F 1.00 \(\rm mol\) of argon is placed in a 0.500-\(\rm L\) container at 21.0 \({\rm \, ^{\circ}C}\) , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, \(a = 1.345 \;\rm (L^2\cdot atm)/mol^2\) and \(b = 0.03219\; \rm L/mol\).

2.05 atm

Explanation:

From ideal gas equation;

PV=nRT

Making P the subject of the formula;

P= nRT/V

Substituting values;

P= 1 × 0.0821 × 294/ 0.5

P= 48.27 atm

Real gas pressure;

(P + an^2/v^2) . (V - nb) = nRT

(P + 1.345 × 1^2/0.5^2) . (0.5 - 1 × 0.03219) = 1 × 0.0821 × 294

(P + 5.38) . (0.46781) = 24.1374

(P + 5.38) = 24.1374/0.46781

(P + 5.38) = 51.5966

P= 51.5966 - 5.38

P= 46.22 atm

Difference = Pideal - Preal

Difference = 48.27 - 46.22

Difference = 2.05 atm

precipitation

explanation:

when it goes from cloud to rain/back to earth in the form of rain

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error