An air bubble at the bottom of a lake 43.5m deep has a volume of 1.00cm/cubed. If the temperature at the bottom is 5.5 degrees celcius and at the top 21.0 degrees celcius, what is the volume of the bubble just before it reaches the surface.
Are we ignoring pressure effects here? These are more substantial than temperature effects.
yes
and how do i convert Celsius to kelvin
K = C + 273
Do you have the volumetric coefficient of thermal expansion for air? Otherwise, we will have to use the ideal gas law which requires we make some assumptions.
i think well have to use ideal gas law
Then pressure is definitely being accounted for. First, let's find the pressure at the bottom of the lake. \[P_b = \rho g h + P_{atm}\]where \(P_{atm} = 101 kPa = 1 atm = 14.7 psi\) Now, from ideal gas law, \[\left [ PV \over n RT \right ]_{bottom} = \left [ PV \over nRT \right ] _{t o p}\]n remains constant so it cancels out. We just calculated the pressure at the bottom, the pressure at the top will be equal to \(P_{atm}\). We know the volume at the bottom and the two temperatures. Leaving the volume at the top the only unknown.
what do i plug in for pressure at the bottom of the lake
can you show me what to plug in
You have to solve for it It is\[P_{Bottom} = \rho gh + P_{atm}\]
but what do i put in for density
and is the g gravity 9.8
Depends on what units you want. Looks like we are in SI, so density would be 1000 kg/m^3 for water, and g is definitely 9.8.
right right
h would be 43.5
Indeed. The height of the water column above the bubble.
right
p.s. your a genius man
what do i put for nRT
8.314
for R. T must be in kelvin.
how did you get that?
The n cancels out. because it does not change.
oh ok
help me on the other one
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