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A bubble with a volume of 1.00 cm3 forms at the bottom of a lake that is 20.0 m deep. The temperature at the bottom of the lake is 10.0°C. The bubble rises to the surface where the water temperature is 35.0°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?

2007-06-28 11:54:09 · 3 answers · asked by Anonymous in Science & Mathematics Physics

3 answers

You haven't said if you are taking pressure into account? I'll assume you are not as you made no reference to it.

PV =nRT

V = nRT/P (nR/P = constant)

Convert to Kelvin and rearrange you get

Volume = 308/283 = 1.088cm^3

2007-06-28 12:00:59 · answer #1 · answered by Anonymous · 0 0

I'd guess it is about both pressure and temperature, otherwise why mention the depth?
So we have a pressure change as well, and V1/V2 = (kT1/P1) / k(T2/P2), or V2 = V1T2P1/(T1P2).
T1 = 10 + 273 = 283
T2 = 35 + 273 = 308
P2 = 1 atm = 101325 Pa
P1 = 1 atm + g (m/s^2)*20 m*1000 kg/m^3
= 101325 Pa + 196000 Pa = 297325 Pa (assuming g=9.8)
Then V2 = 1*308*297325/(283*101325) = 3.194 cm^3.

2007-06-29 12:16:24 · answer #2 · answered by kirchwey 7 · 0 0

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2017-01-23 06:40:10 · answer #3 · answered by Anonymous · 0 0

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