What is the compressibility factor (Z) for 0.02 mole of a van der Waal

(d) (0.1+(1000xx(0.02)^(2))/(V^(2)))V=20xx0.02 =0.1V^(2)-0.4V+0.4=0 =V^(2)-4V+4=0 implies" "V=2L Z=(PV)/(nRT)=(0.1xx2)/(20xx0.02)=0.5

What is the compressibility factor (Z) for 0.02 mole of a van der

Van Der Waals Equation - an overview

SOLVED: The compression factor Z reveals information about

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What is the compressibility factor (Z) for 0.02 mole of a van der Waal

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