An air bubble of volume 15 cc is formed at a depth of 50 m in a lake. If the temperature of the bubble while rising remains constant then what will be the volume of the bubble at the surface?(Given: g=10 m/s2 & atm. pressure = 1.0 * 100000 Pa)
* Physics - Elena, Sunday, April 8, 2012 at 2:05pm
p1•V1 = p2•V2,
ρ•g•h• V1 = p2•V2,
V2 = ρ•g•h• V1/ p2,
ρ = 1000 kg/m^3 is the density of water,
p is the atm. pressure
* Physics - elena plz read it, Monday, April 9, 2012 at 12:04am
i am not getting the answer which is 90cc
V2 = (ρ•g•h+p)• V1/ p =
= (1000•10•50 + 100000) •15/100000 = 90 cm^3
ρ = 1000 kg/m^3 is the density of water,
p is the atm. pressure
To find the volume of the air bubble at the surface, we can use the equation p1•V1 = p2•V2, where p1 is the initial pressure, V1 is the initial volume, p2 is the final pressure, and V2 is the final volume.
In this case, the initial pressure is the atmospheric pressure, given as 1.0 * 100000 Pa. The initial volume is 15 cc.
The final pressure at the surface is also the atmospheric pressure, so p2 = 1.0 * 100000 Pa.
The density of water, ρ, is 1000 kg/m^3. The acceleration due to gravity, g, is 10 m/s^2.
The depth of the air bubble, h, is 50 m.
Using these values, we can calculate the final volume of the bubble at the surface:
V2 = ρ•g•h• V1 / p2
= (1000 kg/m^3)•(10 m/s^2)•(50 m)•(15 cc) / (1.0 * 100000 Pa)
= 0.075 m^3 = 75 cc
Therefore, the volume of the bubble at the surface is 75 cc, not 90 cc.