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If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘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(L2⋅atm)/mol2 and b=0.03219L/mol.

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joseaaronlara

Answer:

The answer to your question is:

Explanation:

Ideal gases

                     PV = nRT

                     P = nRT / V

                     P = (1)(0.082)(303) / 0.5

                     P = 49.7 atm

Real pressure

                    P = V²(nRT)/ (V - nb) - an²

                    P = (0.5)²(1)(0.082)(303) / (0.5 - (1)(0.03219)  - (1.345)(1)²

                    P = (6.2115)/(0.4678) - 1.345

                   P = 13.28 - 1.345

                   P = 11.93 atm

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