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Van der Waals gas

by Outrageous
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Outrageous
#1
Nov12-12, 11:54 AM
P: 375
[P + (n2a/V2)](V - nb) = nRT
What is the P represent? Is the pressure of ideal gas or the pressure we can measure from the real gas?
Consider now it is the pressure of ideal gas but how do we get the value of P from an experiment of real gas?
What if it is the pressure of real gas then why do we need to put (n2a/V2)?

one more thing, Pv=RT,
P= pressure exerted by the ideal gas ?
v= volume of the container / volume occupied by gas?

Thank you.
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Philip Wood
#2
Nov12-12, 05:45 PM
PF Gold
P: 956
p and V in V der W's equation are the measurable quantities: pressure on container walls and volume of container.

Your version of V der W's equation is written for n moles of gas. The corresponding version of the ideal gas equation is pV = nRT. In your version, v is the molar volume, in other words, v = V/n.
Chestermiller
#3
Nov12-12, 06:56 PM
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PF Gold
Chestermiller's Avatar
P: 5,254
Quote Quote by Outrageous View Post
[P + (n2a/V2)](V - nb) = nRT
What is the P represent? Is the pressure of ideal gas or the pressure we can measure from the real gas?
Consider now it is the pressure of ideal gas but how do we get the value of P from an experiment of real gas?
What if it is the pressure of real gas then why do we need to put (n2a/V2)?

one more thing, Pv=RT,
P= pressure exerted by the ideal gas ?
v= volume of the container / volume occupied by gas?

Thank you.
P is supposed to be an approximation to the pressure of a real gas. P can be measured by various methods. Have you ever used a pressure gage to measure your tire pressure? That's one way. You can also use a pressure transducer connected to a digital readout. For a real gas, the van der Waals equation is, in most regions of the parameters, more accurate than the ideal gas law.

Outrageous
#4
Nov12-12, 08:22 PM
P: 375
Van der Waals gas

So P is the pressure exerted by the van der Waals gas.
Then why do we need to do adjustment? I mean the term (a/v^2)?
If P is the pressure of ideal gas then P=(RT)/(v-b) then -(a/v^2) because the pressure of ideal gas is lower than the pressure of van der Waals gas.that is why we minus.
Am I correct? But if I am right then how to measure pressure of ideal gas?
Thank you
Chestermiller
#5
Nov12-12, 08:55 PM
Sci Advisor
HW Helper
Thanks
PF Gold
Chestermiller's Avatar
P: 5,254
Quote Quote by Outrageous View Post
So P is the pressure exerted by the van der Waals gas.
Then why do we need to do adjustment? I mean the term (a/v^2)?
If P is the pressure of ideal gas then P=(RT)/(v-b) then -(a/v^2) because the pressure of ideal gas is lower than the pressure of van der Waals gas.that is why we minus.
Am I correct? But if I am right then how to measure pressure of ideal gas?
Thank you
You realize that the van der Waals equation is just an emperical fit to the P-V-T behavior of a real gas beyond the ideal gas region. In the ideal gas limit, the van der Waals equation approaches the ideal gas equation. Your question about how to measure the pressure of an ideal gas is very puzzling. You measure the pressure of any gas using an appropriate pressure gage. An ideal gas is not a special kind of substance. Every real gas approaches ideal gas behavior and can be regarded as an ideal gas in a certain range of P, V, and T.
Outrageous
#6
Nov12-12, 10:31 PM
P: 375
I understand already. For a system of real gas, I can measure P and v of that system .
Then we have to modify the P and v because it is real gas that deviate from the behaviour of ideal gas. so in order to use the relationship of Pv=RT, the volume have to be v-b. At the same time there is intermolecular forces between gas atoms, so the pressureof real gas we measured(which is lower) has to plus (a/v^2).

Correct?
Philip Wood
#7
Nov13-12, 02:40 AM
PF Gold
P: 956
Yes! I reckon you've got it.

It's usual, incidentally to use a small 'p' for pressure. Small letters are used for 'intensive' quantities (those that remain the same if we partition the system into two halves with a wall and look only at one half), such as pressure (p), density (ρ), molar volume (v) Capital letters are used for 'extensive' quantities which halve when we halve the system, such as volume (V), entropy (S), internal energy (U). Unfortunately there are exceptions, such as temperature (T), which is intensive.
Studiot
#8
Nov13-12, 04:32 AM
P: 5,462
Outrageous
Then we have to modify the P and v because it is real gas that deviate from the behaviour of ideal gas. so in order to use the relationship of Pv=RT, the volume have to be v-b. At the same time there is intermolecular forces between gas atoms, so the pressureof real gas we measured(which is lower) has to plus (a/v^2).
Yes we modify V ( the total volume as measured by the outside world) by realising that the molecules take up space.
To compensate for this we subtract a constant b (or nb for n moles) from the measured volume.

We also need to increase the pressure due to the difference between the attractive forces felt by gas molecules within the body of the gas and those felt by the those at the container boundary.

This is a surface effect, similar to surface tension.
It is not experienced in an unconfined gas or a gas of infinite extent.

So the VDW equation has a sound theoretical basis and it is possible to derive it from simpler mechanical principles.
Yes it is found, empirically, to fit some gas data better than the ideal gas law.
But it is not, of itself empirical.

@Philip Wood
I don't think that there is any concensus as to symbol capatilisation.

I prefer capitals simply because many formulae in thermodynamics have exponents, some complicated.

In fluid dynamics we wish to distinguish between velocity and volume so v and V are pretty common.

Some use θ for temperature to leave T or t free as the symbol for time.

Some quantities are extensive and some are intensive, and your distinction is good, simple and practical.

However some are neither so the use of capitilisation to distinguish is limited.
For example the area A or a of a piston is neither and unaffected by the nature of the quantities on either side.

go well
Philip Wood
#9
Nov13-12, 06:27 AM
PF Gold
P: 956
Area is a good example of a quantity that refuses to be classified as intensive or extensive. One might argue that, from a thermodynamic point of view, it's not usually regarded as a system variable for a fluid, but I'm not going to do so!

I once read that V der W, like the gent he was, gave credit to Laplace, who, some hundred years earlier, had reasoned that for a given fluid there should be a pressure deficit proportional to the square of the density. As you say, the V der W equation is not an empirical equation.
Outrageous
#10
Nov16-12, 12:11 AM
P: 375
Got the answer for my question already .
Thanks


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