Abubble of air is rising up through the ocean. when it is at a depth of 20.0 m below the surface, where the temperature is 5.00°c, its volume is 0.90 cm3. what is the bubble's volume (in cm3) just before it hits the ocean surface, where the temperature is 20.0°c? assume the average density of sea water is 1,025 kg/m3.

The bubble's volume = 2830 cm³

Explanation:

Step 1 : Data given

Depth of 20.0 m below the surface

Temperature is 5.00°C = 278 Kelvin

Volume = 0.90 cm³

Before it hits the ocean surface, its temperature is 20.0°C = 293 Kelvin

Density sea water = 1025 kg /m³

Step 2:

Let's assume that the pressure of the air in the bubble is the same as the pressurein the surrounding water.

Let's consider d as the deepth of the ocean and ρ is tge density of the water

p1 = p0 +pgd

⇒ p0 = atmospheric pressure

Since p1V1 = nRT1  we can calculate the numberof moles as:

n = p1V1/RT1 = (p0+pgf)*V1/RT1

 ⇒ V1 = the volume of the bubble at the bottom of the ocean

 ⇒ T1 = the temperature at the bottom of the ocean

At the surface of the ocean, the pressure = p0

The volume of the bubble is:

V2= nRT2/p0

V2=(T2/T1)*((p0+pgd)/p0) *V1

V2= (293/278) * ((101325 + 1025*9.81 *20)/101325)*0.9

V2 =1.054 * (101325+201.105)/101325)0.9

V2 = 1.054 * 1.002*0.9

V2 =2.83 L = 2830 cm³

The bubble's volume = 2830 cm³

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong