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(III) The specific heat per mole of potassium at low temperatures is given by \(C_{V}=a T+b T^{3}, \quad\) where \(a=2.08 \mathrm{~mJ} / \mathrm{mol} \cdot \mathrm{K}^{2}\) and \(b=2.57 \mathrm{~mJ} / \mathrm{mol} \cdot \mathrm{K}^{4}\). Determine (by integration) the entropy change of \(0.15 \mathrm{~mol}\) of potassium when its temperature is lowered from \(3.0 \mathrm{~K}\) to \(1.0 \mathrm{~K}\).

Short Answer

Expert verified
The entropy change for 0.15 mol of potassium is approximately 3.96 mJ/K.

Step by step solution

01

Understand the Formula for Entropy Change

The change in entropy \( \Delta S \) can be found using the formula \( \Delta S = \int \frac{C_V}{T} \, dT \), where \( C_V = aT + bT^3 \). We will substitute the expression for \( C_V \) and perform the integration between the limits \( T_1 = 3.0 \) K and \( T_2 = 1.0 \) K.
02

Set Up the Integral

We need to calculate the integral \( \Delta S = \int_{3}^{1} \frac{aT + bT^3}{T} \, dT \). Upon simplification, this becomes \( \Delta S = \int_{3}^{1} (a + bT^2) \, dT \).
03

Perform the Integration

Now, integrate each term separately:\[\Delta S = \int_{3}^{1} a \, dT + \int_{3}^{1} bT^2 \, dT\]The first integral \( \int a \, dT = aT \) evaluated from 3 to 1 gives:\[-a(1) + a(3) = a(3-1)\]The second integral \( \int bT^2 \ dT = \frac{bT^3}{3} \) evaluated from 3 to 1 gives:\[-\frac{b(1)^3}{3} + \frac{b(3)^3}{3} = \frac{b}{3}(27 - 1) = \frac{26b}{3}\]
04

Calculate the Specific Change in Entropy

Substitute the values of \(a\) and \(b\) to find the change in entropy:\[\Delta S = a(3-1) + \frac{26b}{3}\]\[a = 2.08 \, \text{mJ/mol K}^2, \quad b = 2.57 \, \text{mJ/mol K}^4\]\[\Delta S = 2.08 \times 2 + \frac{26 \times 2.57}{3} = 4.16 + \frac{66.82}{3}\]\[\Delta S = 4.16 + 22.27 = 26.43 \, \text{mJ/mol K}\]
05

Adjust for Moles of Potassium

Since the system involves 0.15 mol of potassium, scale \( \Delta S \) by 0.15:\[\Delta S_{total} = 0.15 \times 26.43 = 3.9645 \, \text{mJ/K}\]

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Key Concepts

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

Specific Heat
Specific heat is a vital concept in thermodynamics, describing the amount of heat needed to change an object's temperature by a certain degree. It helps us understand how different materials respond to heat energy. In our exercise, specific heat is denoted as \( C_V \) and expressed as a function of temperature: \( C_V = aT + bT^3 \). Here, \( a \) and \( b \) are constants specific to potassium, indicating how its heat capacity varies with temperature, especially at low temperatures.

This exercise uses specific heat to calculate changes in entropy at low temperatures. Specific heat tells us two things:
  • The linear term \( aT \) indicates a straightforward relationship where specific heat increases directly with temperature.
  • The cubic term \( bT^3 \) shows how the heat capacity ramps up dramatically at higher temperatures, albeit still low in our case.
By understanding these components, we can predict how a substance like potassium behaves energetically as the temperature changes.
Low Temperature Physics
Low temperature physics explores the behaviors and properties of materials as they approach absolute zero. At these temperatures, classical physics gives way to quantum mechanics, revealing unique phenomena.

In the context of this exercise, low temperature physics is crucial because the specific heat of materials like potassium reflects subtle quantum effects. At such low temperatures (from 3 K to 1 K), particles exhibit less thermal motion, and factors such as electronic contributions to heat capacity (reflected in the \( aT \) term) become prominent.

One fascinating result in low temperature physics is how traditional predictions for specific heat (like those from the Dulong-Petit law for metals at room temperature) fail. Instead, models like the Debye or Einstein models offer more accurate predictions, as seen in the polynomial expression of our exercise.
Thermodynamic Integration
Thermodynamic integration is a method used to compute changes in thermodynamic quantities such as entropy, using calculus. In this exercise, we employ thermodynamic integration to calculate the change in entropy, denoted by \( \Delta S \).

The process involves setting up an integral of the form \( \Delta S = \int \frac{C_V}{T} \, dT \), which represents the total change in entropy as the temperature varies. For our problem, the integral involves the specific heat equation \( C_V = aT + bT^3 \).
  • The integral first simplifies by substituting \( C_V \), breaking into components that can be separately integrated.
  • By applying the limits of the temperature change (from 3 K to 1 K), we obtain a numerical value for \( \Delta S \).

Thermodynamic integration offers a powerful approach to tackling changes in entropy, especially in the context of non-uniform heat capacities across a temperature range.

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Most popular questions from this chapter

A car engine whose output power is 155 hp operates at about \(15 \%\) efficiency. Assume the engine's water temperature of \(95^{\circ} \mathrm{C}\) is its cold-temperature (exhaust) reservoir and \(495^{\circ} \mathrm{C}\) is its thermal "intake" temperature (the temperature of the exploding gas-air mixture). \((a)\) What is the ratio of its efficiency relative to its maximum possible (Carnot) efficiency? (b) Estimate how much power (in watts) goes into moving the car, and how much heat, in joules and in kcal, is exhausted to the air in \(1.0 \mathrm{~h}\).

(II) The burning of gasoline in a car releases about \(3.0 \times 10^{4}\) kcal/gal. If a car averages \(38 \mathrm{~km} /\) gal when driving \(95 \mathrm{~km} / \mathrm{h},\) which requires \(25 \mathrm{hp},\) what is the efficiency of the engine under those conditions?

(II) A four-cylinder gasoline engine has an efficiency of 0.22 and delivers 180 \(\mathrm{J}\) of work per cycle per cylinder. The engine fires at 25 cycles per second. (a) Determine the work done per second. \((b)\) What is the total heat input per second from the gasoline? (c) If the energy content of gasoline is 130 \(\mathrm{MJ}\) per gallon, how long does one gallon last?

Refrigeration units can be rated in "tons." A 1 -ton air conditioning system can remove sufficient energy to freeze 1 British ton \((2000\) pounds \(=909 \mathrm{kg})\) of \(0^{\circ} \mathrm{C}\) water into \(0^{\circ} \mathrm{C}\) ice in one 24 -h day. If, on a \(35^{\circ} \mathrm{C}\) day, the interior of a house is maintained at \(22^{\circ} \mathrm{C}\) by the continuous operation of a 5 -ton air conditioning system, how much does this cooling cost the homeowner per hour? Assume the work done by the refrigeration unit is powered by electricity that costs \(\$ 0.10\) per kWh and that the unit's coefficient of performance is 15\(\%\) that of an ideal refrigerator. \(1 \mathrm{kWh}=3.60 \times 10^{6} \mathrm{J}\)

Two 1100 -kg cars are traveling 75 \(\mathrm{km} / \mathrm{h}\) in opposite directions when they collide and are brought to rest. Estimate the change in entropy of the universe as a result of this collision. Assume \(T=15^{\circ} \mathrm{C}\) .

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