Please help me with this homework problem I don\'t understand! A typical tank us

Question

Please help me with this homework problem I don't understand! A typical tank used for scuba diving (fig) has a volume of 11 L (about 0.4 ft3) and maximum gauge pressure of 2.10 x 107 Pa (about 3000 lb/in2 gauge). The empty tank contains 11.0L of air at 21oC and at 1atm. The air is hot when it comes out of the compressor, when the tank is filled, the temperature is 42oC and the gauge pressure is 2.10 x 107 Pa. What mass of air was added? (air is a mixture of 78% N2, 21% O2 and 1% miscellaneous gases, its average molecular mass is 28.8 g/mol

Explanation / Answer

PV = nRT and you want n after the air has been added. Solve for n:

n = PV/RT

and consider that both V and R are constant. So the number of moles of gas present is directly proportional to pressure and inversely proportional to temperature. The starting number of moles is

ninit = (11 l)/(24.14 l/mol) = 0.456 mol     [24.14 is molar volume at 21 celsius]

Absolute pressure is gauge pressure plus atmospheric = 2.10 * 107 Pa + 101325 Pa = 2.11 * 107 Pa so the pressure increase is (2.11 * 107)/(101325) = 208.2

Absolute temperature is celsius plus 273.2 so the temperature increase is 315.2/294.2 = 1.07

So the final number of moles nfinal = ninit * 208.2 / 1.07 = 88.58 mol

and since we are given molar mass of 28.8 g/mol for air, the mass of air added is 88.58 * 28.8 = 2551 g or 2.551 kg

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