Question

a (Pa m6 mol2) b (m3-mol-) 2.65 x 10-5 Hydrogen Н, NH3 0.0245 Nonideal Gases 3.71 x 10-5 Ammonia 0.4225 You have two 1.00 L containers at 298 K. One contains 1.00 mol of hydrogen gas while the second contains 1.00 mol of ammonia gas. (a) Calculate the pressure in each container according to the ideal gas law. (b) Calculate the actual pressure in each container.

Answer 1

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for ideal gas PV = nRT - P(IL) - Il mol) (0-0821 L-atm.) (2986) molk p = 240 ¢7 atm o latm = 101325 Pa. PH₂ = PNH₂ = (240478 101325) Pa = 2048 x 106 Pa. : (P + na) (v=nb) = nRT (for real gas) V2 (Pta ) (w-b) = RT (for 1 mol real gas) = L = 10-3 m3 Pue + 0-0245 ) (10-3-2-65x10-5) = 3-314 * 299 106 . PH₂ = [2.52 x 108/pan lly Poviny + 0-4-225) (102 - 31.1210-5) = 9-314x298 8.314 x 298 + 0.4 225 10-6 PNH3 = [2.15 x 106] Pa