A2.0cm3 air bubble is released from the sandy bottom of a warm, shallow sea, where the gauge pressure is 1.2atm . 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?

V2 = P1×V1/P2

V2 = 4.4cm3

Explanation:

Applying Boyle's law

P1×V1 = P2×V2

V2 = P1×V1/P2

Where;

V1 = initial volume of air bubble = 2.0cm3

V2 = final volume of air bubble

P1 = initial pressure of air bubble = 1.2 + 1 atm = 2.2 atm

(Absolute pressure = guage pressure + atmospheric pressure)

P2 = final pressure of air bubble = 1 atm

V2 = 2.0× 2.2/1.0

V2 = 4.4cm3

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

d. platinum would have the greatest temperature increase because it has the lowest specific heat.