Enthelpy change of an isothermal process

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Homework Help Overview

The discussion revolves around the enthalpy change of an isothermal process, particularly focusing on why it is considered to be zero for ideal gases and the implications for real gases.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between enthalpy change, internal energy, and work done during an isothermal process. Questions arise regarding the conditions under which enthalpy change is zero and the definitions of work in this context.

Discussion Status

Several participants have provided insights into the reasoning behind the enthalpy change being zero for ideal gases, while also noting that this may not hold for real gases. There is an ongoing examination of the definitions and assumptions related to pressure, volume, and temperature changes.

Contextual Notes

There is a mention of differing behaviors between ideal and real gases, as well as the implications of using different equations to describe the system's properties.

sachin123
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what is the enthalpy change of an isothermal process?(Why is it 0?)
 
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delta H=delta U+work done on the system(PV work)
For an isthermal expansion of a gas,delta U is 0.Since the gas expands,W is not zero.Then how can delta H be 0?
 
Δ(pv) isn't pv work, as both pressure and volume are changing. its best to change this to nRΔT. Now its easy to see the because ΔT=0, Δ(pv) is also 0.
 
sachin123 said:
what is the enthalpy change of an isothermal process?(Why is it 0?)
It is 0 if you are dealing with an ideal gas but not necessarily for a real gas.

AM
 
delta H= nC(p)ΔT where C(p) is molar heat capacity at constant pressure. Since in isothermal process ΔT is zero. Thr4 enthalpy change is zero and the term 'pv' is not work. They are pressure and volume of system respectively...
 

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