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Air is compressed in a piston–cylinder device from 90 kPa and 20°C to 650 kPa in a reversible isothermal process. Determine the entropy change of air. The gas constant of air is R = 0.287 kJ/kg·K. (You must provide an answer before moving on to the next part.)

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luisongonz

Answer:

-0.5674 [tex]\frac{kJ}{kgK}[/tex]

Explanation:

The entropy change of an ideal gas can be calculated by:

[tex]s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})[/tex]

You can review its deduction on van Wylen 6 Edition, section 8.10.

This is an isothermal process, so the entropy change will be calculated as:

[tex]s_{2}-s_{1}=-Rln(\frac{P_{2}}{P_{1}})[/tex]

[tex]s_{2}-s_{1}=-0.287*ln(\frac{650}{90})=-0.5674[/tex][tex]\frac{kJ}{kgK}[/tex]

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