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.
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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.
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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.
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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: \[ \Delta H = \Delta E + P \Delta V \]
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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.
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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.