Prove that two reversible adiabatic paths can never cross. Assume that the energy of the system under consideration is a function of temperature only. (Hint. Suppose that two such paths can intersect, and complete a cycle with the two paths plus one isothermal path. Consider the changes accompanying each stage of the cycle and show that they conflict with the Kelvin statement of the Second Law.)

Short Answer

Expert verified
Assuming that two reversible adiabatic paths intersect and constructing a hypothetical cycle with them and an isothermal path leads to a contradiction of the Kelvin-Planck statement of the Second Law, showing such paths cannot cross.

Step by step solution

01

Assume two reversible adiabatic paths intersect

Begin by supposing that two reversible adiabatic paths, path A and path B, do cross each other at some point in the P-V diagram where P is the pressure and V is the volume.
02

Create a cycle with an isothermal path

Construct a closed cycle by connecting the end points of path A and B with an isothermal path (constant temperature). Since we are considering the energy as a function of temperature only, the internal energy remains constant on the isothermal path.
03

Examine the adiabatic paths

On both adiabatic paths (A and B), there is no heat transfer, as per the definition of an adiabatic process. This implies that any work done on or by the system will result in a change in the internal energy. Since internal energy is only a function of temperature and there's no heat transfer, the temperature does not change, which leads to the conclusion that no work is done in an adiabatic process that is also reversible. This conflicts with the premise that these paths could have a common intersection.
04

Consider the Kelvin-Planck statement of the Second Law

According to the Kelvin-Planck statement, it's impossible to construct an engine that, operating in a cycle, will convert all absorbed heat into an equivalent amount of work. If the closed cycle created from the intersection of the two adiabatic paths and the isothermal path did no work, it would imply that the engine described is possible, contradicting the Second Law.
05

Conclude the proof by contradiction

The assumption that reversible adiabatic paths can cross leads to a contradiction with the Second Law of Thermodynamics. Therefore, two reversible adiabatic paths cannot intersect.

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Key Concepts

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

Second Law of Thermodynamics
The Second Law of Thermodynamics is a fundamental principle that provides a direction to various processes in nature. It states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible.

Entropy, in simple terms, can be thought of as the measure of disorder or randomness within a system. This law has profound implications in thermodynamics, one of which is the Kelvin-Planck statement that was referred to in the exercise. According to this statement, no process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of that heat into work. There must always be some waste heat released into the environment.

In the context of the exercise, invoking the Second Law of Thermodynamics led to a contradiction when assuming the intersection of two reversible adiabatic paths. The law implies that a cycle created in such a way would lead to an impossible engine that violates this very law by fully converting heat into work without any waste, a contradiction which therefore proves that two reversible adiabatic paths cannot intersect.

Understanding the Second Law is crucial not only for grasping thermodynamics as a whole but particularly in the analysis of the feasibility and efficiency of thermal systems. It underscores the unattainability of perfect energy conversion and the inevitable increase of entropy, shaping how engineers design engines and refrigerators amongst other systems.
P-V diagram
A P-V diagram, also known as a Pressure-Volume diagram, is a very useful tool in the study of thermodynamics for visualizing the state of a system. On this graph, pressure (P) is usually denoted on the y-axis and volume (V) on the x-axis. Each point on the diagram represents a unique state of the system with a specific pressure, volume, and temperature.

The utility of a P-V diagram becomes apparent when we consider the thermodynamic processes a system undergoes. These are depicted as paths on the diagram that connect different states. For instance, isothermal processes (at constant temperature) often appear as hyperbolic curves, while adiabatic processes (where no heat is transferred) appear steeper as compared to isothermal ones, because of the rapid change in pressure for a given change in volume.

In the case of the original exercise, the P-V diagram was essential in illustrating why two reversible adiabatic paths cannot cross. Without a graphical representation, it would be challenging to intuitively grasp the idea that intersecting paths would violate a fundamental law of thermodynamics. In essence, P-V diagrams are a powerful means to communicate and analyze the relationships between pressure, volume, and temperature, and the work done by or on a system during various processes.
Isothermal Process
An isothermal process is a thermodynamic process in which the temperature of the system remains constant. This is a key concept in understanding how energy is transferred within a thermodynamic system. During an isothermal process in an ideal gas, any work done by the system or on the system is accompanied by a corresponding heat transfer, so that the internal energy of the system remains unchanged.

On a P-V diagram, an isothermal process for an ideal gas is represented by a hyperbolic curve. This is because, according to Boyle's Law, the pressure of the gas is inversely proportional to its volume at constant temperature. Hence, as the volume increases, the pressure decreases, and vice versa, along a smooth continuous curve without abrupt changes in direction.

