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Show that DeltaH=DeltaU+Deltan((g)),RT...

Show that `DeltaH=DeltaU+Deltan_((g)),RT`

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The energy change taking place at constant pressure and at a constant temperature is called enthalpy change `(triangleH)`.
Mathematically `triangleH=triangleU+P. triangleU` When `triangleU`= Internal energy change.
Enthalpy is state function. Thus, the magnitude of enthalpy change depends only upon the enthalpies in the initial and the final states.
`triangleH=[H_("Products")-H_("reactants")]`
For gaseous reaction, `triangleH=triangleU+trianglenRT`
Consider `PV_(1)=n_(1) RT`
`PV_(2)=n_(2) RT`
`PV_(2)=n_(2) RT`
`PV_(2)-PV_(1)=n_(2) RT-n_(1) RT`
`P(V_(2)-V_(1))=(n_(2)-n_(1)) RT`
`P xx triangleV=triangle n_(g) RT`
`triangleH=triangleU+trianglen_(g) RT`
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