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A scuba diver at a depth of 43.0 m below the surface of the sea off the shores of Panama City, where the temperature is 4.0°C, releases an air bubble with volume 17.0 cm3. The bubble rises to the surface where the temperature is 19.0°C. What is the volume of the bubble immediately before it breaks the surface? The specific gravity for seawater is 1.025.
so (p1v1/t1)=(p2v2/t2)
v2=v1(p1/p2)(t2/t1)
p1=101 kPa+ (roe)gh where roe is 1.025 x10^3 Kpa
what is g here?
and i am not sure what units this all should be in
this is what i did...
v2=.17m^3(44176kPa/101kpa)(292.15K/277.15K)
should .17m be in 17 cm^3 or what and what about he kPa can it stay that way
i am confused
so (p1v1/t1)=(p2v2/t2)
v2=v1(p1/p2)(t2/t1)
p1=101 kPa+ (roe)gh where roe is 1.025 x10^3 Kpa
what is g here?
and i am not sure what units this all should be in
this is what i did...
v2=.17m^3(44176kPa/101kpa)(292.15K/277.15K)
should .17m be in 17 cm^3 or what and what about he kPa can it stay that way
i am confused