Why is this important in the exercise? The role of the isothermal process in creating a cycle on a P-V diagram helped demonstrate that two reversible adiabatic paths could not intersect. Any deviation, such as a crossing of paths, would suggest a process capable of producing work without transferring heat into or out of a system, which directly contradicts the principles governed by isothermal and adiabatic processes, therefore highlighting their adherence to the Second Law of Thermodynamics.

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Most popular questions from this chapter

Suppose that \(S\) is regarded as a function of \(p\) and \(\mathrm{T}\). Show that \(T \mathrm{dS}=\) \(C_{\mathrm{p}} \mathrm{d} T-\alpha T V \mathrm{~d} p .\) Hence, show that the energy transferred as heat when the pressure on an incompressible liquid or solid is increased by \(\Delta p\) is equal to \(-\alpha T V \Delta p .\) Evaluate \(q\) when the pressure acting on \(100 \mathrm{~cm}^{3}\) of mercury at \(0^{\circ} \mathrm{C}\) is increased by \(1.0\) kbar. \(\left(\alpha=1.82 \times 10^{-4} \mathrm{~K}^{-1} .\right)\)

To calculate the work required to lower the temperature of an object, we need to consider how the coefficient of performance changes with the temperature of the object. (a) Find an expression for the work of cooling an object from \(\mathrm{T}\), to \(\mathrm{T}_{\mathrm{r}}\) when the refrigerator is in a room at a temperature \(\mathrm{T}_{\mathrm{h}}\). Hint. Write \(d w=d y l c\left(T\right.\) relate \(d q\) to \(\mathrm{dT}\) through the heat capacity \(\mathrm{C}_{p}\), and integrate the resulting expression. Assume that the heat capacity is independent of temperature in the range of interest. (b) Use the result in part (a) to calculate the work needed to freeze \(250 \mathrm{~g}\) of water in a refrigerator at \(293 \mathrm{~K}\). How long will it take when the refrigerator operates at \(100 \mathrm{~W}\) ?

Given that \(s_{m}^{\circ}=29.79 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) for bismuth at \(100 \mathrm{~K}\) and the following tabulated heat capacities data (D.G. Archer, \(J\). Chem. Eng. Data 40 , 1015 (1995)), compute the standard molar entropy of bismuth at \(200 \mathrm{~K}\). $$ \begin{array}{lccccccl} \text { TIK } & 100 & 120 & 140 & 150 & 160 & 180 & 200 \\ C_{\mathrm{Am}} /\left(\mathrm{K} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) & 23.00 & 23.74 & 24.25 & 24.44 & 24.61 & 24.89 & 25.11 \end{array} $$ Compare the value to the value that would be obtained by taking the heat capacity to be constant at \(24.44 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) over this range.

Consider a perfect gas contained in a cylinder and separated by a frictionless adiabatic piston into two sections \(\mathrm{A}\) and \(\mathrm{B}\). All changes in \(\mathrm{B}\) is isothermal; that is, a thermostat surrounds \(\mathrm{B}\) to keep its temperature constant. There is \(2.00 \mathrm{~mol}\) of the gas in each section. Initially, \(T_{\mathrm{A}}==T_{\mathrm{B}}=300 \mathrm{~K}, V_{\mathrm{A}}=\) \(V_{\mathrm{B}}\) \(=2.00 \mathrm{dm}^{3}\). Energy is supplied as heat to Section A and the piston moves to the right reversibly until the final volume of Section B is \(1.00 \mathrm{dm}^{3}\). Calculate (a) \(\Delta S_{\mathrm{A}}\) and \(\Delta S_{\mathrm{B}}\), (b) \(\Delta A_{\mathrm{A}}\) and \(\Delta \mathrm{A}_{\mathrm{B}}\), (c) \(\Delta G_{\mathrm{A}}\) and \(\Delta G_{\mathrm{B}}\), (d) AS of the total system and its surroundings. If numerical values cannot be obtained, indicate whether the values should be positive, negative, or zero or are indeterminate from the information given. (Assume \(C_{\mathrm{v}, \mathrm{m}}=20 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\).)

The adiabatic compressibility, \(\kappa_{S}\), is defined like \(\kappa_{T}\) (eqn 2.44) but at constant entropy. Show that for a perfect gas \(p \gamma \kappa_{\mathrm{S}}=1\) (where \(\gamma\) is the ratio of heat capacities).

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