Physical significance of vdP (Thermodynamics)

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The term vdP in thermodynamics represents the work associated with displacing a system's volume V against a pressure difference dP. It highlights the distinction between enthalpy (dH) and heat transfer (dQ), where at constant pressure, vdP equals zero, leading to dH equaling dQ. The discussion emphasizes the confusion arising from not differentiating between internal and external pressures in reversible thermodynamics. Additionally, during adiabatic free expansion, enthalpy remains unchanged despite the absence of heat transfer, as indicated by the equation dH = dQ + vdP. Understanding the physical significance of vdP is crucial for analyzing energy changes and their impact on properties like entropy.
Shivanand
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The difference between Enthalpy and Heat transfer is the term vdP as
dH-dQ=vdP
What is the physical significance of the term vdP? :rolleyes:
Wherever an article describes the difference between enthalpy and heat transfer, it is stated that heat transfer and enthalpy are same at constant pressure as vdP=0
Why does everyone avoid dealing with the term? :confused:

And during adiabatic free expansion of a gas, what happens to the enthalpy? Heat transfer is absent, so dH = [STRIKE]dQ[/STRIKE] + vdP
Is the term related to the entropy change in this process? :approve:
 
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The problem with the Vdp term as related to enthalpy, arrises from a lack of distinction in (reversible) thermodynamics between the internal and external pressure.
The attachment in https://www.physicsforums.com/showthread.php?t=88987&referrerid=219693 , (post#7) might help you out. The Vdp term stands for work needed or obtained from displacing a system with volume V over a pressure-difference dp, and fits with the formulation of the first law as dH=Q+We, where We is the usefull work done on the system.
 
Shivanand said:
The difference between Enthalpy and Heat transfer is the term vdP as
dH-dQ=vdP
What is the physical significance of the term vdP? :rolleyes:
Wherever an article describes the difference between enthalpy and heat transfer, it is stated that heat transfer and enthalpy are same at constant pressure as vdP=0

VdP is the contribution of the energy of a system with volume V whne the pressure changes by dP. If you keep the pressure constant then dP=0 and dH=dQ.
 
Zeppos10 said:
The problem with the Vdp term as related to enthalpy, arrises from a lack of distinction in (reversible) thermodynamics between the internal and external pressure.
The attachment in https://www.physicsforums.com/showthr...errerid=219693 , (post#7) might help you out. The Vdp term stands for work needed or obtained from displacing a system with volume V over a pressure-difference dp, and fits with the formulation of the first law as dH=Q+We, where We is the usefull work done on the system
I will check your link and reply soon. Thanks for your help :)
 
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A. Neumaier said:
VdP is the contribution of the energy of a system with volume V whne the pressure changes by dP. If you keep the pressure constant then dP=0 and dH=dQ.

How does the energy manifest itself and how would it be measurable? It is known that we can measure internal energy (dU) and also work (pdV). What are the possible ways in which the term Vdp affects the properties (like entropy) ?
 
Shivanand said:
How does the energy manifest itself and how would it be measurable? It is known that we can measure internal energy (dU) and also work (pdV). What are the possible ways in which the term Vdp affects the properties (like entropy) ?
One meusures energy differences, using the first law, keeping some of the terms constant, observing the shift in the others, and converting it to energy changes.
Similarly for entropy changes.
 
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