An air bubble has a volume of 1.70 cm³ when it is released by a submarine 115 m below the surface of a lake. what is the volume of the bubble when it reaches the surface? assume the temperature and the number of air molecules in the bubble remain constant during its ascent.

V = 20.67 cm³

Explanation:

In this case, let's apply the Boyle's law which is:

P1V1 = P2V2

Where P1 and V1 would be condition in the water, and P2 and V2 would be the condition at the surface.

By logic, at the surface, pressure should be equals to 1 atm or 1.01x10^5 N/m²

We know the volume of the bubble at first which is 1.70 cm³ and we need to calculate V2. We know how much is P2, but we don't know the value of P1, which is the pressure of the bubble below the sea; this can be calculated using Pascal's principle which is the following expression:

P1 = Po + dgh

Where:

Po: innitial pressure, which we can assume is 1 atm

d = density of the substance, in this case, water (1000 kg/m³)

g = gravity (9.8 m/s²)

h = distance of the bubble from the surface (115 m)

Now replacing this data in the boyle's law we have the following:

P1V1 = P2V2

V2 = P1V1/P2

V2 = (Po + dgh) * V1 / P2

Replacing the data we have:

V2 = (1.01x10^5 + 1000*9.8*115) * 1.7 / 1.01x10^5

V2 = 2,087,600 / 1.01x10^5

V2 = 20.67 cm³

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6