Calculate the pressure exerted by ar for a molar volume of 0.590 l⋅mol−1 at 295 k using the van der waals equation of state. the van der waals parameters a and b for ar are 1.355 bar⋅dm^6⋅mol^(−2) and 0.0320 dm^3⋅mol^(−1), respectively.

the pressure P= 40.03 bar

Explanation:

The Van der Waals equation states that

P= R*T/(V-b) - a/V²

where

T= absolute temperature = 295 K

V= molar volume = 0.590 L/mol

R= ideal gas constant = 0.082 atm*L/mol*K

a= van der Waals parameter = 1.355 bar⋅dm⁶/mol² * (0.987 atm/bar) * (1 L²/dm⁶) = 1.337 atm ⋅L²/mol²

b= van der Waals parameter = 0.032 dm³/mol *(1 L/dm³) = 0.032 L/mol

then replacing values

P= R*T/(V-b) - a/V² = 0.082 atm*L/mol*K*295 K/( 0.590 L/mol-0.032 L/mol) -  1.337 atm ⋅L²/mol²/(0.590 L/mol)² = 39.51 atm

P= 39.51 atm / (0.987 atm/bar) = 40.03 bar

in chemistry, adhesion and cohesion are attractive forces that exist between molecules. they hold the molecules together from each other. they are different in that adhesive force exists between molecules of different substance while cohesive force happens between molecules of same substance.

adhesion between water and glass causes water molecules at the edge of water and glass interface to climb up the wall in a narrow glass tube. this is called capillary action. cohesion is what gives glass its shape and causes water to form a droplet.