Question

e consists of 49.09. N 300 and 18.09.CO.ID occupies 5.36 liters at 45.0 oC and a total pressure of 3.40 and 18.0% CO2. The gas mixture a. Calculate the total number of mols present b. Calculate the partial pressure of each gas component and sho the total pressure. sure of each gas component and show that they add up to c. If the gas mixture is heated to 100.0 °C at constant volume of 5.36 liters, calculate the new total pressure. 4. Calculate the density (g/L) of ammonia gas (NH) in a container of 2.50 liters at a temperature of 35 °C and a pressure of 965 Torr?

Answer 1

- & 49%N2 33%ą 18%.co. - Vol of winture = 5:36 Ltr - Temperaturo = 45% = 273+45= 318k to Assuning the percentage given as weight. No of moles = given coulighet i Molar mass no of uoles of Ne=) nu = 4 = 1.75 no of moles of a =) na = 33 = 1.031 ni of moles of Co, 3 neo = 18 = 0.409 (9) Total no' of miles = 0.409+1.75+ 1.03) = 3.190 Mole fraction = no ofwsles of 1 component Total notes I Mole fraction & Total nules of N = 1.75 = 0.848 3.190

U Mole fraction of 0 = 1:03!= 0.323 de frailion of co= 0.409 = 0.128 3.190 Mole fraction of the 3.190 b) Partial Pressure = nule fra Partial Pressure of Pressure = mole fraction & Pressure total u Bessure of us = 0.548 X 3.40 1 = 1.863 atre Partial Pressure of = 0.323 x 3.40 421 me = 1.098 atm Partial Pressure of c = 0.435 alu If we add all 3 Partial Pressure 186371098 +0.435 = 3396 afu I which is nearly equal to P Total = 3.40 To According to Gay lussoe's loud Pressure is directly propektionelto temperature [coustaut Volume 7 - P& T . 7 => P, = 3.40 ahu, h = ? I T = 318 K , Ty = 373 K yorum => € = 37273. 373x3. Yo 318 New total Pressure 12 = 3.988 atm]