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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 \(b = 0.03219\; \rm L/mol\).

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Answer:

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

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