How Do You Calculate ΔH and Determine Volume Changes in Thermodynamics?

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SUMMARY

The discussion centers on calculating the change in enthalpy (ΔH) for a thermodynamic process at constant pressure, given the internal energy change (ΔU) of -225 kJ and expansion work of +52 kJ. The relationship between these variables is clarified using the equation ΔU = q + w, leading to q = -277 kJ. The correct formula for ΔH is ΔH = ΔU + PΔV, resulting in ΔH = -173 kJ. The volume increases due to the positive work done by the system.

PREREQUISITES
  • Understanding of thermodynamic concepts such as internal energy (ΔU) and enthalpy (ΔH).
  • Familiarity with the first law of thermodynamics and the equation ΔU = q + w.
  • Knowledge of pressure-volume work (PΔV) and its implications in thermodynamic processes.
  • Basic algebra skills for manipulating thermodynamic equations.
NEXT STEPS
  • Study the relationship between internal energy and enthalpy in detail.
  • Learn about the implications of work done on or by a system in thermodynamics.
  • Explore examples of constant pressure processes and their calculations.
  • Investigate the significance of heat transfer (q) in thermodynamic systems.
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Students studying thermodynamics, educators teaching thermodynamic principles, and professionals involved in chemical engineering or physical chemistry.

deviljay
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Hi All!
I ran across a homework problem ans am confused by the question.
here it is:
For a certain process at constant pressure, ∆U = − 225 kJ and +52 kJ of expansion work is
done. What is ∆H for this process and does the volume increase or decrease?

I know that ∆U=q+w and therefore q=-277 kJ, also I know that change in enthalpy = ∆U-P∆V.
However I am all confused on the relationship between these variables.

I appreciate any feedback. Thank you for your time.
 
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delta U = delta H - work ( done on the system or by the system )
delta U , internal energy can be gained or lost
delta H , is negative if heat is evolved and positive if heat is absorbed.
work = PdeltaV and is negative if done by the system , as in expansion
and is positive if done on the system.
So the double negative gave you the + 52 kJ
 
deviljay said:
Hi All!
I ran across a homework problem ans am confused by the question.
here it is:
For a certain process at constant pressure, ∆U = − 225 kJ and +52 kJ of expansion work is
done. What is ∆H for this process and does the volume increase or decrease?

I know that ∆U=q+w and therefore q=-277 kJ, also I know that change in enthalpy = ∆U-P∆V.
However I am all confused on the relationship between these variables.

I appreciate any feedback. Thank you for your time.

correction to my previous reply: there is no double negative giving a + 52 kJ
delta U = delta H - work,PV so in the above problem it should be
-225 kJ = -173 kJ - 52 kJ
 
Last edited:

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