plz help.Conceptual doubt.

H(enthaly)=E+PV

Differentiating,we get,H = E + (PV)

     =nC_vdT  + PV + VP

     =nC_vdT + nRT + VP

For isothermal process,dT=0,T=0

hence, H = 0 + 0 + VP  =  VP

BUT FOR ISOTHERMAL PROCESSES, IT IS SAID  U=0,H=0

EXPLAIN!!!!!!!!!!!!!!!

heya!!

delH is d enthalpy chaneg at constant pressure

so delH=delE+P(delV)

and itz not

delH=delE+p(delv)+v(delP)

hope u get it :)

WHAT ABOUT AT CONSTANT TEMPERATURE?

PLZ ELABORATE . CONCEPT IS NOT CLEAR

enthalpy calculated at constant pressure

so u cant take that del(pv)

instead itz P(delV)

since pressure is constant

at constant temp

delH=delE+P(delv)

so

delE=0

itz at constant T,constnt P thus V shud b constant

PV=nRT

so Pdelv=0

It is Zero when enthalpy is calulated for constant pressure

Enthalpy is defined by

where P and V are the pressure and volume, and U is internal energy. Enthalpy is then a precisely measurable state variable, since it is defined in terms of three other precisely definable state variables. It is somewhat parallel to the first law of thermodynamics for a constant pressure system

It is a useful quantity for tracking chemical reactions. If as a result of an exothermic reaction some energy is released to a system, it has to show up in some measurable form in terms of the state variables. An increase in the enthalpy H = U + PV might be associated with an increase in internal energy which could be measured by calorimetry, or with work done by the system, or a combination of the two.

The internal energy U might be thought of as the energy required to create a system in the absence of changes in temperature or volume. But if the process changes the volume, as in a chemical reaction which produces a gaseous product, then work must be done to produce the change in volume. For a constant pressure process the work you must do to produce a volume change ΔV is PΔV. Then the term PV can be interpreted as the work you must do to "create room" for the system if you presume it started at zero volume.