How Does Pressure and Temperature Affect an Air Bubble's Volume at Sea Depth?

[cutegirl1980]

At depth of 20 m below the surface of the sea, where the temperature is 5.0ºC, a diver exhales an air bubble having a volume of 1 cm3.

a. If the surface temperature of the sea is 22.0ºC, what is the volume of the air bubble just before it breaks the surface? (Start hint: pressure beneath the surface of a liquid is Psurface + ρgh)
b. If a 250 mm steel welding rod is lowered from the surface to the diver working at the 20m depth, what is the change in length of the rod when it achieves thermal equilibrium?


kind of having problem with how to approach part B, thanks

 

[Doc Al]

Hint: What's the thermal expansion coefficient for steel?

 

[wais]

for pointing out the hint for part A

For part A, we can use the ideal gas law to find the volume of the air bubble just before it breaks the surface. The ideal gas law states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. We can rearrange this equation to solve for V: V = (nRT)/P.

Since we know the initial volume (1 cm3) and temperature (5.0ºC = 278.15 K), we just need to find the pressure at 20 m below the surface. The hint provided tells us that the pressure beneath the surface of a liquid is Psurface + ρgh, where Psurface is the pressure at the surface, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth.

We can find the pressure at the surface by using the ideal gas law again, but this time with the surface temperature of 22.0ºC = 295.15 K. Plugging in the values, we get Psurface = (nRT)/V = (1 mol * 0.0821 L atm/mol K * 295.15 K)/1 cm3 = 24.3 atm.

Now we can find the pressure at 20 m below the surface: P = 24.3 atm + (1000 kg/m3 * 9.8 m/s2 * 20 m) = 24.3 atm + 19600 Pa = 25.3 atm.

Finally, we can plug in all the values to find the volume of the air bubble just before it breaks the surface: V = (1 mol * 0.0821 L atm/mol K * 278.15 K)/25.3 atm = 0.91 cm3. So the volume of the air bubble will decrease as it rises to the surface due to the decrease in pressure and increase in temperature.

For part B, we can use the thermal expansion equation: ΔL = αLΔT, where ΔL is the change in length, α is the coefficient of linear expansion, L is the initial length, and ΔT is the change in temperature.

We are given the initial length (250 mm) and the change in temperature (22.0