm{C}\). Calculate the enthalpy change for dissolving the salt in \(\mathrm{J} / \mathrm{g}\) and \(\mathrm{kJ} / \mathrm{mol}\). Assume that the specific heat capacity of the solution is \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that no heat is transferred to the surroundings or to the calorimeter.

Assuming gasoline is pure \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\), predict the signs of \(q\) and \(w\) for the process of combusting gasoline into \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\).

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \text { (unbalanced) } $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{~kJ}\). If \(2.50 \times 10^{2} \mathrm{~mL} \mathrm{~N}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50 \mathrm{~atm}\) and \(4.50 \times\) \(10^{2} \mathrm{~mL} \mathrm{O}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50\) atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

One of the components of polluted air is NO. It is formed in the high- temperature environment of internal combustion engines by the following reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta H=180 \mathrm{~kJ} $$ Why are high temperatures needed to convert \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to \(\mathrm{NO}\) ?

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