Chapter 4: Problem 73
When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its temperature, the change in entropy is (a) \(C_{\mathrm{V}} \ln 2\) (b) \(C_{\mathrm{P}} \ln 2\) (c) \(R \ln 2\) (d) \(\left(C_{\mathrm{V}}-R\right) \ln 2\)
Short Answer
Expert verified
\((C_{\mathrm{V}} - R) \ln 2\)
Step by step solution
01
- State the entropy change formula for an ideal gas
The entropy change \((\Delta S)\) for an ideal gas when volume and temperature change can be calculated using the formula: \[\Delta S = nC_{\mathrm{V}} \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{V_2}{V_1}\right)\] where \((T_1)\) and \((T_2)\) are the initial and final temperatures, \((V_1)\) and \((V_2)\) are the initial and final volumes, \((C_{\mathrm{V}})\) is the molar heat capacity at constant volume, \((R)\) is the gas constant, and \((n)\) is the number of moles of the gas.
02
- Apply the given change in volume and temperature
According to the given conditions, the volume is compressed to half its initial volume \((V_2 = \frac{1}{2}V_1)\) and the temperature is increased to twice its initial temperature \((T_2 = 2T_1)\). Substituting these values into the entropy change formula gives us: \[\Delta S = nC_{\mathrm{V}} \ln\left(\frac{2T_1}{T_1}\right) + nR \ln\left(\frac{\frac{1}{2}V_1}{V_1}\right)\]
03
- Simplify the entropy change expression
Simplify the expression using properties of logarithms: \[\Delta S = nC_{\mathrm{V}} \ln(2) + nR \ln\left(\frac{1}{2}\right)\] Note that \(\ln\left(\frac{1}{2}\right) = \ln(2^{-1}) = - \ln(2)\). Substituting this into the equation, we get: \[\Delta S = nC_{\mathrm{V}} \ln(2) - nR \ln(2)\]
04
- Combine like terms
When combining terms with the same logarithmic part, the expression becomes: \[\Delta S = n(C_{\mathrm{V}} - R) \ln(2)\] Since it's given that the change involves one mole of gas (\(n = 1\)), we can remove the \((n)\) term: \[\Delta S = (C_{\mathrm{V}} - R) \ln(2)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Heat Capacity
Molar heat capacity, denoted as either \(C_P\) for constant pressure or \(C_V\) for constant volume, is a physical property that represents the amount of heat required to change the temperature of one mole of a substance by one degree Celsius (or Kelvin). For an ideal gas, the molar heat capacity under constant volume \(C_V\) is particularly important as it indicates how much energy is needed to raise the temperature of the gas when the volume is held fixed.
Understanding \(C_V\) can also provide insights into the degrees of freedom of gas particles—that is, the number of ways in which they can store energy. Typically, for a monoatomic ideal gas, \(C_V\) is lower than for a diatomic gas because monoatomic gases have fewer degrees of freedom. However, for the purpose of the entropy change calculation in our exercise, we only need to know that entropy changes with temperature are directly proportional to \(C_V\), which leads us to see that as temperature changes, so does the entropy.
Understanding \(C_V\) can also provide insights into the degrees of freedom of gas particles—that is, the number of ways in which they can store energy. Typically, for a monoatomic ideal gas, \(C_V\) is lower than for a diatomic gas because monoatomic gases have fewer degrees of freedom. However, for the purpose of the entropy change calculation in our exercise, we only need to know that entropy changes with temperature are directly proportional to \(C_V\), which leads us to see that as temperature changes, so does the entropy.
Gas Constant
The gas constant, symbolized by \(R\), is a fundamental parameter in the field of thermodynamics. It relates the energy scale to the temperature scale and appears in the ideal gas law \(PV = nRT\), where \(P\) stands for pressure, \(V\) for volume, \(n\) for number of moles, and \(T\) for temperature. It plays a role in various equations and is the same for all ideal gases, being defined as 8.314462618 J/(mol\(\cdot\)K).
In the context of entropy change, the gas constant is used when the volume of the gas is changing. It characterizes the relationship between the shift in entropy and the shift in volume of the gas. For instance, when a gas expands, there is an increase in entropy, while a compression results in a decrease in entropy. Therefore, in the given exercise, the gas constant \(R\) is crucial to calculate the part of entropy change that arises from the volume change.
In the context of entropy change, the gas constant is used when the volume of the gas is changing. It characterizes the relationship between the shift in entropy and the shift in volume of the gas. For instance, when a gas expands, there is an increase in entropy, while a compression results in a decrease in entropy. Therefore, in the given exercise, the gas constant \(R\) is crucial to calculate the part of entropy change that arises from the volume change.
Volume and Temperature Change
In thermodynamics, when dealing with an ideal gas, changes in volume and temperature are key facilitators of entropy change. Temperature measures the average kinetic energy of the particles, while volume relates to the space the gas occupies. As an ideal gas expands, the same number of particles occupy a larger space, which corresponds to an increase in the disorder or randomness, and thus, an increase in entropy.
Similarly, when the temperature of a gas increases, the kinetic energy of the particles goes up, and the particles tend to disperse more, resulting in a higher entropy. The exercise presents both a change in volume (compression to half the original volume) and temperature (increase to twice the original temperature). These simultaneous changes must be accounted for to correctly determine the total entropy change. The interplay of such changes underpins many processes in natural and industrial settings, emphasizing the importance of understanding the principles governing the behavior of ideal gases.
Similarly, when the temperature of a gas increases, the kinetic energy of the particles goes up, and the particles tend to disperse more, resulting in a higher entropy. The exercise presents both a change in volume (compression to half the original volume) and temperature (increase to twice the original temperature). These simultaneous changes must be accounted for to correctly determine the total entropy change. The interplay of such changes underpins many processes in natural and industrial settings, emphasizing the importance of understanding the principles governing the behavior of ideal gases.
Properties of Logarithms
The properties of logarithms are essential tools for algebraic manipulation in many areas of mathematics, including calculus and thermodynamics. A logarithm, at its core, is an exponent that indicates the power to which a base must be raised to produce a given number. Mathematically, this relationship is expressed as \( b^y = x \Rightarrow y = \log_b(x) \).
Some useful properties that help simplify complex expressions include the logarithm of a product \( \log_b(xy) = \log_b(x) + \log_b(y) \), the logarithm of a quotient \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \), and the logarithm of a power \( \log_b(x^y) = y \cdot \log_b(x) \). These properties are brilliantly showcased in the solution to the exercise, where they are used to simplify the entropy change equation, ultimately reducing it to a form \( (C_V - R) \ln(2) \), rendering the calculation more manageable and cognitively digestible.
Some useful properties that help simplify complex expressions include the logarithm of a product \( \log_b(xy) = \log_b(x) + \log_b(y) \), the logarithm of a quotient \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \), and the logarithm of a power \( \log_b(x^y) = y \cdot \log_b(x) \). These properties are brilliantly showcased in the solution to the exercise, where they are used to simplify the entropy change equation, ultimately reducing it to a form \( (C_V - R) \ln(2) \), rendering the calculation more manageable and cognitively digestible.