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Chapter 5: Problem 34

(a) Under what condition will the enthalpy change of a process equal the amount of heat transferred into or out of the system? (b) During a constant-pressure process, the system releases heat to the surroundings. Does the enthalpy of the system increase or decrease during the process? (c) In a constant-pressure process, \(\Delta H=0\). What can you conclude about \(\Delta E, q,\) and \(w ?\)

Short Answer

Expert verified
(a) The enthalpy change of a process will equal the amount of heat transferred into or out of the system under the condition of constant pressure, represented as: \(\Delta H = q_P\). (b) In a constant-pressure process where the system releases heat to the surroundings, the enthalpy of the system decreases as \(\Delta H\) will be negative. (c) If \(\Delta H = 0\) in a constant-pressure process, this implies \(q_P = 0\). Based on the first law of thermodynamics, we can conclude that the change in internal energy (\(\Delta E\)) is equal to the negative of the work done by the system (\(w\)), i.e., \(\Delta E = -w\).

Step by step solution

01

Condition for enthalpy change equal to heat transferred

For the enthalpy change to equal the amount of heat transferred into or out of the system, the process needs to happen at a constant pressure. Mathematically, this can be represented as: \[ \Delta H = q_P \] Where \(\Delta H\) is the change in enthalpy, and \(q_P\) is the heat transferred at constant pressure. #b#
02

Enthalpy Change during Constant-Pressure Process

In a constant pressure process, the enthalpy change is given by: \[ \Delta H = q_P \] Since the system is releasing heat to the surroundings, the heat transfer \(q_P\) will be negative, which implies that \(\Delta H\) will also be negative. Therefore, the enthalpy of the system decreases during the process. #c#
03

Conditions for Constant-Pressure Process with no Enthalpy Change

If \(\Delta H = 0\) during a constant-pressure process, the heat transferred at constant pressure must also be zero, i.e., \(q_P = 0\). Now, we can use the first law of thermodynamics which states that the change in internal energy (\(\Delta E\)) is equal to the heat added to the system (\(q\)) minus the work done by the system (\(w\)): \[ \Delta E = q - w \] Under constant pressure, \(q_P = q\), so in this case, \(q = 0\). This simplifies the equation to: \[ \Delta E = -w \] Since \(\Delta H = 0\), we can conclude that the change in the internal energy (\(\Delta E\)) is equal to the negative of the work done by the system (\(w\)).

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Key Concepts

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

Constant-Pressure Process
In thermodynamics, a constant-pressure process is one where the pressure remains unaltered throughout the process. This is particularly significant because, under these conditions, the change in enthalpy (\( \Delta H\) can be directly equated to the heat exchanged, denoted as \( q_P \).
  • Equational Representation: The enthalpy change at constant pressure is mathematically expressed as \( \Delta H = q_P \). This means that under constant pressure, we don't have to account separately for heat and enthalpy change, because they are equivalent.
  • Practical Implications: Many real-world processes, such as boiling water at atmospheric pressure, occur under constant pressure, making this concept vital for understanding everyday thermodynamics.
When analyzing a constant-pressure process, recognizing that any change in enthalpy directly correlates with the heat exchange simplifies the calculation of heat and energy changes in the system.
First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle stating that energy cannot be created or destroyed, only transferred or transformed. When applied to constant-pressure processes, this law plays a pivotal role in evaluating the energy dynamics involved.
  • Formula: The law is expressed as \( \Delta E = q - w \), where \( \Delta E \) represents the change in internal energy, \( q \) is the heat added to the system, and \( w \) is the work done by the system.
  • Connection to Enthalpy: At constant pressure, the heat exchanged \( q \) corresponds to \( q_P \), making it straightforward to determine changes in internal energy. If \( q_P \) is known, and the work done is measured, one can easily find the internal energy change.
This law ensures that all energy changes in a system are balanced by the heat introduced or taken from the system and the work completed by or on the system. Hence, it reinforces the underpinning of energy conservation in physical and chemical processes.
Internal Energy Change
Internal energy change is a central theme in thermodynamics, representing the total change in energy within a system. This concept is essential when analyzing processes at constant pressure.
  • Calculation: For a process where \( \Delta H = 0 \) and \( q_P = 0 \), we apply the simplified equation from the first law: \( \Delta E = -w \). This indicates that any change in internal energy is countered by the work done by the system.
  • Significance of Zero Enthalpy Change: When the enthalpy remains unchanged, it signifies a perfect balance between heat exchange and work done. Such a scenario implies that the system transforms energy primarily through work without any net external heat transfer.
Understanding the nuances of internal energy change, especially under constant pressure, provides deeper insights into the energy interactions within a system, aiding in comprehensive energy management and analysis.

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Most popular questions from this chapter

Two solid objects, A and B, are placed in boiling water and allowed to come to the temperature of the water. Each is then lifted out and placed in separate beakers containing \(1000 \mathrm{~g}\) of water at \(10.0^{\circ} \mathrm{C}\). Object A increases the water temperature by \(3.50^{\circ} \mathrm{C} ; \mathrm{B}\) increases the water temperature by \(2.60{ }^{\circ} \mathrm{C}\). (a) Which object has the larger heat capacity? (b) What can you say about the specific heats of \(\mathrm{A}\) and \(\mathrm{B}\) ?

In a thermodynamic study, a scientist focuses on the properties of a solution in an apparatus as illustrated. A solution is continuously flowing into the apparatus at the top and out at the bottom, such that the amount of solution in the apparatus is constant with time. (a) Is the solution in the apparatus a closed system, open system, or isolated system? (b) If the inlet and outlet were closed, what type of system would it be?

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is produced by plants as follows: $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g) \\ \Delta H=5645 \mathrm{~kJ} \end{aligned} $$ About \(4.8 \mathrm{~g}\) of sucrose is produced per day per square meter of the earth's surface. The energy for this endothermic reaction is supplied by the sunlight. About \(0.1 \%\) of the sunlight that reaches the earth is used to produce sucrose. Calculate the total energy the sun supplies for each square meter of surface area. Give your answer in kilowatts per square meter \(\left(\mathrm{kW} / \mathrm{m}^{2}\right.\) where \(\left.1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}\right).\)

Suppose that the gas-phase reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow\) \(2 \mathrm{NO}_{2}(g)\) were carried out in a constant-volume container at constant temperature. (a) Would the measured heat change represent \(\Delta H\) or \(\Delta E\) ? (b) If there is a difference, which quantity is larger for this reaction? (c) Explain your answer to part (b).

(a) What is the electrostatic potential energy (in joules) between an electron and a proton that are separated by \(230 \mathrm{pm}\) ? (b) What is the change in potential energy if the distance separating the electron and proton is increased to \(1.0 \mathrm{nm}\) ? (c) Does the potential energy of the two particles increase or decrease when the distance is increased to \(1.0 \mathrm{nm}\) ?
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