difference between one mole of a gas and van der waal equation
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pressure, as follows;
At fairly high pressures
a/V^2 may be neglected in comparison with P.
The Vander Waals equation becomes
P (V – b) = RT
PV – Pb = RT
PV = RT + Pb or PV > RT
Therefore: at high pressures, PV > RT. This explains the raising parts of the isotherms, at high pressures, plotted between Z vs P.
Note: By dividing with RT on each side, the above equation can be written as:

where z= compressibility factor (Z= PV/nRT)(where n=1)
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OTHER ANSWERS

Mohit Guru, studied at Hindu Vidya Peeth, Sonipat (2013)
Answered Nov 28, 2018 · Author has 60answers and 9.7k answer views
At higher pressure value of wanderwals constant a becomes zero as due to greater pressure the molecules come closer to each other and hence the repulsive force dominates
So as a=0
(P+an^2 \ V^2 )(v-nb) =nRT
P (v-nb) = nRT
PV - Pnb =nRT
Dividing throughout by RT
Z-pnb/RT = 1
Z= 1+pnb/RT
As n=1
Z=1 +Pb/RT
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Priyam Study Centre, M.Sc. Chemistry & Inorganic Chemistry, Guru Ghasidas Vishwavidyalaya
Answered Nov 16, 2018 · Author has 210answers and 66.9k answer views
Let us consider two hypothetical cases to show the size effect and attraction effect on the pressure of the gas:
(a)For the real gas a=0 (that is no inter molecular attraction ) but b≠0 (size considers):
We have Van der Waals Equation,
(P + an²/V² )(V - nb) = nRT but a=0
Hence, P = nRT/(V-nb) ) 〉Pi
Since, Pi = nRT/V only.
It means that the molecular size effect (repulsive interaction) creates higher pressure then that observed by the ideal gas where molecules have no volume.
(a)For the real gas a≠0 (existence of attractive force) but b=0 (no size consider):
We have Van der Waals Equation,
(P + an²/V² )(V - nb) = nRT but b=0(no size).
Hence, P = nRT/Vーan²/V²ㄑPi
Since, Pi=nRT/V only.
Thus, inter molecular attraction effect reduces the pressure of the real gases.