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(a) Does the entropy of the surroundings increase for spontancous processes? (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of \(\Delta S_{\text {gar }}\) ? (c) During a certain reversible process, the surroundings undergo an entropy change, \(\Delta S_{\text {sury }}=-78 \mathrm{~J} / \mathrm{K}\). What is the entropy change of the system for this process?

Short Answer

Expert verified
(a) For spontaneous processes, the entropy change of the surroundings (ΔS_sur) may increase or decrease, depending on the entropy change of the system, ΔS_sys. (b) Since the given process is spontaneous and the entropy of the system decreases (ΔS_sys < 0), ΔS_sur > 0 and |ΔS_sur| > |ΔS_sys|. (c) The entropy change of the system for the reversible process is ΔS_sys = +78 J/K.

Step by step solution

01

(a) Entropy and Spontaneity

According to the second law of thermodynamics, for a spontaneous process, the total entropy change of the universe, which consists of the system and its surroundings, must be greater than zero, ΔS_uni > 0. Therefore, for a spontaneous process, the entropy change of the surroundings, ΔS_sur, may increase or decrease, depending on the entropy change of the system, ΔS_sys.
02

(b) Sign and Magnitude of ΔS_gar

Since the given process is spontaneous and the entropy of the system decreases (ΔS_sys < 0), the entropy of the surroundings must increase (ΔS_sur > 0) to make ΔS_uni > 0. Moreover, ΔS_sur must be larger in magnitude compared to the decrease in entropy of the system (|ΔS_sur| > |ΔS_sys|) to maintain a positive value for ΔS_uni.
03

(c) Entropy Change of the System in a Reversible Process

In a reversible process, the entropy change of the universe is equal to zero (ΔS_uni = 0) because the process can proceed in both directions. Since we are given the entropy change of the surroundings as ΔS_sur = -78 J/K, the entropy change of the system must be of equal magnitude and opposite in sign to make the total entropy change equal zero. Thus, the entropy change of the system for this process is ΔS_sys = +78 J/K.

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Key Concepts

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

Second Law of Thermodynamics
The second law of thermodynamics is a fundamental principle of physics that governs the direction of processes. It states that the total entropy of an isolated system can never decrease over time, and it always aims to increase, reaching a state of thermodynamic equilibrium. This concept helps us understand why certain processes occur naturally while others do not.
The law essentially tells us that spontaneous processes are those that lead to an increase in the entropy of the universe, which is the sum of the entropies of the system and its surroundings. In simple terms, processes happen naturally when the total disorder or randomness in the universe increases.
  • The formula for this principle is \( \Delta S_{\text{uni}} = \Delta S_{\text{sys}} + \Delta S_{\text{sur}} > 0 \).
  • If the total entropy change of the universe is greater than zero, the process is spontaneous.
It is crucial in predicting whether a chemical reaction or physical process will occur under given conditions.
Spontaneous Processes
Spontaneous processes are those that occur naturally without any external intervention. These processes transform a system from a state of higher energy to lower energy and tend to happen because they increase the total entropy of the universe.
Even if the entropy of the system decreases during a spontaneous process, the entropy of the surroundings might increase enough to ensure that the total entropy (system + surroundings) is still positive. In essence, for a process to be spontaneous, the entropy lost by the system must be less than the entropy gained by the surroundings.
  • This means that \( \Delta S_{\text{uni}} = \Delta S_{\text{sys}} + \Delta S_{\text{sur}} > 0 \) even if \( \Delta S_{\text{sys}} < 0 \).
  • The surroundings need to gain more entropy than the system loses, ensuring that the overall process is natural and feasible.
Hence, in spontaneous processes, the universe moves towards a more disordered state.
Reversible Processes
Reversible processes are a theoretical concept where all changes are carried out infinitely slowly, allowing the system and surroundings to remain in equilibrium at each step. In reality, most processes are irreversible, but understanding reversible processes helps us understand the limits of efficiency and the second law of thermodynamics.
For a reversible process, the total entropy change of the universe is zero, meaning that there is no net change in entropy. Unlike spontaneous processes, a reversible process can go forward or backward without affecting the total entropy of the universe.
  • The entropy change relationship is \( \Delta S_{\text{uni}} = \Delta S_{\text{sys}} + \Delta S_{\text{sur}} = 0 \).
  • For example, if the entropy of the surroundings decreases by a certain amount, the entropy of the system increases by the same amount, and vice versa.
This concept is valuable in determining the maximum work that can be extracted from a system or performed on it.
Entropy Change of the Universe
The entropy change of the universe is key in understanding the progress of thermodynamic processes. It combines the entropy changes of the system and its surroundings and determines whether a process can occur spontaneously.
This concept is described mathematically as \( \Delta S_{\text{uni}} = \Delta S_{\text{sys}} + \Delta S_{\text{sur}} \). For a process to be spontaneous, the entropy change must be positive, reflecting an overall increase in disorder or randomness in the universe.
  • If \( \Delta S_{\text{uni}} > 0 \), the process is spontaneous.
  • If \( \Delta S_{\text{uni}} = 0 \), the process is reversible, meaning it can occur in both directions without any net change.
Therefore, by calculating the entropy change of the universe, one can predict the feasibility of a process and the direction in which it will naturally occur.

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Most popular questions from this chapter

(a) What sign for \(\Delta S\) do you expect when the volume of \(0.200\) mol of an ideal gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from an initial volume of \(10.0 \mathrm{~L}\) ? (b) If the final volume is \(18.5 \mathrm{~L}\) calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.

Indicate whether each statement is true or false. (a) \(\Delta S\) for an isothermal process depends on both the temperature and the amount of heat reversibly transferred. (b) \(\Delta S\) is a state function. (c) The second law of thermodynamics says that the entropy of the system increases for all spontaneous processes.

A certain reaction has \(\Delta H^{\circ}=+23.7 \mathrm{~kJ}\) and \(\Delta 5^{\circ}=+52.4 \mathrm{~J} / \mathrm{K}\). (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Indicate whether cach statement is true or false. (a) The second law of thermodynamics says that entropy is conserved. (b) If the entropy of the system increases during a reversible process, the entropy change of the surroundings must decrease by the same amount. (c) In a certain spontaneous process the system undergoes an entropy change of \(4.2 \mathrm{~J} / \mathrm{K}\); therefore, the entropy change of the surroundings must be \(-4.2 \mathrm{~J} / \mathrm{K}\).

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of \(\mathrm{SO}_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at \(298 \mathrm{~K}\) ? (b) In principle, is this reaction a feasible method of removing \(\mathrm{SO}_{2}\) ? (c) If \(\mathrm{P}_{\mathrm{so}_{2}}=\mathrm{P}_{\mathrm{A}_{2} \mathrm{~s}}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at \(298 \mathrm{~K}\). (d) Would you expect the process to be more or less effective at higher temperatures?

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