Anonymous asked in Science & MathematicsChemistry · 1 month ago

Chemistry Help??!!?

Please help me, I'm on my last attempt! 

Calculate the change in entropy when the pressure of 5.75 g of helium gas is decreased from 320.0 kPa to 40.0 kPa while the temperature decreases from 423 K to 273 K. Assume ideal behavior.

1 Answer

  • 1 month ago

    Since entropy is a state variable, the entropy change may be path-independent so long as the process is quasi-static.

    So let's say you first decrease the pressure by a factor of 8 without changing T, but then reduce the temperature without changing V or p.

    During that first process, the gas will do work and must therefore absorb heat to achieve its expansion (as temperature is not changing).  The work done is the integral of p delta-v

    = RT ln(p1/p2)

    = (8.314 J/molK)(423 K) ln(8)

    Therefore, the gas must absorb 7.313 kJ/mol of heat during this first process, and as this heat is all absorbed at 423K, the entropy change is Q/T = 17.288 J/(mol K).

    But now the gas is cooled at constant pressure to 273 K.  For a monatomic gas (helium), the molar specific heat is (5/2)R

    = 20.785 J/(molK).  To find the entropy change, we integrate 

    dQ/T = (20.785 J/(molK))* dT/T

    = (20.785 J/molK)*ln(273/423) = -9.102 J/molK.

    So the overall entropy change is +8.2 J/molK, arising mainly from the injection of heat needed to achieve the volume increase.

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