Given that:Hydrogen temperature = 300C = 303KLoad lifting = 400KgAtmospheric pressure = 13.6 × 103 × 0.75 × 9.81 = 1.00 × 105 N/m2 = 1.00 barAtmospheric Temperature = 270C = 300KThe mass that can be lifted due to buoyancy force,So the mass of air displaced by balloon(ma) = Mass of balloon hydrogen gas (mb) + load lifted ...(i)Since PV = mRT; ma= PaVa/RTa; R = 8314/29 = 287 KJ/Kgk For Air; 29 = Mol. wt of air= 1.00 × 105 × V/ 287 × 300 = 1.162V Kg ...(ii)Mass of balloon with hydrogenmb = PV/RT = 1.00 × 105 × V/ (8314/2 × 300) = 0.08V Kg ...(iii)Putting the values of (ii) and (iii) in equation (i)1.162V = 0.08V + 400V = 369.67 m3But we know that the volume of a balloon (sphere) = 4/3Πr3322 = 4/3Πr3r = 4.45 m ......ANSQ. 42: Manometer measure the pressure of a tank as 250cm of Hg. For the density of Hg 13.6 × 103
