Why 'Enthalpy change' (Delta H) is equal to 'Heat transfer at constant pressure' (Qp)? [closed]








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Why 'Enthalpy change' (∆H) is equal to 'Heat transfer at constant pressure' (Qp)?


∆H = ∆U + ∆pV, here only expansion work done by the sustem is added. If non expansion work is done on the system then it should be,
∆H ≠ Qp.
What's wrong? Explain.




In,Thermodynamics,Work is defined as   

$W$= $-\int\_{V\_1}^{V\_2}pdV$. So, if $dV$=0, there is no work done on the system,as $V\_1=V\_2$.
  

Coming to the enthalpy,you are right.  

$\Delta$H=$\Delta$U + $\Delta$($PV$).But if you apply the product rule then,  

$\Delta$H=$\Delta$U + $P$$\Delta$V + $V$$\Delta$P.  

At, constanst presurre,$\Delta$P=$0$. so at constant pressure,$\Delta$H=$\Delta$U + $P$$\Delta$V.  
But,according,to first law of Thermodynamics,  
$Q$= $\Delta$U - $W$.At,constant pressure, the $p$ in the integral comes out.So,  
$W$= $-p\int\_{V\_1}^{V\_2}dV$= -$p$($V\_2-V\_1$) = -$p$$\Delta$V.So, first law in constant presuure becomes,  
$Q\_p$= $\Delta$U - (-$p$$\Delta$V)= $\Delta$U+$p$$\Delta$V= $\Delta$H. Thus, it is justified why Enthalpy change equals Heat change at constant pressure.


 2

Enthalpy is a intensive property of the material that has nothing to do with the specific process that the material is subjected to.
It just so happens in a constant pressure condition, involving only P-V work, that the change in enthalpy is equal to the heat added.


If non expansion work is done on the system, then ∆H≠Qp. It should be mentioned while writing '∆H = ∆U + p∆V =Qp', that 'it is applicable when only p-V work is done'.


Thanks to @ChesterMiller


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