Enthelpy change of an isothermal process

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The enthalpy change of an isothermal process is zero for an ideal gas because the temperature remains constant, leading to no change in internal energy (ΔU) and thus no change in enthalpy (ΔH). Although work is done on the system during gas expansion, the relationship ΔH = ΔU + work indicates that the internal energy change is zero. For real gases, the enthalpy change may not be zero due to interactions that deviate from ideal behavior. The equation AM delta H = nC(p)ΔT reinforces that with ΔT being zero, the enthalpy change remains zero. Therefore, in ideal conditions, the enthalpy change in an isothermal process is indeed zero.
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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