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What is the entropy change for \(3.20 \mathrm{~mol}\) of an ideal monatomic gas undergoing a reversible increase in temperature from \(380 \mathrm{~K}\) to \(425 \mathrm{~K}\) at constant volume?

Short Answer

Expert verified
The entropy change is approximately 4.45 J/K.

Step by step solution

01

Identify the Formula for Entropy Change

For an ideal gas undergoing a temperature change at constant volume, the entropy change \( \Delta S \) can be calculated using the formula: \[ \Delta S = nC_v \ln \frac{T_2}{T_1} \] where \( n \) is the number of moles of gas, \( C_v \) is the molar heat capacity at constant volume, \( T_1 \) is the initial temperature, and \( T_2 \) is the final temperature. For a monatomic ideal gas, \( C_v = \frac{3}{2}R \), where \( R \) is the ideal gas constant \( 8.314 \frac{J}{mol \cdot K} \).
02

Substitute Values into Formula

Given: \( n = 3.20 \) mol, \( T_1 = 380 \) K, \( T_2 = 425 \) K, and \( C_v = \frac{3}{2}R = \frac{3}{2} \times 8.314 \frac{J}{mol \cdot K} \). Substitute these values into the formula for entropy change: \[ \Delta S = 3.20 \times \frac{3}{2} \times 8.314 \ln \left( \frac{425}{380} \right) \]
03

Calculate the Natural Logarithm

Calculate the natural logarithm \( \ln \left( \frac{425}{380} \right) \): \[ \ln \left( \frac{425}{380} \right) = \ln(1.1184) \approx 0.1114 \]
04

Compute the Entropy Change

Insert the calculated natural logarithm into the formula and evaluate: \[ \Delta S = 3.20 \times \frac{3}{2} \times 8.314 \times 0.1114 \] Perform calculations: \( \frac{3}{2} \times 8.314 = 12.471 \), then \( 3.20 \times 12.471 \times 0.1114 \approx 4.45 \) J/K.

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Key Concepts

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

Ideal Gas
An ideal gas is a theoretical concept that is used to simplify the complex behavior of gases. Ideal gases are hypothetical and help us understand the properties and behavior of real gases under various conditions. There are a few key assumptions about ideal gases that make them different from real gases:
  • Gas molecules have no volume, meaning they are point particles.
  • There are no intermolecular forces between the gas molecules, except during collisions.
  • All collisions between gas molecules and the walls of the container are perfectly elastic, meaning that there is no loss of energy.
These assumptions allow scientists to use the ideal gas law, \[PV = nRT\]where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. While no real gas perfectly fits these assumptions, many gases at high temperatures and low pressures behave very similarly to an ideal gas.
Molar Heat Capacity
The molar heat capacity of a substance is an important property that tells us how much energy is required to raise the temperature of one mole of that substance by one degree Kelvin. For gases, molar heat capacity can be defined at constant volume \(C_v\) or at constant pressure \(C_p\). This distinction is crucial because gases can expand, affecting their energy requirements.For a monatomic ideal gas, which means the gas particles are single atoms, the molar heat capacity at constant volume is given by:\[C_v = \frac{3}{2} R\]Here, \(R\) is the ideal gas constant. This value arises from the analysis of the energy contributions from the translational motion of the gas particles. It's important to note that for diatomic or polyatomic gases, \(C_v\) would be higher due to additional degrees of freedom such as rotational and vibrational motions.
Reversible Process
A reversible process is a theoretical concept in thermodynamics that represents an ideal scenario where the system changes in such a way that the system and its surroundings can be returned to their original states by infinitesimally small modifications. In simpler terms, it can reverse direction at any time from its current state by an infinitely small change. Reversible processes are important in thermodynamics because they are the most efficient way for a system to use energy. Practically, all real processes are irreversible due to friction, turbulence, or other forms of wasting energy. However, studying reversible processes helps us understand the upper limit of efficiency. In the context of this exercise, the increase in temperature of the ideal gas is considered reversible, which allows us to use a specific formula for the entropy change.
Temperature Change
Temperature change in a gas affects its kinetic energy, and subsequently, its entropy. Entropy is a measure of disorder or randomness in a system. When the temperature of a gas changes, the particle movement within the gas also changes, which impacts the system's entropy.In thermodynamics, especially when dealing with ideal gases, the relation between temperature and entropy change is crucial. As seen in this exercise, the formula used is:\[\Delta S = n C_v \ln \left( \frac{T_2}{T_1} \right)\]Where \(\Delta S\) is the change in entropy, \(n\) is the number of moles, \(C_v\) is the molar heat capacity at constant volume, \(T_1\) is the initial temperature, and \(T_2\) is the final temperature. The logarithmic term \(\ln \left( \frac{T_2}{T_1} \right)\) helps us understand that the entropy change isn't linear, and small temperature changes can result in significant entropy differences.

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Most popular questions from this chapter

An ideal gas undergoes a reversible isothermal expansion at \(77.0^{\circ} \mathrm{C},\) increasing its volume from \(1.30 \mathrm{~L}\) to \(3.40 \mathrm{~L}\). The entropy change of the gas is \(22.0 \mathrm{~J} / \mathrm{K}\). How many moles of gas are present?

As a sample of nitrogen gas \(\left(\mathrm{N}_{2}\right)\) undergoes a temperature increase at constant volume, the distribution of molecular speeds increases. That is, the probability distribution function \(P(v)\) for the molecules spreads to higher speed values, as suggested in Fig. \(19-8 b\). One way to report the spread in \(P(v)\) is to measure the difference \(\Delta v\) between the most probable speed \(v_{P}\) and the rms speed \(v_{\text {rms }}\). When \(P(v)\) spreads to higher speeds, \(\Delta v\) increases. Assume that the gas is ideal and the \(\mathrm{N}_{2}\) molecules rotate but do not oscillate. For \(1.5 \mathrm{~mol}\), an initial temperature of \(250 \mathrm{~K},\) and a final temperature of \(500 \mathrm{~K}\), what are (a) the initial difference \(\Delta v_{i},\) (b) the final difference \(\Delta v_{f}\) and (c) the entropy change \(\Delta S\) for the gas?

A \(500 \mathrm{~W}\) Carnot engine operates between constant-temperature reservoirs at \(100^{\circ} \mathrm{C}\) and \(60.0^{\circ} \mathrm{C}\). What is the rate at which energy is (a) taken in by the engine as heat and (b) exhausted by the engine as heat?

Suppose \(1.0 \mathrm{~mol}\) of a monatomic ideal gas initially at \(10 \mathrm{~L}\) and \(300 \mathrm{~K}\) is heated at constant volume to \(600 \mathrm{~K},\) allowed to expand isothermally to its initial pressure, and finally compressed at constant pressure to its original volume, pressure, and temperature. During the cycle, what are (a) the net energy entering the system (the gas) as heat and (b) the net work done by the gas? (c) What is the efficiency of the cycle?

How much energy must be transferred as heat for a reversible isothermal expansion of an ideal gas at \(132^{\circ} \mathrm{C}\) if the entropy of the gas increases by \(46.0 \mathrm{~J} / \mathrm{K} ?\)

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