Question

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WORK-OUT PROBLEM # 1 : Use Van der Waals equation to calculate pressure when 6.0374 g of Htg) are confined to the volume of 1.00 L at 298 K. Is the gas in repulse or attractive region of the molecule-molecule interaction? bar-dm me excluded volume factor-0.0265 mol dm , atomic mass of hydrogen -1.0079) (attractive factor-0.24523 mol ess v-n b)(Pt Tru Tr (nk)=(P V-nb Tu (nR) an V-nb v 2 C.O3149 5.901nl 0079 (J.48k) (5, 91 ml8.314-10 )-(11) IL)-(5.99m)o.o P LILP +2) 148.407 8,7978 efat ut gwther defuies nirming Conpchton etror 0.S41A6S 7A8r P=176.M69 bar P= 167.611 bur (4 4

Need help finding the correct answer as well as stating whether the gas is in the repulsive or attractive region of the molecule-molecule interaction.

Answer 1

Data given. gram of H₂ = 6.0374g. V= 1.00L = 1.odm3 K= 290k a = 0.24 52 bar amb mola R = 0.00314 bar dm3 b= 0.0265 dm? mal mass of Hydrogen = 1.0079 k.mod Solution Vander waad's equation (p+ me ) (V-nb) = nRT divided by (v.nb) in each sides P + human =

Putting the value. na 6.0374 g = 2.995 mol 241.0079 g/mol (2.995 mareul X 0.08314 bar dm? x2 98 módok 1dm3 – 2.995 mol x 0.0265 dm? - (2.995m) 40 245 2 bardme 2 (12m3) 2 Mol2 mol P = 174.203 bardm3 1 dm3 – 0.0794 dm3/ 0.97 mol? 40.2 452 berr ams mo? I amb P = 74.203 bar am – 2. 1994 bar. 0.9206 ams PF 90.603 bar – 2.1994 bar (P = 78.4036 bar