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Van der Waals gasby Outrageous
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#1
Nov1212, 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. 


#2
Nov1212, 05:45 PM

PF Gold
P: 963

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. 


#3
Nov1212, 06:56 PM

Mentor
P: 5,402




#4
Nov1212, 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)/(vb) 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 


#5
Nov1212, 08:55 PM

Mentor
P: 5,402




#6
Nov1212, 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 vb. 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? 


#7
Nov1312, 02:40 AM

PF Gold
P: 963

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. 


#8
Nov1312, 04:32 AM

P: 5,462

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 


#9
Nov1312, 06:27 AM

PF Gold
P: 963

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. 


#10
Nov1612, 12:11 AM

P: 375

Got the answer for my question already .
Thanks 


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