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23. The Second Law of Thermodynamics

Entropy Equations for Special Processes

Physics

23. The Second Law of Thermodynamics

Entropy Equations for Special Processes

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6:33 minutes

Problem 57b

Textbook Question

A general theorem states that the amount of energy that becomes unavailable to do useful work in any process is equal to T_L∆S, where T_L is the lowest temperature available and ∆S is the total change in entropy during the process. Show that this is valid in the specific cases of: the free adiabatic expansion of an ideal gas.

Verified step by step guidance

Understand the problem: The goal is to verify the theorem T_LΔS = unavailable energy for the specific case of the free adiabatic expansion of an ideal gas. In this process, the gas expands without exchanging heat (Q = 0) and without doing work (W = 0). The entropy change (ΔS) must be calculated to confirm the theorem.

Step 1: Recall the first law of thermodynamics, which states ΔU = Q - W. For a free adiabatic expansion, Q = 0 and W = 0, so ΔU = 0. This means the internal energy of the gas remains constant during the process.

Step 2: Use the fact that for an ideal gas, internal energy depends only on temperature. Since ΔU = 0, the temperature of the gas remains constant during the free adiabatic expansion. This is an isothermal process.

Step 3: Calculate the entropy change (ΔS) for the gas. For an isothermal process, the entropy change of an ideal gas is given by the formula:

Δ S = n R ln(V

_f/V_i), where n is the number of moles of gas, R is the universal gas constant, and V_f and V_i are the final and initial volumes, respectively.

Step 4: Since the process is adiabatic and no work is done, the energy unavailable for useful work is equal to T_LΔS. Substitute the expression for ΔS into the theorem:

T

_LΔS = T_LnRln(V_f/V_i). This confirms the validity of the theorem for this specific case.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Entropy

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it quantifies the amount of energy in a physical system that is not available to do work. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, which implies that processes tend to move towards a state of maximum entropy.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the surroundings. In such processes, any change in the internal energy of the system is solely due to work done on or by the system. For an ideal gas undergoing free adiabatic expansion, the gas expands into a vacuum without doing work on the surroundings, leading to a change in entropy without heat transfer.

Ideal Gas Behavior

An ideal gas is a theoretical gas composed of many particles that are in constant random motion and interact only through elastic collisions. The behavior of an ideal gas is described by the ideal gas law, which relates pressure, volume, and temperature. In the context of free adiabatic expansion, ideal gas behavior simplifies the analysis, as it allows us to assume that the gas expands freely without any external pressure acting against it.

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Related Practice

Textbook Question

At low temperature the specific heat of diamond varies with absolute temperature T according to the Debye equation C_V = 1.88 x 10³ (T/T_D)³ Jmol^-1 K^-1 where the Debye temperature for diamond is T_D = 2230 K. Determine the entropy change of 1.00 mol of diamond when it is heated at constant volume from 4 K to 40 K. [Hint: dS = nC_VdT/T, where n is the number of moles.]

Multiple Choice

3 moles of an ideal gas are in the left side of an hourglass-shaped container, separated by a thin barrier. The right side is completely empty, but the volume of the left and right sides are equal. The barrier is suddenly removed, and the gas freely expands into the vacuum. What is the change in entropy?

The specific heat per mole of potassium at low temperatures is given by C_V = aT + bT³, where a = 2.08 mJ/mol·K² and b = 2.57 mJ/mol· K⁴. Determine (by integration) the entropy change of 0.15 mol of potassium when its temperature is lowered from 5.0 K to 1.0 K.

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Textbook Question

A general theorem states that the amount of energy that becomes unavailable to do useful work in any process is equal to T_L∆S, where T_L is the lowest temperature available and ∆S is the total change in entropy during the process. Show that this is valid in the specific cases of a falling rock that comes to rest when it hits the ground.

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Textbook Question

Determine the work available in a 2.2-kg block of copper at 490 K if the surroundings are at 290 K. Use results of Problem 57.

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