A bubble with a volume of 1.00 cm^3 forms at the bottom of a lake that is 30 m deep. The temperature at the bottom of the lake is 10°C. The bubble rises to the surface where the water temperature is 25°C. Assume that the bubble is small enough that its temperature always matches that of its surroundings. What is the volume of the bubble just before it breaks the surface of the water? Ignore surface tension. Units in cm^3
2007-12-04 06:35:08 · 1 answers · asked by Eli D 1 in Science & Mathematics ➔ Physics
First, let's raise our bubble from the murky depths to the surface.
Atm Pressure= 101.325 kPa
Water Pressure at 30m = Atm Pressure +30m *1000 kg/m^3 *9.81m/sec^2
=101.325 + 294.3 kPa
=395.6 kPa
Expansion of Bubble due to pressure change=
Water Pressure 30m/Atm Pressure= 395.6 kPa/101.3 kPa
=3.905
.....................................................
Thermal Expansion-- Use Perfect Gas law.
(273.15+25)/(273.15+10) =1.053
Therefore volume=
= 1.00 cm^3*1.053*3.905
=4.11 cm^3
2007-12-04 08:08:18 · answer #1 · answered by Frst Grade Rocks! Ω 7 · 1⤊ 0⤋