Calculating the work done on a decomposed solid

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SUMMARY

The discussion focuses on calculating the work done during the decomposition of 1.0 mol of CaCO3 (s) at 800 degrees Celsius under two conditions: with a piston and open to the atmosphere. The work done at constant external pressure is calculated using the equation w = -P_externalΔV. The participant expresses confusion regarding the volume changes during decomposition and the implications of the pressure conditions, specifically whether the 1.0 atm refers to the pressure exerted by the piston or the pressure during the reaction. Understanding the reaction of carbonate decomposition is essential for solving the problem.

PREREQUISITES
  • Understanding of gas laws and pressure-volume relationships
  • Familiarity with the decomposition reaction of calcium carbonate
  • Knowledge of thermodynamic work calculations
  • Basic principles of chemistry related to states of matter
NEXT STEPS
  • Study the decomposition reaction of calcium carbonate (CaCO3) to understand volume changes
  • Learn about the ideal gas law and its application in work calculations
  • Investigate the concept of constant external pressure in thermodynamics
  • Explore the differences in work calculations between closed and open systems
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Chemistry students, educators, and professionals involved in thermodynamics and reaction kinetics, particularly those studying gas behavior and work calculations in chemical processes.

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Homework Statement



A sample consisting of 1.0 mol CaCO3 (s) was heated to 800 degrees Celsius, when it decomposed. The heating was carried out in a container fitted with a piston that was initially resting on the solid. (1) Calculate the work done during complete decomposition at 1.0atm. (2) What work would be done if insead of having a piston the container was open to the atmosphere?

Homework Equations



For (2), wconstant external pressure= -pexternal\DeltaT

The Attempt at a Solution


I'm just not clear about what's happening in this situation...when the CaCO3 is decomposing, I assume the volume is changing...but is the volume larger than when it was in its solid form, or smaller (due to the presence of the piston). Is the pressure on the piston changing? To what does the 1.0 atm refer to- is that a constant pressure on the piston, or only refer to the pressure at which the decomposition takes place, and then th pressure changes?
Once I understand what's actually happening with regard to this decomposition, I can select the correct equations, so I don't really need anyone to 'solve' this problem for me- I just am having difficulty understanding what's changing, what's staying the same, in what direction the change is occurring, etc.

Thank you very much for your time and effort =)
 
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Start with the reaction of carbonate decomposition, that should give you a hint.
 

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