A scuba diver floats 25.0 m below the surface of the sea where the water tempera

Question

A scuba diver floats 25.0 m below the surface of the sea where the water temperature is 5 degree C (density of sea water = 1030 kg/m^3). The scuba diver brings a small device that can blow soap bubbles containing helium. What is the pressure underwater at a depth of 25.0m? The diver blows a Helium bubble of volume 10 ml (1 ml = 1 cm^3). The bubble rises to the surface where the temperature is 20 degree C. What is the volume of the bubble when it reaches the surface? Calculate the mass of Helium inside the bubble (molar mass of He is 4 g/mol)

Explanation / Answer

Given: temperature of water at 25 m below surface (T1) = 5+273 = 278 K ; density of water () = 1030 Kg/m3

(a) Pressure at a depth of 25 m from surface (P1) = *g*25 = 1030*9.8*25 = 252350 Pa

(b) Here initial volume (V1) = 10 ml ; final pressure (P) = 105 Pa ; and final temperature (T) = 273+20 = 293 K

Hence using gas equation we have P*V/T = P1*V1/T1   where V is the final volume

So V = P1*V1*T/T1*P = 252350*10*293/100000*278 = 26.59 ml

(c) Now we have the gas equation PV = nRT where n is the number of moles of gas and R is the gas constant = 0.082 L atm K1 mol1

Hence n = PV/RT = 1*26.59*10-3/0.082*293 = 1.1*10-3 moles

So the mass of Helium = 4*0.0011 = 0.0044 g

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