Textbook Question
(a) Does the entropy of the surroundings increase for spontaneous processes?
Ch.19 - Chemical Thermodynamics
Chapter 19, Problem 28
(a) What sign for Δ𝑆 do you expect when the pressure on 0.600 mol of an ideal gas at 350 K is increased isothermally from an initial pressure of 0.750 atm?
(b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process.
(c) Which of the following statements about this process are true? (i) The entropy change you calculated will be the same for at any other constant temperature. (ii) The value of Δ𝑆 you calculated is valid only if the compression is done irreversibly. (iii) If the number of moles of gas being compressed were decreased by a factor of three, the entropy change would increase by a factor of three.
Verified step by step guidance1
(a) Sign of ΔS: When the pressure on a gas is increased isothermally, the volume decreases. Since entropy (ΔS) is related to the disorder or randomness of a system, a decrease in volume at constant temperature typically leads to a decrease in entropy. Therefore, we expect ΔS to be negative.
(b) Calculating ΔS: Use the formula for entropy change for an ideal gas under isothermal conditions: \( \Delta S = nR \ln \left( \frac{P_1}{P_2} \right) \), where \( n \) is the number of moles, \( R \) is the ideal gas constant (8.314 J/mol·K), \( P_1 \) is the initial pressure, and \( P_2 \) is the final pressure. Substitute the given values: \( n = 0.600 \) mol, \( P_1 = 0.750 \) atm, \( P_2 = 1.20 \) atm.
(c) Evaluating statements: (i) True, because entropy change for an isothermal process depends only on the initial and final states, not on the path or specific temperature, as long as the temperature is constant. (ii) False, the calculated ΔS is valid for a reversible process; for an irreversible process, the entropy change would be different. (iii) False, if the number of moles is decreased by a factor of three, the entropy change would decrease by a factor of three, not increase.
Additional Consideration: Remember that the natural logarithm function \( \ln \) will yield a negative value when the argument is less than 1, which aligns with our expectation of a negative ΔS for compression.
Final Thought: Always check the units and ensure consistency, especially when using the ideal gas constant \( R \). Here, we used \( R = 8.314 \) J/mol·K, which is appropriate for calculations involving entropy in joules.>
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Entropy (ΔS)
Entropy is a measure of the disorder or randomness in a system. In thermodynamics, a positive change in entropy (ΔS > 0) indicates an increase in disorder, while a negative change (ΔS < 0) suggests a decrease in disorder. For an ideal gas, changes in entropy can be influenced by factors such as temperature, pressure, and volume, particularly during processes like expansion or compression.
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Entropy in Thermodynamics
Ideal Gas Law
The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of an ideal gas through the equation PV = nRT. This law is essential for understanding how changes in pressure and volume affect the behavior of gases. In the context of the question, it helps predict how the entropy of the gas changes when the pressure is altered while maintaining a constant temperature.
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01:15Ideal Gas Law Formula
Isothermal Process
An isothermal process occurs at a constant temperature, meaning that any heat added to or removed from the system does not change its temperature. For an ideal gas undergoing isothermal compression or expansion, the internal energy remains constant, and the work done on or by the gas directly affects its entropy. Understanding this concept is crucial for calculating the entropy change during the given process.
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Related Practice
Textbook Question
(b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of ΔSsurr?
Textbook Question
(c) During a certain reversible process, the surroundings undergo an entropy change, ΔSsurr = -78 J/K. What is the entropy change of the system for this process?
Textbook Question
For the isothermal expansion of a gas into a vacuum, ΔE = 0, q = 0, and w = 0. (b) Explain why no work is done by the system during this process.
Textbook Question
(a) What is the difference between a state and a microstate of a system?
Textbook Question
(b) As a system goes from state A to state B, its entropy decreases. What can you say about the number of microstates corresponding to each state?