A diver releases an air bubble of volume 2.0 cm³ from a depth of 15 m below the surface of the lake, where the temperature in 7.0°C. What is the volume of the bubble when it reaches just below the surface of the law, where the temperature is 20°C.
I know that the answer is 5.1 cm³, but what are the steps to reach the answer?
2007-06-25 14:14:22 · 2 answers · asked by Concerned 1 in Science & Mathematics ➔ Physics
P1V1/T1 = P2V2/T2
at 15 m depth>>
P1 = 1atm+ hdg (air+water column)
P1 = 1.01*10^5 + (15*1000*9.8) = (1.01+1.47)*10^5 N/m^2
P1 = 2.48*10^5
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P2 = 1 atm (air column) = 1.01*10^5 N/m^2
V2 = (P1/P2) * (T2/T1) V1
V2 = (2.48*10^5/1.01*10^5) * [(273+20)/(273+7)] (2cm^3)
V2 = (2.455) * [1.046] (2cm^3)
V2 = 5.135 cm^3
2007-06-25 14:38:04 · answer #1 · answered by anil bakshi 7 · 0⤊ 0⤋
Although the previous poster showed you the proper way of performing the calculation for an isothermal upwelling, I doubt that the bubble will rise slowly enough for the air inside to respond instantaneously to the temperature change of its surroundings. In the adiabatic case, which your example implies, your volume will be less than 5.1 cm^3.
2007-06-25 21:47:29 · answer #2 · answered by Jeff 3 · 1⤊ 0⤋