A system undergoes a process in which the entropy change is \(+5.51 \mathrm{JK}^{-1}\). During the process, \(1.50 \mathrm{~kJ}\) of heat is added to the system at \(300 \mathrm{~K}\). The correct information regarding the process is (a) the process thermodynamically reversible. (b) the process is thermodynamically irreversible. (c) the process may or may not be thermodynamically reversible. (d) the process must be isobaric.

Short Answer

Expert verified
The process is thermodynamically irreversible because the actual entropy change is greater than the entropy change calculated for a reversible process.

Step by step solution

01

Analyze the Thermodynamic Process

To determine whether the process is reversible or irreversible, we must calculate the entropy change if the process were reversible. The entropy change in a reversible process when heat is added or removed can be calculated using the formula \(\Delta S = \frac{Q}{T}\), where \(\Delta S\) is the entropy change, \(Q\) is the heat added to the system, and \(T\) is the absolute temperature at which the process occurs.
02

Calculate the Theoretical Reversible Entropy Change

Using the formula for the entropy change in a reversible process, we calculate \(\Delta S_{rev} = \frac{Q}{T}\). Here, \(Q = 1.50 \mathrm{~kJ} = 1500 \mathrm{~J}\) (since 1 kJ = 1000 J) and \(T = 300 \mathrm{~K}\). Thus, \[\Delta S_{rev} = \frac{1500 \mathrm{~J}}{300 \mathrm{~K}} = 5 \mathrm{JK}^{-1}\].
03

Compare the Calculated Entropy Change with Actual Entropy Change

The actual entropy change is given as \(\Delta S = +5.51 \mathrm{JK}^{-1}\). Comparing this with the calculated entropy change for a reversible process \(\Delta S_{rev} = 5 \mathrm{JK}^{-1}\), we see that \(\Delta S > \Delta S_{rev}\). In a reversible process, the actual entropy change would match the calculated entropy change for a reversible process. Since \(\Delta S > \Delta S_{rev}\), the extra entropy indicates that additional entropy has been produced in the system or its surroundings, thus making the process irreversible.

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Key Concepts

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

Entropy Change
Entropy is often referred to as a measure of disorder or randomness in a system. When a thermodynamic process occurs, such as the transfer of heat, the entropy of the system can change. The given problem states that the entropy change is +5.51 JK-1, which signifies an increase in entropy. This rise suggests that the system's disorder increased during the process. Entropy change is a central concept in thermodynamics because it helps determine the direction of heat flow and the efficiency of energy conversions.

Understanding entropy change is crucial not only in theoretical thermodynamics but also in practical applications like engine design, refrigeration, and the study of environmental processes. A positive entropy change, as observed in our example, often signifies that heat has been absorbed by the system, contributing to the increased disorder.
Reversible and Irreversible Processes
Thermodynamic processes are categorized as either reversible or irreversible. A reversible process is an idealized or hypothetical scenario in which a system undergoes changes infinitely slowly, allowing for the exact reversal of the process without any net change in the system and its environment. In reality, reversible processes are not attainable, but they provide a useful benchmark.

On the other hand, irreversible processes occur spontaneously and cannot return both the system and the environment to their original states without external work. Irreversibility in a process usually involves factors like friction, turbulence, inelastic collisions, and natural heat flow from a hot body to a cooler one. The problem at hand specifies that the actual entropy change is greater than what would be calculated for a reversible process, indicating that the process is indeed irreversible.
Heat Transfer Thermodynamics
Heat transfer in thermodynamics is the movement of thermal energy from one object or system to another. It occurs through three primary mechanisms: conduction, convection, and radiation. In a thermodynamic process, understanding heat transfer is vital for predicting how energy will move and how it will affect the entropy of the systems involved.

