Question

Derive relationship between $H\,and\,U$ .

Answer 1

Hint: In order to answer this question, to derive the relation between $H\,and\,U$ , first of all we should know that what is the exact term is instead of $H\,and\,U$ . Then we should go through their chemical properties, and just compare the similarities to relate them.

Complete answer:
In the given question, we have to derive the relation between enthalpy \[(H)\] and internal energy $(U)$ .
Let ${H_1}$ be the enthalpy of the system in the initial state and ${H_f}$ be the enthalpy of the system in a final state. Let ${U_i}\,and\,{V_i}$ be the internal energy and volume in the initial state and ${U_f}\,and\,{V_f}$ be the internal energy and volume in a final state.
Now, as we know that,
$H = U + PV$
Therefore,
${H_i} = {U_i} + P{V_i}$ …..(i)
${H_f} = {U_f} + P{V_f}$ …..(ii)
Subtracting equation (i) from (ii), we have:
$\begin{gathered}
  {H_f} - {H_i} = ({U_f} + P{V_f}) - ({U_i} + P{V_f}) \\
   \Rightarrow {H_f} - {H_i} = ({U_f} - {U_i}) + P({V_f} - {V_i}) \\
   \Rightarrow \Delta H = \Delta U + P\Delta V\,\,\,......(iii) \\
\end{gathered} $
Here, $\Delta U\,and\,P\Delta V$ are the change in internal energy and work energy respectively.
Hence, equation (iii) is the relationship between $H\,and\,U$ .

Note:
Heat effects measured at constant pressure indicate changes in enthalpy of a system and not in changes of internal energy of the system. Using calorimeters operating at constant pressure, the enthalpy change of a process can be measured directly.