May 2011 - Career Point

⎛ n ⎞ ⎛ n ⎞Correction term ∝ ⎜ ⎟ ⎜ ⎟⎝ V ⎠ ⎝ V ⎠2nor Correction term = a...( 3)2VWhere a is the proportionality constant and is ameasure of the forces of attraction between themolecules. Thus2np i = p + a...(4)2VThe unit of the term an 2 /V 2 will be the same as that ofthe pressure. Thus, the SI unit of a will be Pa m 6 mol –2 .It may be conveniently expressed in kPa dm 6 mol –2 .When the expressions as given by Eqs (1) and (4) aresubstituted in the ideal gas equation p i V i = nRT, weget⎛2⎞⎜n ap + ⎟ (V – nb) = nRT ...(5)2⎝ V ⎠This equation is applicable to real gases and is knownas the van der Waals equation.Values of van der Waals Constants :The constants a and b in van der Waals equation arecalled van der Waals constants and their valuesdepend upon the nature of the gas. TheyVan Der Waals ConstantsaGas 6 2kPa dm mol –H 2HeN 2O 2Cl 2NONO 2H 2 OCH 4C 2 H 6C 3 H 8C 4 H 10 (n)C 4 H 10 (iso)C 5 H 12 (n)COCO 221.7643.457140.842137.802657.903135.776535.401553.639228.285556.173877.8801466.1731304.0531926.188150.468363.959bdm3 mol –10.026 610.023 700.039 130.031 830.056 220.027 890.044 240.030 490.042 780.063 800.084 450.122 60.114 20.146 00.039 850.042 67are characteristics of the gas. The values of theseconstants are determined by the critical constants ofthe gas. Actually, the so-called constant vary to someextent with temperature and this shows that the vander Waals equation is not a complete solution of thebehaviour of real gases.Applicability of the Van Der Waals Equation :Since the van der Waals equation is applicable to realgases, it is worth considering how far this equationcan explain the experimental behaviours of realgases. The van der Waals equation for 1 mole of agas is⎛ ⎞⎜ap + ⎟ (V 2m – b) = RT ..(i)⎝ V m ⎠At low pressure When pressure is low, the volume issufficiently large and b can be ignored in comparisonto V m in Eq. (i). Thus, we have⎛ ⎞⎜a⎟ap + V 2m = RT or pV m + =RT⎝ V m ⎠V maor Z = 1 –...(ii)VmRTFrom the above equation it is clear that in the lowpressure region, Z is less than 1. On increasing thepressure in this region, the value of the term(a/V m RT) increase as V is inversely proportional to p.Consequently, Z decreases with increase of p.At high pressure When p is large , V m will be smalland one cannot ignore b in comparison to V m .2However, the term a / V m may be considerednegligible in comparison to p in Eq. (i) Thus,pbp(V m – b) = RT or Z = 1 + ...(iiii)RTHere Z is greater than 1 and increases linearly withpressure. This explains the nature of the graph in thehigh pressure region.A high temperature and low pressure Iftemperature is high, V m will also be sufficiently large2and thus the term a / V m will be negligibly small. Atthis stage, b may also be negligible in comparison toV m . Under these conditions, Eq. (i) reduces to anideal gas equation of state:pV m = RTHydrogen and helium The value of a is extremelysmall for these gases as they are difficult to liquefy.Thus, we have the equation of state as p(V m – b) = RT,obtained from the van der Waals equation by2ignoring the term a / V m . Hence, Z is always greaterthan 1 and it increases with increase of p.The van dar Waals equation is a distinctimprovement over the ideal gas law in that it givesqualitative reasons for the deviations from idealbehaviour. However, the generality of the equation islost as it contains two constants, the values of whichdepend upon the nature of the gas.XtraEdge for IIT-JEE 32 MAY 2011