A 1.6 air bubble is released from the sandy bottom of a warm, shallow sea, where the gauge pressure is 1.6 . The bubble rises slowly enough that the air inside remains at the same constant temperature as the water. What is the volume of the bubble as it reaches the surface?
(1.6+ 1.0)*1.6)/1.0
Since gauge pressure=p-p_0, you must add p_0 back to the gauge pressure and then multiply by the volume and then divide by atmospheric pressure.
My physics teacher says that 1 atm is added.
To find the volume of the bubble as it reaches the surface, we can use the physics principle known as Boyle's Law. Boyle's Law states that the pressure and volume of a gas are inversely proportional when temperature is kept constant.
Given:
Initial pressure of the air bubble, P1 = 1.6 atm
Final pressure of the air bubble at the surface, P2 = 1 atm (atmospheric pressure)
Initial volume of the air bubble, V1 = 1.6 cc (cm^3)
Using Boyle's Law, we can set up the following equation:
P1 * V1 = P2 * V2
Substituting the given values:
1.6 atm * 1.6 cc = 1 atm * V2
Simplifying the equation:
2.56 cc = V2
Therefore, the volume of the bubble as it reaches the surface is 2.56 cc.
pV=nRT (bottom of ocean)
pV=nRT (top of ocean)
n,R,T are all constants so those cancel out
pV(bottom)= pV(top)
pV(bottom)/p(top)=V(top)
1.6*1.6/1.0=V
V=2.56cm3