/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 If a gas at constant temperature... [FREE SOLUTION] | 91Ó°ÊÓ

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If a gas at constant temperature and pressure expands, then its (a) internal energy decreases (b) entropy increases and then decreases (c) internal energy increases (d) internal energy remains constant

Short Answer

Expert verified
(d) internal energy remains constant.

Step by step solution

01

Analyzing the Problem Statement

The problem asks about what happens when a gas expands at constant temperature (isothermal process) and pressure. We need to consider thermodynamic principles to determine the correct answer.
02

Applying the First Law of Thermodynamics

The first law of thermodynamics states that the change in internal energy (\(\Delta U\)) is equal to the heat added to the system (\(Q\)) minus the work done by the system (\(W\)): \(\Delta U = Q - W\). During an isothermal expansion, the temperature remains constant, suggesting there is no change in internal energy: \(\Delta U = 0\).
03

Considering Work Done Due to Expansion

In an isothermal expansion, although the gas does work on the surroundings, compensating heat is supplied to keep the temperature constant. This means that the net change in internal energy is zero.
04

Evaluating Entropy Changes

For an isothermal expansion, the entropy of the gas increases as the gas spreads into a larger volume. Entropy, a measure of disorder, generally does not decrease spontaneously in isolated systems, so option (b) is incorrect.
05

Finalizing the Conclusion

Given that the internal energy remains constant in an isothermal process, and entropy increases (without later decreasing), the answer can be deduced from understanding these thermodynamic concepts.

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Key Concepts

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

Internal Energy
In the context of a gas expanding isothermally (at constant temperature), understanding internal energy is crucial. Internal energy, denoted by \( U \), is the total energy contained within a system. It comprises the kinetic energy of molecules due to their motion and the potential energy resulting from intermolecular forces. For an isothermal process, the temperature remains unchanged, and so does the internal energy.According to the first law of thermodynamics, the change in internal energy \( \Delta U \) is given by \( \Delta U = Q - W \), where \( Q \) is the heat added to the system and \( W \) is the work done by the system. During isothermal expansion, any work done by the gas is exactly offset by heat absorbed from the surroundings. This balance ensures that the internal energy stays constant, emphasizing option (d) in the original exercise. Hence, the internal energy of the system does not change because the temperature, which dictates molecular kinetic energy, remains constant.
Entropy
Entropy is a key concept in thermodynamics, representing the degree of disorder or randomness in a system. During an isothermal expansion, a gas moves to occupy a larger volume, which naturally increases disorder. Thus, entropy increases.The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. This means spontaneous processes, like the expansion of gas into a vacuum or larger area, invariably lead to an increase in entropy. Since the temperature remains steady in an isothermal process, entropy will rise without any subsequent decrease, contrary to some misconceptions.Moreover, entropy change \( \Delta S \) due to heat transfer is calculated using \( \Delta S = \frac{Q}{T} \), where \( T \) is the constant temperature. This highlights that as the gas takes in heat to balance the work done, entropy continuously increases.
First Law of Thermodynamics
The first law of thermodynamics plays a central role in understanding energy changes in thermodynamic processes. In summary, it states that energy cannot be created or destroyed, only transferred or converted. The formal representation is: \( \Delta U = Q - W \).During an isothermal expansion, even though the gas does work (\( W \)) on its surroundings, the first law assures us that the total change in internal energy (\( \Delta U \)) remains zero, as the work done is exactly compensated by the heat absorbed (\( Q \)).In practical terms:
  • Work done by the system means the system loses energy in terms of work output.
  • Heat absorbed suggests the system gains energy in equivalence to the external work.
In essence, the first law confirms the conservation of energy, such that during an isothermal process, all changes between heat added and work done even out to maintain consistent internal energy.

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Most popular questions from this chapter

The work done by a system is 10 joule, when 40 joule heat is supplied to it. What is the increase in internal energy of system? (a) \(30 \mathrm{~J}\) (b) \(50 \mathrm{~J}\) (c) \(40 \mathrm{~J}\) (d) \(20 \mathrm{~J}\)

Which one of the following statements is false? (a) temperature is a state function (b) work is a state function (c) change in the state depends upon initial and final state (d) work appears at the boundary of the system

The factor of \(\Delta \mathrm{G}\) values is important in metallurgy. The \(\Delta\) G values for the following reactions at \(800^{\circ} \mathrm{C}\) are given as \(\mathrm{S}_{2}(\mathrm{~s})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_{2}(\mathrm{~g}) ; \Delta \mathrm{G}=-544 \mathrm{~kJ}\) \(2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{ZnS}(\mathrm{s}) ; \Delta \mathrm{G}=-293 \mathrm{~kJ}\) \(2 \mathrm{Zn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{s}) ; \Delta \mathrm{G}=-480 \mathrm{~kJ}\) the \(\Delta \mathrm{G}\) for the reaction, \(2 \mathrm{ZnS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{~g})\) will be (a) \(-357 \mathrm{~kJ}\) (b) \(-731 \mathrm{~kJ}\) (c) \(-773 \mathrm{~kJ}\) (d) \(-229 \mathrm{~kJ}\)

Which of the following relations are correct? (a) \(\mathrm{H}=\mathrm{G}+\mathrm{TS}\) (b) \(\mathrm{E}=\mathrm{H}+\mathrm{PV}\) (c) \(\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}\) (d) \(\mathrm{q}_{\mathrm{v}}=\mathrm{q}_{\mathrm{p}}-\Delta \mathrm{n}_{(\mathrm{g})} \mathrm{RT}\)

The enthalpy and entropy change for a chemical reaction are \(-2.5 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\) and \(7.4 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respec- tively. The reaction at \(298 \mathrm{~K}\). is (a) non-spontaneous (b) spontaneous (c) reversible (d) irreversible

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