Calculating Work W: when pdV or Vdp?

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In summary, -An isothermal process is the same as an isentropic process, with the exception of heat addition (or removal)-A reversible isothermal process is the same as a reversible isentropic process, with the exception of heat removal-A reversible adiabatic process is the same as a reversible isentropic process, with the exception of heat addition
  • #1
dries
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If you have to calculate the amount of energy needed to perform a compression of the amount of energy is gained by expanding. Then I don't know when to use the integral
W = int pdV
and when
W = int Vdp

thanks in advance
 
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  • #2
If you compress something does the volume change?
 
  • #3
the process is isentropical (so yes)
 
  • #4
If the volume changes why are you contemplating Vdp?
 
  • #5
using the isentropic equation
pV^k = constant = p1V1^k
you can write V in function of p so you get another integral

int (p1/p)^(1/k)*V1 dp
=
p1^(1/k)*V1 * int p^(-1/k) dp

An analog conversion is possible for the pdV, but i found both as methode to calculate the required energy. Now I wan't to know when i have to use which one

(thanks for the fast responses allready)
 
  • #6
PdV states the volume change at constant pressure.

So you use this for thermodynamic processes carried out at constant pressure eg the free expansion of a gas against the atmosphere, most chemical reactions, carried out in an open beaker.

VdP states the change in pressure at constant volume so you would use this when say inflating a bicycle tyre or gas cylinder, whose volume does not change.
 
  • #7
P or V don't have to be constants as shown in previous formule of isentropic change, they can be dependent of each other. Imagine a compressor, both volume and pressure changes when the air is being compressed.
 
  • #8
P or V don't have to be constants as shown in previous formule of isentropic change, they can be dependent of each other. Imagine a compressor, both volume and pressure changes when the air is being compressed.

Are you telling me or asking me?
 
  • #9
PdV is the reversible work done on a system by changing the volume.

VdP is the change in enthalpy for a process which is both reversible and adiabatic (or isentropic).

More info http://en.wikipedia.org/wiki/Isentropic_process#Isentropic_flow".

So, if your process occurs in a http://en.wikipedia.org/wiki/Thermodynamic_system#Closed_system" (where there is no mass flow in or out), you use PdV.

If the process occurs in an http://en.wikipedia.org/wiki/Thermodynamic_system#Open_system" (where there is mass flow coming in or out), you use vdP.
 
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  • #10
Thanks a loth for the valuable reply!
 
  • #11
Good morning, jack action

I am worried about your statements.

Can we apply them to a real calculation, say the work done by the gas in expanding

100 litres of neon at 0o C and 10 bar (atmosphere) of pressure, expanded to 1 bar (atmosphere) of pressure

1) Via a reversible isothermal process

2) via a reversible adiabatic process

3) via a sudden (nonreversible) adiabatic process

Are the above systems open or closed?

I make the answers

1) 232 850 Joules

2) 91410 Joules

3) 54740 Joules
 
  • #12
I think I'm right, but I'm open to discussion. My experience is based on engine processes.

For example, when you compress the air inside the cylinder with the valve closed, that is a closed system. In this case, the work input needed by the piston will be PdV (which is CvdT for an isentropic process as shown in my previous link).

But if you achieve the same compression with a turbine which has open ends with continuous airflow, that is an open system. In this case, the work input needed by the turbine will be VdP (which is CpdT for an isentropic process as shown in my previous link). In this case, you also need to take http://en.wikipedia.org/wiki/Stagnation_temperature" into account, since the fluid is in motion at the inlet and outlet.

Work required by an open system is then larger than work required by a closed system since Cp > Cv.

The OP did say the process was isentropic but was not more specific. You're talking about an isothermal process, which I would treat as an isentropic process with heat addition (or removal), which is described in my previous links for both closed and open systems.

As for determining if the free expansion of a gas against the atmosphere is an open or closed system, I haven't really thought of it, but I would risk an answer by saying it is a closed system. The system would be the entire atmosphere where no fluid comes in or out.
 
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  • #13
Hello, jack, the OP was not very clear but seemed to want to know when to use PdV and when to use VdP in calculating the work term in thermodynamic processes.

This was particularly muddled

the amount of energy needed to perform a compression of the amount of energy is gained by expanding.


I was trying to lead him towards a sensible understanding and statement before he seemed to start arguing the toss.

However the question, as I have posed it, is a reasonable one that troubles many and deserves an airing.

You're talking about an isothermal process,

Actually I was talking about three different types of process (for good reasons) and put my results where my mouth is.
I will happily display the calculations ( they each only a few lines long) and discuss those reasons, but I wondered if you got the same values?

For example, when you compress the air inside the cylinder with the valve closed, that is a closed system. In this case, the work input needed by the piston will be PdV (which is CvdT for an isentropic process as shown in my previous link).

But if you achieve the same compression with a turbine which has open ends with continuous airflow, that is an open system. In this case, the work input needed by the turbine will be VdP (which is CpdT for an isentropic process as shown in my previous link). In this case, you also need to take stagnation temperature into account, since the fluid is in motion at the inlet and outlet.

One of the keys to getting thermodynamic calculations right is to correctly define the system boundaries.

Another is clearly to use the correct equations.

You are talking about the flow version of the first law or if you like a modified Bernoulli in you turbine.

I look forward to your reply
 
  • #14
Studiot said:
I am worried about your statements.

Studiot said:
I was trying to lead him towards a sensible understanding and statement before he seemed to start arguing the toss.

However the question, as I have posed it, is a reasonable one that troubles many and deserves an airing.

You will have to show your work and explain more as I don't understand what you are trying to prove or what worries you about my statements.
 

FAQ: Calculating Work W: when pdV or Vdp?

What is the formula for calculating work when pressure and volume are changing?

The formula for calculating work when pressure and volume are changing is W = pdV, where W is work, p is pressure, and dV is the change in volume.

How do you calculate work when pressure and volume are changing at a constant rate?

If pressure and volume are changing at a constant rate, the formula for calculating work becomes W = Vdp, where W is work, V is volume, and dp is the change in pressure.

Can you provide an example of calculating work using the formula W = pdV?

Yes, for example, if a gas in a piston is compressed from 10 L to 5 L at a constant pressure of 2 atm, the work done would be W = (2 atm)(5 L - 10 L) = -10 L*atm. The negative sign indicates that work is being done on the system.

Is it possible for work to be negative when calculating using the formula W = Vdp?

Yes, it is possible for work to be negative when using the formula W = Vdp. This would occur if the system is doing work on its surroundings, rather than the surroundings doing work on the system.

How does the sign convention for work differ between the two formulas W = pdV and W = Vdp?

The sign convention for work differs between the two formulas. In W = pdV, a negative sign indicates work is being done on the system, whereas in W = Vdp, a negative sign indicates work is being done by the system. This is because the volume and pressure are changing in opposite directions in these two formulas.

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