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Josephjenkins620
26.11.2019 •
Chemistry
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.
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Ответ:
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
Ответ: