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In the accompanying cylinder diagram a chemical process occurs at constant temperature and pressure. (a) Is the sign of \(w\) indicated by this change positive or negative? (b) If the process is endothermic, does the internal energy of the system within the cylinder increase or decrease during the change and is \(\Delta E\) positive or negative? [Sections \(5.2\) and \(5.3\) ]

Short Answer

Expert verified
a) The sign of work (w) during the change cannot be determined without knowing the change in volume during the chemical process. b) If the process is endothermic, the internal energy change (ΔE) will depend on the work done by the system. If the work is positive, ΔE will also be positive (increase in internal energy), and if the work is negative, ΔE will be negative (decrease in internal energy).

Step by step solution

01

(1) Recall the First Law of Thermodynamics

The first law of thermodynamics states that the change in internal energy (ΔE) of a system is equal to the heat absorbed (Q) by the system minus the work done (W) by the system, mathematically represented as: \[ ΔE = Q - W\]
02

(2) Determine the sign of work (W) during the change

The term "constant temperature and pressure" implies an isobaric (constant pressure) process. With the pressure being constant, the sign of the work done will depend on the change in volume (ΔV) of the cylinder. If the volume increases (expansion), the work done by the system will be positive, and if the volume decreases (compression), the work done will be negative. Since there is no information about the change in volume, we cannot determine the sign of work (w) during this change.
03

(3) Analyze if the process is endothermic

If a process is endothermic, it means that the system absorbs heat from the surroundings during the chemical process. Therefore, in an endothermic process, the heat absorbed (Q) is positive.
04

(4) Determine internal energy change (ΔE) for an endothermic process

We know from the first law of thermodynamics that ΔE = Q - W. Since the process is endothermic, Q has a positive value. As the work (W) done by the system could be positive or negative depending on the change in volume, which is not specified in the question, we can only conclude that if the work is positive, the internal energy will increase (ΔE > 0) and if the work is negative, the internal energy will decrease (ΔE < 0). In summary: a) We cannot determine the sign of work without knowing the change in volume during the chemical process. b) If the process is endothermic, the internal energy change (ΔE) will depend on the work done by the system. If the work is positive, ΔE will also be positive (increase in internal energy), and if the work is negative, ΔE will be negative (decrease in internal energy).

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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 fundamental concept in thermodynamics. It's the total energy contained within a system due to both the kinetic and potential energies of its molecules.
In the context of the First Law of Thermodynamics, internal energy (ΔE) is the energy change resulting from a chemical or physical process.
The law is expressed as:
\[ΔE = Q - W\]
  • \(Q\) represents the net heat added to the system. Positive if heat is absorbed.
  • \(W\) signifies the net work done by the system. Positive if the system does work on surroundings.
The internal energy change depends on these two factors and provides insight into how energy is transformed within the system itself.
For example, if a system absorbs more heat than the work it does, its internal energy increases.
The Nature of an Isobaric Process
An isobaric process is where the pressure remains constant throughout the entire process. This occurs in everyday applications such as heating air in a piston engine or boiling water in an open pot.
When pressure is fixed, any heat added or removed from the system typically results in a change in volume. This relationship is crucial to understanding work in an isobaric process.
The work done by the system can be calculated by:
\[W = PΔV\]
  • \(P\) is the constant pressure.
  • \(ΔV\) is the change in volume. Positive if the volume expands.
Work is positive if the system expands, meaning energy is expended to push against external pressure. If the system contracts, the work is done on the system and work is negative. Understanding these relationships is vital in thermodynamics, especially when predicting the effects of a process on a system's energy.
Characteristics of an Endothermic Process
In an endothermic process, a system absorbs heat from its surroundings. This characteristic marks a fundamental difference from exothermic processes, where heat is released.
The absorption of heat means that the system draws in energy, increasing the likelihood of an increase in internal energy, provided work done doesn't offset this. Key points to note include:
  • Endothermic reactions are typically identified by a temperature drop in surroundings or a requirement for external heat application.
  • Common examples include melting, boiling, and photosynthesis.
  • If \(Q\) is positive, heat is absorbed, contributing to a potential increase in internal energy.
In summary, the reaction's endothermic nature indicates that heat being absorbed influences the change in internal energy. How it interacts with work done by or on the system will determine the net change in internal energy.

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Most popular questions from this chapter

The Sun supplies about \(1.0\) kilowatt of energy for each square meter of surface area \(\left(1.0 \mathrm{~kW} / \mathrm{m}^{2}\right.\), where a watt \(\left.=1 \mathrm{~J} / \mathrm{s}\right)\). Plants produce the equivalent of about \(0.20 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per square meter. Assuming that the sucrose is produced as follows, calculate the percentage of sunlight used to produce sucrose. $$ \begin{array}{r} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g) \\ \Delta H=5645 \mathrm{~kJ} \end{array} $$

(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) \(H\) is a state function, but \(q\) is not a state function. Explain. (c) For a given process at constant pressure, \(\Delta H\) is positive. Is the process endothermic or exothermic?

(a) Write an equation that expresses the first law of thermodynamics in terms of heat and work. (b) Under what conditions will the quantities \(q\) and \(w\) be negative numbers?

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{CO}_{2}(s)\) or \(1 \mathrm{~mol} \mathrm{CO} 2(g)\) at the same temperature, (b) \(2 \mathrm{~mol}\) of hydrogen atoms or \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), (c) \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) and \(0.5 \mathrm{~mol} \mathrm{O}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(g)\) at \(25^{\circ} \mathrm{C}\), (d) \(1 \mathrm{~mol} \mathrm{~N}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{~N}_{2}(g)\) at \(300^{\circ} \mathrm{C} .\)

At \(20^{\circ} \mathrm{C}\) (approximately room temperature) the average velocity of \(\mathrm{N}_{2}\) molecules in air is \(1050 \mathrm{mph}\). (a) What is the average speed in \(\mathrm{m} / \mathrm{s}\) ? (b) What is the kinetic energy (in J) of an \(\mathrm{N}_{2}\) molecule moving at this speed? (c) What is the total kinetic energy of \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) molecules moving at this speed?

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