The method of heat transfer is consequential to entropy. When heat is transferred to a system, it usually leads to an increase in the entropy of that system. However, the total entropy change of the system and its environment must be considered to determine the overall entropy change of the universe, which, according to the second law of thermodynamics, must always increase or remain constant for a spontaneous process.
Calculation of Entropy
Calculating entropy involves using specific thermodynamic formulas. The fundamental formula \(\Delta S = \frac{Q}{T}\) allows us to calculate the entropy change, \(\Delta S\), when a reversible, isothermal process involves the transfer of heat, \(Q\), at a constant temperature, \(T\). Here, temperature must be in Kelvin to ensure accuracy as it is an absolute scale.

The exercise provides us with a theoretical framework to calculate entropy under reversible conditions. However, the observed higher actual entropy change suggests that the irreversible process involved additional entropy production, which is common in natural processes. When calculating entropy for such processes, additional factors contributing to irreversibility must be accounted for, making the calculation more complex than the idealized reversible scenario.

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Most popular questions from this chapter

Oxygen gas weighing \(64 \mathrm{~g}\) is expanded from 1 atm to \(0.25\) atm at \(30^{\circ} \mathrm{C}\). What is the entropy change, assuming the gas to be ideal? \((\ln 4=1.4, R=8.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol})\) (a) \(23.24 \mathrm{~J} / \mathrm{K}\) (b) \(34.86 \mathrm{~J} / \mathrm{K}\) (c) \(46.48 \mathrm{~J} / \mathrm{K}\) (d) \(11.62 \mathrm{~J} / \mathrm{K}\)

A quantity of \(1.6 \mathrm{~g}\) helium gas is expanded adiabatically \(3.0\) times and then compressed isobarically to the initial volume. Assume ideal behaviour of gas and both the processes reversible. The entropy change of the gas in this process is \((\ln 3=1.1)\) (a) \(-1.1 \mathrm{cal} / \mathrm{K}\) (b) \(+1.1 \mathrm{cal} / \mathrm{K}\) (c) \(-2.2 \mathrm{cal} / \mathrm{K}\) (d) \(+2.2 \mathrm{cal} / \mathrm{K}\)

Heat liberated by a given amount of an ideal gas undergoing reversible isothermal process is \(1200 \mathrm{cal}\) at \(300 \mathrm{~K}\). What is the Gibbs free energy change of the gas in this process? (a) zero (b) \(+1200\) cal (c) \(-1200\) cal (d) \(4 \mathrm{cal}\)

The solubility of \(\mathrm{NaCl}(\mathrm{s})\) in water at \(298 \mathrm{~K}\) is about 6 moles per litre. Suppose you add 1 mole of \(\mathrm{NaCl}(\mathrm{s})\) to a litre of water. For the reaction: \(\mathrm{NaCl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}\) \(\rightarrow \mathrm{NaCl}(\mathrm{aq})\) (a) \(\Delta G>0, \Delta S>0\) (b) \(\Delta G<0, \Delta S>0\) (c) \(\Delta G>0, \Delta S<0\) (d) \(\Delta G<0, \Delta S<0\)

One mole of a real gas is subjected to heating at constant volume from \(\left(P_{1}\right.\), \(V_{1}, T_{1}\) ) state to \(\left(P_{2}, V_{1}, T_{2}\right)\) state. Then it is subjected to irreversible adiabatic compression against constant external pressure of \(P_{3}\) atm, till the system reaches final state \(\left(P_{3}, V_{2}, T_{3}\right) .\) If the constant volume molar heat capacity of real gas is \(C_{\mathrm{V}}\), then the correct expression for \(\Delta H\) from State 1 to State 3 is (a) \(C_{\mathrm{V}}\left(T_{3}-T_{1}\right)+\left(P_{3} V_{1}-P_{1} V_{1}\right)\) (b) \(C_{\mathrm{V}}\left(T_{2}-T_{1}\right)+\left(P_{3} V_{2}-P_{1} V_{1}\right)\) (c) \(C_{\mathrm{y}}\left(T_{2}-T_{1}\right)+\left(P_{3} V_{1}-P_{1} V_{1}\right)\) (d) \(C_{\mathrm{P}}\left(T_{2}-T_{1}\right)+\left(P_{3} V_{1}-P_{1} V_{1}\right)\)

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