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The temperature of \(1.00 \mathrm{~mol}\) of a monatomic ideal gas is raised reversibly from \(300 \mathrm{~K}\) to \(400 \mathrm{~K}\), with its volume kept constant. What is the entropy change of the gas?

Short Answer

Expert verified
The entropy change of the gas is approximately 3.585 J/K.

Step by step solution

01

Identify the Formula

For a monatomic ideal gas with constant volume, the change in entropy \( \Delta S \) is given by the formula: \[ \Delta S = nC_V \ln\left(\frac{T_2}{T_1}\right) \] where \( n \) is the number of moles, \( C_V \) is the molar heat capacity at constant volume, and \( T_2 \) and \( T_1 \) are the final and initial temperatures, respectively.
02

Determine Values

Given that \( n = 1.00 \) mol, \( T_1 = 300 \) K, and \( T_2 = 400 \) K, we need to find \( C_V \). For a monatomic ideal gas, \( C_V = \frac{3}{2}R \), where \( R \) is the ideal gas constant with a value of \( 8.314 \text{ J/mol K} \).
03

Calculate \( C_V \)

Calculate the molar heat capacity at constant volume \( C_V \) using the formula: \[ C_V = \frac{3}{2} \times 8.314 = 12.471 \text{ J/mol K} \].
04

Plug Values into the Formula

Substitute the values into the entropy change formula: \[ \Delta S = 1.00 \times 12.471 \times \ln\left(\frac{400}{300}\right) \].
05

Solve for the Entropy Change

Calculate the natural logarithm and solve: \[ \ln\left(\frac{400}{300}\right) = \ln\left(\frac{4}{3}\right) \approx 0.287682 \] Therefore, \[ \Delta S = 1.00 \times 12.471 \times 0.287682 \approx 3.585 \text{ J/K} \].

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Key Concepts

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

Monatomic Ideal Gas
When we talk about a monatomic ideal gas, we refer to a gas composed entirely of single atom molecules, like helium, neon, or argon. These gases are often used in physics and chemistry due to their simplicity. They follow the ideal gas law, which assumes that interactions between gas particles and the volume of gas particles themselves are negligible. This makes calculations straightforward.
Because monatomic gases have less structural complexity, they are described well by simple physical laws and equations. Understanding this concept is important when solving problems involving changes in temperature and entropy for a monatomic gas under set conditions.
Heat Capacity at Constant Volume
The heat capacity at constant volume, denoted as \( C_V \), is a crucial concept when studying thermodynamics. \( C_V \) is the amount of heat needed to raise the temperature of a mole of gas by 1 Kelvin while keeping the volume constant.
For a monatomic ideal gas, \( C_V \) can be calculated using the formula: \( C_V = \frac{3}{2} R \). Here, \( R \) is the ideal gas constant with a value of 8.314 J/mol K. This formula reflects the fact that monatomic gases only have translational degrees of freedom in three dimensions—x, y, and z.
In the context of the exercise, once we determine \( C_V \), it helps us calculate the entropy change due to a temperature rise while the gas is kept in a constant volume environment.
Temperature Change
Temperature change is a key factor in many thermodynamic processes. When dealing with ideal gases, temperature changes can significantly affect properties such as pressure, volume, and entropy.
In this case, we deal with a monatomic ideal gas whose temperature rises from 300 K to 400 K. When temperature changes at constant volume, we can compute the change in entropy using the formula \( \Delta S = nC_V \ln\left(\frac{T_2}{T_1}\right) \).
Understanding how temperature changes influence the behavior of gases helps us predict how energy is transferred and transformed. This is valuable both in theoretical studies and practical applications, like engines and cooling systems. Always remember, as temperature increases, the kinetic energy of gas molecules increases, leading to changes in other properties like entropy.

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Most popular questions from this chapter

At very low temperatures, the molar specific heat \(C_{V}\) of many solids is approximately \(C_{V}=A T^{3}\), where \(A\) depends on the particular substance. For aluminum, \(A=3.15 \times 10^{-5} \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}^{4} .\) Find the entropy change for \(4.00 \mathrm{~mol}\) of aluminum when its temperature is raised from \(5.00 \mathrm{~K}\) to \(10.0 \mathrm{~K}\).

In an experiment, \(200 \mathrm{~g}\) of aluminum (with a specific heat of \(900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) at \(100^{\circ} \mathrm{C}\) is mixed with \(50.0 \mathrm{~g}\) of water at \(20.0^{\circ} \mathrm{C}\), with the mixture thermally isolated. (a) What is the equilibrium temperature? What are the entropy changes of (b) the aluminum, (c) the water, and (d) the aluminum- water system?

To make ice, a freezer that is a reverse Carnot engine extracts \(42 \mathrm{~kJ}\) as heat at \(-15^{\circ} \mathrm{C}\) during each cycle, with coefficient of performance \(5.7 .\) The room temperature is \(30.3^{\circ} \mathrm{C}\). How much (a) energy per cycle is delivered as heat to the room and (b) work per cycle is required to run the freezer?

Suppose that a deep shaft were drilled in Earth's crust near one of the poles, where the surface temperature is \(-40^{\circ} \mathrm{C}\), to a depth where the temperature is \(800^{\circ} \mathrm{C}\). (a) What is the theoretical limit to the efficiency of an engine operating between these temperatures? (b) If all the energy released as heat into the low-temperature reservoir were used to melt ice that was initially at \(-40^{\circ} \mathrm{C}\), at what rate could liquid water at \(0^{\circ} \mathrm{C}\) be produced by a 100 MW power plant (treat it as an engine)? The specific heat of ice is \(2220 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} ;\) water's heat of fusion is \(333 \mathrm{~kJ} / \mathrm{kg} .\) (Note that the engine can operate only between \(0^{\circ} \mathrm{C}\) and \(800^{\circ} \mathrm{C}\) in this case. Energy exhausted at \(-40^{\circ} \mathrm{C}\) cannot warm anything above \(-40^{\circ} \mathrm{C}\).)

An air conditioner operating between \(93^{\circ} \mathrm{F}\) and \(70^{\circ} \mathrm{F}\) is rated at \(4000 \mathrm{Btu} / \mathrm{h}\) cooling capacity. Its coefficient of performance is \(27 \%\) of that of a Carnot refrigerator operating between the same two temperatures. What horsepower is required of the air conditioner motor?

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