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Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pressure is 100 kPa and produces 630 kJ of work output. Assume air has constant specific heats evaluated at 300 K.

Determine the entropy change of the air.

Answer :

fahadisahadam

Answer:

The entropy change of the air is [tex]0.240kJ/kgK[/tex]

Explanation:

[tex]T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa[/tex]

[tex]T_{2}[/tex]  is unknown

we can apply the following expression to find [tex]T_{2}[/tex]

[tex]-w_{out} =mc_{v} (T_{2} -T_{1} )[/tex]

[tex]T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }[/tex]

now substitute

[tex]T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K[/tex]

To find entropy change of the air we can apply the ideal gas relationship

Δ[tex]s_{air}[/tex][tex]=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }[/tex]

Δ[tex]s_{air} =[/tex][tex]1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )[/tex]

Δ[tex]s_{air} =0.240kJ/kgK[/tex]

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