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

Filo Student Questions For CBSE , Grade 9

Answered: Question 2: For the following parts,…

Full article: Efficient Phase Equilibrium Calculation for

its mole fraction. Solution : P=KH​⋅X⇒PCO2( g)​=KH​⋅X(CO2​)​⇒0.01=35​×100..

its mole fraction. Solution : P=KH​⋅X⇒PCO2( g)​=KH​⋅X(CO2​)​⇒0.01=35​×100..

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Compressibility factor Z = PV / nRT is plotted against pressure as shown below:What is the correct order for the liquefiability of the gases shown in the above graph? A. CO 2

For $CO$, isotherm is of the type as shown. Near the point compressibility factor $Z$ is?\n \n \n \n \n 1.$\\left( {1 + \\dfrac{b}{V}} \\right)$ 2.$\\left( {1 - \\dfrac{b}{V}} \\right)$3.$\\left( {1 + \\