Question

A.1.5 cm² air bubble is released from the sandy bottom of a warm, shallow sea, where the gauge pressure is 1.2 atm. The bubble rises slowly enough that the air inside remains at the same constant temperature as the water.

 You may want to review (Pages 367-372) 

Part A

What is the volume of the bubble as it reaches the surface?

Prat B

As the bubble rises, is heat energy transferred from the water to the bubble or from the bubble to the water?

Answer 1

Part - A

Given that the gauge pressure, Pg = 1.2 atm

Now, we need to convert pressure gauge to absolute pressure.

As we know that -

Gauge pressure = Absolute pressure - Atmospheric pressure

=> Pg = P -1 atm

So,

Absolute Pressure = pressure gauge + 1 atm

=> P = 1.2 atm + 1 atm
=> P = 2.2 atm at the sandy bottom

We can also make the assumption that the Atmospheric pressure on the surface is 1 atm.

Next we use the Combined gas law to determine the final volume.

(P2V2/T2)=(P1V1/T1)

Since we know temperature is kept constant we can modify the equation:

P2V2=P1V1

=> V2 = (P1V1) / P2

Now we plug in what we know:
P2 =1 atm

P1 =2.2 atm

V1=1.5 cm^3

Therefore -

V2 = (2.2 atm x 1.5 cm^3) / (1atm) = 3.3 cm^3

Hence, volume of the bubble when it reaches the surface, V2 = 3.3 cm^3 = 3.3 x 10^-6 m^3

Part -B

As the bubble rises, the heat energy from water is transferred to the bubble.

Hence, first option is the correct answer.