A bubble 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.

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Chemistry

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17 February, 00:00

A bubble 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.

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Answers (2)

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Payten Arias · Answer 1

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

Reyna Garcia · Answer 2

2.83 cm³

Explanation:

The ideal gas law states that the relationship between pressure (P), volume (V), temperature (T) and the moles (n) are constant:

PV/n*T = constant

So, assuming that the number of moles doesn't vary in this process:

P1*V1/T1 = P2*V2/T2

Where 1 is the state when the bubble under water, and 2 when it hits the surface. The pressure of the bubble is the same as its surroundings, so by the Stevin's theorem, inside a liquid:

P1 = Patm + ρ*g*h

Where Patm is the atmospheric pressure (101325 Pa), ρ is the density (1025 kg/m³), g is the gravity acceleration (9.8 m/s²), and h is the depth (20.0 m), so:

P1 = 101325 + 1025*9.8*20

P1 = 302225 Pa

P2 = Patm = 101325 Pa, T1 = 5.00°C = 278 K, T2 = 20°C = 293 K. So:

(302225*0.90) / 278 = (101325*V2) / 293

978.4263 = 345.8191V2

345.8191V2 = 978.4263

V2 = 2.83 cm³