What is the volume of 9.783 × 10²³ atoms of Kr at 9.25 atm and 512K? A) 7.38 L B) 3.69 L C) 1.85 L D) 15.4 L E) 30.8 L

Question

Quizplus · Accepted Answer

To find the volume of the given number of atoms of Kr at a given pressure and temperature, we need to use the Ideal Gas Law: PV = nRT

Where, P = pressure = 9.25 atm 
V = volume (unknown) 
n = number of moles of Kr atoms = (9.783 × 10^21)/Avogadro's number = 1.622 × 10^1 mol 
R = ideal gas constant = 0.0821 L·atm/mol·K 
T = temperature = 512 K

Substituting these values in the equation above and solving for V, we get:

V = (nRT)/P = (1.622 × 10^1 mol x 0.0821 L·atm/mol·K x 512 K)/9.25 atm 
V = 7.38 L

Therefore, the answer is A) 7.38 L.

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