A. Calculate the pressure of ethanol vapor, at 82 degrees C if 1.000 mole of eth

Question

A. Calculate the pressure of ethanol vapor, at 82 degrees C if 1.000 mole of ethanol gas occupies 30.00L. Use the Van der Waals equation and compare that result with one obtained using the ideal gas law

B. Liquid oxygen was first prepared by heating potassium chlorate in a closed vessel to obtain oxygen at a high pressure. The oxygen was then cooled until it liquefied.

                                             2KClO3 -> 2KCl + 3 O2.

If 70 grams of potassium chlorate reacts in a 2.50L vessel, which was initially evacuated, what pressure of oxygen will be attained when the temperature is finally cooled to 25 degrees C? Use the equation and ignore the volume of the solid product.

Explanation / Answer

A. Calculate the pressure of ethanol vapor, at 82 degrees C if 1.000 mole of ethanol gas occupies 30.00L. Use the Van der Waals equation and compare that result with one obtained using the ideal gas law

T = 82°C = 355K

P = ?

V = 30 L

n = 1

PV = nRT

P = nRT/V = (1*0.082*355)/(30) = 0.970 atm

For Van der Waals:

A = 12.18 L^2 bar/mol^2 = 1.28 J m3 / mol^2

B = 0.08407 L/mol

(P-a(n/v)^2)(V-nb) = nRT

P = nRT/(V-nb) + a(n/v)^2

P = 1*8.314*355/(0.03-1*0.0008407) + 1.28*(1/0.03)^2 = 102641.044348 Pa

P = 102641.044348/101325 = 1.0129 atm

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