Calculating Total Entropy Change for a Three-Step Process

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SUMMARY

The discussion focuses on calculating the total entropy change for a three-step process involving an ideal gas with a molar heat capacity (CV,m) of 7/2R at an initial temperature of 300K. The first step is an isothermal expansion where 10.37 kJ of heat is absorbed, resulting in an entropy change (dS1) of 34.6 J/K. The second step is a reversible adiabatic expansion with no heat exchange (dS2 = 0). The final step, which is neither adiabatic nor isothermal, leads to an unknown heat exchange but results in an entropy change (dS3) of -34.6 J/K, confirming the cyclic nature of the process.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically entropy and heat transfer.
  • Familiarity with ideal gas behavior and the concept of molar heat capacity.
  • Knowledge of reversible and adiabatic processes in thermodynamics.
  • Ability to perform calculations involving heat and entropy changes.
NEXT STEPS
  • Study the principles of thermodynamics, focusing on entropy calculations in cyclic processes.
  • Learn about the implications of adiabatic processes on entropy and heat exchange.
  • Explore the concept of reversible processes and their impact on thermodynamic systems.
  • Investigate the relationship between heat capacity and entropy for different types of gases.
USEFUL FOR

This discussion is beneficial for students and professionals in thermodynamics, particularly those studying heat transfer, entropy calculations, and ideal gas behavior in engineering and physical sciences.

StonieJ
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Question:
Three moles of an ideal gas (CV,m = 7/2R) initially at 300K are taken through a series of three compression/expansion steps:

The gas is expanded isothermally and reversibly at 300K. This process involves 10.37 kJ of heat going into the system.
The gas is then expanded reversibly and adiabatically to a new volume and temperature. This process involves no heat.
The gas is then compressed in a process that is neither adiabatic nor isothermal until it is back to the original state where it started. This process involves an unknown amount of heat, but we do know that for this process ?Ssurroundings = 40 J/K.
What is dStotal for process 3?



I tried solving this by simply saying that the sum of dS for steps 1, 2, and 3 were zero, since the process is cyclic. Therefore:

dS1 + dS2 = -dS3

We are told the reversible heat for step 1, so we can find it by:

dS1 = qrev/T = 10.37 kJ / 300 K = 34.6 J/K

The second step is adiabatic, so q = 0, and therefore dS2 = 0. This results in:

dS3 = -dS1 = -34.6 J/K.

My biggest concern is that I'm not using enough information provided by the question.
 
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