Physical significance of vdP (Thermodynamics)

In summary, The term vdP stands for the contribution of energy when a system with volume V experiences a pressure change of dP. At constant pressure, dP is equal to 0 and the enthalpy and heat transfer are the same. The term is often overlooked due to a lack of distinction between internal and external pressure in thermodynamics. It represents work done on the system and can affect properties such as entropy. Energy and entropy changes can be measured by observing shifts in other terms and converting them.
  • #1
Shivanand
9
0
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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  • #2
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.
 
  • #3
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.
 
  • #4
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 [Broken] , (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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  • #5
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) ?
 
  • #6
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.
 

1. What is the physical significance of the van der Waals equation in thermodynamics?

The van der Waals equation is a modification of the ideal gas law that takes into account the attractive and repulsive forces between gas molecules. This is important in thermodynamics because it allows for a more accurate calculation of the behavior of real gases, which can deviate significantly from ideal gas behavior at high pressures and low temperatures.

2. How does the van der Waals equation relate to intermolecular forces?

The van der Waals equation includes two correction terms that account for intermolecular forces: the van der Waals constant "a" represents the attractive forces between molecules, while the constant "b" represents the volume occupied by the molecules themselves. These terms allow for a more realistic description of gas behavior by taking into account the interactions between molecules.

3. What is the physical significance of the "a" and "b" constants in the van der Waals equation?

The "a" constant in the van der Waals equation represents the strength of the attractive forces between gas molecules. The higher the value of "a", the stronger the intermolecular forces and the more likely the gas is to deviate from ideal behavior. The "b" constant represents the volume that each molecule occupies and is used to correct for the volume excluded by the molecules themselves.

4. How does the van der Waals equation account for the non-ideal behavior of real gases?

The van der Waals equation includes two correction terms, "a" and "b", that account for the attractive and repulsive forces between gas molecules. These forces are not accounted for in the ideal gas law, which assumes that gas particles have no volume and do not interact with each other. By including these correction terms, the van der Waals equation can better describe the behavior of real gases at high pressures and low temperatures.

5. What are some limitations of the van der Waals equation?

The van der Waals equation is more accurate than the ideal gas law, but it still has some limitations. It does not account for other factors that can affect gas behavior, such as molecular shape, molecular polarity, and the presence of impurities. Additionally, the constants "a" and "b" are specific to each gas and can vary with temperature and pressure, making the equation less generalizable. Finally, the van der Waals equation does not accurately predict the behavior of gases at extremely high pressures or low temperatures.

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