What is the compressibility factor (Z) for 0.02 mole of a van der Waal
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(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
atm. 6. What is the compressibility factor (Z) 0.02 mole of a van der Waals' gas pressure of lam. Assume the size of gas molecules is negligible. Given : RT = 20
⏩SOLVED:What is the compressibility factor (Z) for 0.02 mole of a
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