Question

How many grams of Kr are in a 1.93 L cylinder at 84.7 °C and 7.63 atm?

Answer 1

Assume that Kr behaves as an ideal gas and apply the ideal gas law.

P*V = n*R*T where P = pressure exerted by Kr gas = 7.63 atm; V = volume occupied by Kr gas = volume of the cylinder = 1.93 L and T = temperature of the Kr gas in Kelvin scale = 84.7°C ≡ (84.7 + 273) K = 357.7 K.

Plug in values and obtain n

(7.63 atm)*(1.93 L) = n*(0.082 L-atm/mol.K)*(357.7 K)

====> n = (7.63*1.93)/(0.082*357.7) mol = 0.502052 mole.

Atomic mass of Kr = 83.798 g/mol.

Mass of Kr corresponding to 0.502052 mole = (0.502052 mole)*(83.798 g/mol) = 42.07095 g ≈ 42.0709 g (ans).