Math Problem Statement

For a reversible process, the following equation describes the entropy changes of the universe, system, and surroundings. ΔSuniv = ΔSsys + ΔSsurr = 0 For an irreversible process, the following equation applies. ΔSuniv = ΔSsys + ΔSsurr > 0 For each of the following processes, predict whether ΔSsurr is positive, negative, or zero. (a) Isothermally expanding a gas, ΔSsys > 0:

positive negative zero

(b) Reacting hydrogen and oxygen gas, ΔSsys < 0:

positive negative zero

Solution

For these questions, we need to predict the sign of ΔSsurr\Delta S_{\text{surr}} (the entropy change of the surroundings) based on the entropy change of the system, ΔSsys\Delta S_{\text{sys}}, and whether the process is reversible or irreversible.

  1. Isothermal expansion of a gas (ΔSsys>0\Delta S_{\text{sys}} > 0)

    • When a gas expands isothermally, the entropy of the system increases because the molecules have more space to occupy, leading to increased disorder.
    • Since ΔSsys>0\Delta S_{\text{sys}} > 0, if this process is carried out in a reversible manner, ΔSsurr\Delta S_{\text{surr}} would be negative (to ensure ΔSuniv=0\Delta S_{\text{univ}} = 0).
    • For an irreversible process, ΔSuniv>0\Delta S_{\text{univ}} > 0, so while ΔSsurr\Delta S_{\text{surr} } might still be negative, it would be smaller in magnitude than ΔSsys\Delta S_{\text{sys}} to result in a positive ΔSuniv\Delta S_{\text{univ}}.

    Answer: ΔSsurr\Delta S_{\text{surr}} is likely negative.

  2. Reaction between hydrogen and oxygen gas (ΔSsys<0\Delta S_{\text{sys}} < 0)

    • In this reaction, hydrogen and oxygen combine to form water, which is a more ordered (lower entropy) state than the gaseous reactants, so ΔSsys<0\Delta S_{\text{sys}} < 0.
    • For this exothermic reaction, heat is released to the surroundings, increasing the entropy of the surroundings.
    • Thus, ΔSsurr\Delta S_{\text{surr}} is positive because the surroundings gain entropy as they absorb the released heat.

    Answer: ΔSsurr\Delta S_{\text{surr}} is positive.

Let me know if you'd like further details or have questions.

Related Questions:

  1. How does temperature affect ΔSsurr\Delta S_{\text{surr}} during an exothermic reaction?
  2. What would happen to ΔSuniv\Delta S_{\text{univ}} if both ΔSsys\Delta S_{\text{sys}} and ΔSsurr\Delta S_{\text{surr}} were negative?
  3. In a reversible process, can ΔSsys\Delta S_{\text{sys}} be non-zero while ΔSuniv\Delta S_{\text{univ}} remains zero?
  4. How does the entropy change of a system compare between reversible and irreversible expansions?
  5. Can an endothermic reaction ever have a positive ΔSsurr\Delta S_{\text{surr}}?

Tip: For isothermal processes, remember that any heat transferred impacts both system and surroundings, influencing their entropy changes in opposite directions.

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Entropy
Irreversible and Reversible Processes

Formulas

ΔSuniv = ΔSsys + ΔSsurr
ΔSuniv = 0 for reversible processes
ΔSuniv > 0 for irreversible processes

Theorems

Second Law of Thermodynamics

Suitable Grade Level

Undergraduate level (Chemistry or Physics)