# The compression factor (compressibility factor) for one mole of a van der Waals gas at `0^(@)C` and 100 atm pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the van der Waals constant a.

32.8 k+

1.6 k+

Answer

Text Solution

Z=(PV)/(nRT), i.e., 0.5=(100xxV)/(1xx0.082xx273)" or "V=0.1119" L"

(P+(a)/(V^(2)))(V-b)=RT

for 1 mol

Neglecting b,

(P+(PV)/(V^(2)))V=RT " or "PV+(a)/(V)=RT" or " (PV)/(RT)+(a)/(VRT)=1

or

" " a=(1-(PV)/(RT))VRT=(1-0.5)0.1119xx0.082xx273=1.252" atm "L^(2)mol^(-2)

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