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entropy of expansion

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plu:
I am trying to calculate the entropy change involved in the expansion of 1 mole of O2 gas at an initial temperature of 120 K and pressure of 4 atm to a final temperature of 90 K and pressure of 1 atm.  The process is stated to be adiabatic, so dSsurr = 0.  I used the following equation to calculate dSsys:

dSsys = nRln(V2/V1) + nCpln(T2/T1)

Cp of O2 is given to be 28.2 J/K.mol .  Then, using the ideal gas law, V1 and V2 can be found to be 2.46 L and 7.39 L respectively.  My final answer is thus 1.03 J/K.mol .  However, the solution that my instructor gives has the answer as 3.41 J/K.mol and uses this equation:

dSsys = nRln(P1/P2) + nCpln(T2/T1)

This suggests that there is an inverse relation between P and V during the expansion, but this cannot be true because T is not constant.  Which solution is correct?

Donaldson Tan:
Since entropy is a state variable, then we can divide the pathway into 2 consercutive path.

The first path involves constant temperature, ie. 120K, 4atm to 120K, 1atm

The second path involves constant pressure, ie. 120K, 1atm to 90K, 1atm.


--- Quote from: plu on May 23, 2006, 11:55:07 AM ---dSsys = nRln(P1/P2) + nCpln(T2/T1)

--- End quote ---

The expression in bold represents the first path.

The underlined expression represents the second path.

Hope this clears things up for you.

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