Question

Using the expressions for reversible (1) and irreversible expansion $W_{rev} = -nRT \ln \left[ \frac{V_{final}}{V_{initial}} \right]$ (1) $W_{irrev} = -p_{ex} \Delta V$ (2) prove the Clausius inequality $dS \ge \frac{dq_{irrev}}{T}$ In an isolated system, where $dq = 0$, what does this require of the entropy change in a spontaneous process?

          Using the expressions for reversible (1) and irreversible expansion
$W_{rev} = -nRT \ln \left[ \frac{V_{final}}{V_{initial}} \right]$ (1)
$W_{irrev} = -p_{ex} \Delta V$ (2)
prove the Clausius inequality $dS \ge \frac{dq_{irrev}}{T}$
In an isolated system, where $dq = 0$, what does this require of the entropy change
in a spontaneous process?

Using the expressions for reversible (1) and irreversible expansion
Wrev = -nRT ln[ (Vfinal)/(Vinitial)] (1)
Wirrev = -pexΔ V (2)
prove the Clausius inequality dS ≥(dqirrev)/(T)
In an isolated system, where dq = 0, what does this require of the entropy change
in a spontaneous process?

Added by Michael S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

04/04/2023

Step 1

Step 1: The Clausius inequality states that the change in entropy of a system is greater than or equal to the heat added to the system divided by the temperature. Show more…

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Using the expressions for reversible (1) and irreversible expansion $$W_{rev} = -nRT \ln \left[ \frac{V_{final}}{V_{initial}} \right]$$ (1) $$W_{irrev} = -p_{ex} \Delta V$$ (2) prove the Clausius inequality $$dS \ge \frac{dq_{irrev}}{T}$$ In an isolated system, where dq = 0, what does this require of the entropy change in a spontaneous process? 5 marks