What is the volume occupied by a mixture of 0.203 mol of N₂ and 0.203 mol of O₂ gases at 1.16 atm and 44.7°C? A) 2.74 $\times$ 10² L B) 9.13 L C) 0.642 L D) 1.28 L E) 4.57 L

Question

Quizplus · Accepted Answer

The correct answer is B because the volume occupied by the mixture of gases can be calculated using the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Substituting the given values into the equation, we get: V = (0.203 mol + 0.203 mol) * 0.0821 L*atm/(mol*K) * (44.7°C + 273.15) K / 1.16 atm = 9.13 L.

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