7.41 grams Helium = 1.851 moles of Helium PV = nRT T = PV/nR = 3.50 atm x 72.8L/1.851 mol x 0.08205L atm mol-1 K-1 = 1.68 x 103 K =1.68 x 10^3 - 273 = 1.41 x 10^3 °C Solution 7.41 grams Helium = 1.851 moles of Helium PV = nRT T = PV/nR = 3.50 atm x 72.8L/1.851 mol x 0.08205L atm mol-1 K-1 = 1.68 x 103 K =1.68 x 10^3 - 273 = 1.41 x 10^3 °C.