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Is change in enthalpy defined only at constant pressure?

I know Q-W=del(U) And Q at constant pressure equals del(H) (enthalpy change).

Can Q (heat given to the system) be used interchangeably with del(H) in the first equation?

Is del(H) =nCp del(T)? How?

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Is change in enthalpy defined only at constant pressure?

No. The definition of $\Delta H$ is $\Delta H = \Delta U + \Delta (PV)$ Simple as that.

I know Q-W=del(U) And Q at constant pressure equals del(H) (enthalpy change).

Can Q (heat given to the system) be used interchangeably with del(H) in the first equation?

No. Only if the applied pressure is held constant during the change.

Is del(H) =nCp del(T)? How?

No. Only for an ideal gas. For real gases, this equation is not correct outside the limit of ideal gas behavior. For an ideal gas, $\Delta U=nCv\Delta T$ and $\Delta (PV)=nR\Delta T$, so $\Delta H=n(C_V+R)\Delta T$. And then $C_p=(\partial H/\partial T)_P=C_v+R$, so $\Delta H=nC_P\Delta T$.

For a real gas, liquid, or solid, $$dH=nC_PdT+\left[V-T\left(\frac{\partial V}{dT}\right)_P\right]dP$$ Note that the term in brackets is zero for an ideal gas.

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Is change in enthalpy defined only at constant pressure?

No, enthalpy-change is not meant only for isobar processes.

Can Q (heat given to the system) be used interchangeably with del(H) in the first equation?

No, not always. Only when the non-compression work done by the system is zero, then only $\delta Q$ be interchanged with $\mathrm dH\;.$ See my answer to this ChemSE post.

Is del(H) =nCp del(T)?

Yes, indeed.

How?

Check my answer to this ChemSE post.

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