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Unformatted text preview: 16. We assume that the pressure of the air in the bubble is essentially the same as the
pressure in the surrounding water. If d is the depth of the lake and ρ is the density of
water, then the pressure at the bottom of the lake is p1 = p0 + ρgd, where p0 is
atmospheric pressure. Since p1V1 = nRT1, the number of moles of gas in the bubble is
n = p1V1/RT1 = (p0 + ρgd)V1/RT1, where V1 is the volume of the bubble at the bottom of the lake and T1 is the temperature
there. At the surface of the lake the pressure is p0 and the volume of the bubble is V2 =
nRT2/p0. We substitute for n to obtain
V2 = T2 p0 + ρ gd
277 K ( )( ) 1.013 × 105 Pa + 0.998 × 103 kg/m3 9.8 m/s 2 ( 40 m ) = 1.0 × 102 cm3 . 5 1.013 × 10 Pa ( 20 cm )
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
- Spring '08