Question

Consider a process carried out at constant pressure, where the only work allowed is pressure, volume (W =POV) under these conditions the expression: ∆E =QP +W , ∆E = QP + POV, QP = ∆E +POV. Arrange the equation at constant P and V with the enthalpy change

Answer 1

In a process occurring at constant pressure, we can identify some important relationships involving enthalpy, heat, work, and the change in internal energy (∆E). Let's go through the relationships step by step and focus on how they connect to enthalpy.

1. **Work Done at Constant Pressure:**
The work done by the system at constant pressure is given by:
\[
W = P \Delta V
\]
Here, \(W\) is the work, \(P\) is the pressure, and \(\Delta V\) is the change in volume.

2. **First Law of Thermodynamics:**
The first law of thermodynamics can be expressed as:
\[
\Delta E = Q + W
\]
where \(\Delta E\) is the change in internal energy, \(Q\) is the heat added to the system, and \(W\) is the work done by the system.

3. **At Constant Pressure:**
When dealing with processes at constant pressure, the heat added to the system is denoted as the change in enthalpy:
\[
Q_P = \Delta H
\]
Thus, we can rewrite the first law at constant pressure as:
\[
\Delta E = \Delta H - P \Delta V
\]
Rearranging this gives:
\[
\Delta H = \Delta E + P \Delta V
\]

4. **Additions to the Equation:**
Now, you want to express everything in terms of enthalpy change:
- From \(Q_P = \Delta H\), replace \(Q_P\) in the first law reformulated2 to highlight the enthalpy.

5. **Final Expression:**
At constant pressure and including the enthalpy change, we summarize as:
\[
\Delta H = \Delta E + P \Delta V
\]
or alternatively:
\[
\Delta E = \Delta H - P \Delta V
\]

In conclusion, at constant pressure, the enthalpy change (\(\Delta H\)) is related to the change in internal energy (\(\Delta E\)) and the pressure-volume work done by the system. The expression highlights the connection between enthalpy, internal energy, and work in processes carried out at constant pressure.