Pressure, Volume & Temperature: Homework Solutions

  • Thread starter Thread starter ar202
  • Start date Start date
  • Tags Tags
    Pressure Volume
AI Thread Summary
The discussion revolves around calculating the pressure changes in a gas bottle and the volume of an air bubble exhaled by a diver. Initially, the gas bottle holds air at 232 BAR and 30°C, and after cooling to 8°C, the pressure is calculated to be 215 BAR using the combined gas law. The diver, working at a depth of 40 meters, experiences an increased pressure, which is determined to be 492800 Pa. As the diver exhales a 0.5-liter bubble, the volume of the bubble when it reaches the surface is estimated to be 2.2 liters. Participants seek confirmation on their calculations and understanding of the pressure dynamics involved.
ar202
Messages
45
Reaction score
0

Homework Statement



A 12 litre gas bottle is initially charged with air to 232 BAR at 30 C. What will be the new pressure in the bottle once it has cooled to 8 C?

A diver breathes from the gas bottle above, is working at a depth of 40m, and at a temperature of 8 C.

If the diver exhales a bubble of air with a volume of 0.5 litres, what will the volume of the bubble be when it reaches the surface?

Homework Equations



not sure...

The Attempt at a Solution



I know for the first part that using the combined gas law

(P1/T1) x T2 = 215 BAR

P = pressure
T = temperature in kelvin

However I'm stuck on where to go with the next part.

Any pointers/hints in the right direction would be really appreciated. thanks
 
Physics news on Phys.org
Since you know the diver's depth you can determine the pressure at that depth. You are also given the temperature and volume of the air bubble at that depth. As the bubble rises it will expand to a larger volume. You will need to determine a pressure and temperature at the surface to find the new volume.
 
Have ended up with the volume of the bubble at surface = 2.2L

anyone think this is correct?
 
ar202 said:
Have ended up with the volume of the bubble at surface = 2.2L

anyone think this is correct?

Not me.

What is a pressure at 40 meters? What is a pressure at the surface?
 
Pressure at 40m = 1000(p) x 9.81(g) x 40(h) + Atmospheric pressure (100'000) = 492800

Pressure at surface = 100'000 Pa

...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top