Derive a relationship between ΔH and ΔE for a process in which the temperature of a fixed amount of an ideal gas changes.

Verified step by step guidance

1. Start with the first law of thermodynamics: $\mathrm{\Delta }E=q+w$, where $\mathrm{\Delta }E$ is the change in internal energy, $q$ is the heat exchanged, and $w$ is the work done.
2. For an ideal gas, the work done by the system during expansion or compression at constant pressure is given by $w=-P_{\text{ext}}\mathrm{\Delta }V$, where $P_{\text{ext}}$ is the external pressure and $\mathrm{\Delta }V$ is the change in volume.
3. The enthalpy change $\mathrm{\Delta }H$ is defined as $\mathrm{\Delta }H=\mathrm{\Delta }E+P\mathrm{\Delta }V$.
4. Substitute the expression for work into the first law equation: $\mathrm{\Delta }E=q-P_{\text{ext}}\mathrm{\Delta }V$.
5. Combine the expressions for $\mathrm{\Delta }E$ and $\mathrm{\Delta }H$ to derive the relationship: $\mathrm{\Delta }H=q-P_{\text{ext}}\mathrm{\Delta }V+P\mathrm{\Delta }V=q+(P-P_{\text{ext}})\mathrm{\Delta }V$.