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Consider 2.00 moles of Argon, an ideal gas, at a density of 5.00 g/L and a pressure of 2.00 atm. What is the closest value to the temperature (in K) of this gas?

Respuesta :

Answer:

194.73 K

Explanation:

Use the Ideal Gas Law:    PV = n RT

   R = .082057366  L Atm / k-mole

2 moles Argon weigh in at 79.896 gm

  79.896 gm / 5 gm/L = 15.9792 Liters

Ideal Gas Law:

(2)(15.9792) = 2 (.082057366) ( K)

194.73 K

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