A balloon at the surface had a volume of 7 cubic meters. If the balloon is taken to depth of 40 meters, what is its new volume?

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To answer this question, we need two pieces of information. First, how does pressure affect volume? Second, what is the pressure at a depth of 40 meters? If we assume that the air in the balloon is an ideal gas, we can use the ideal gas law to relate temperature,...

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To answer this question, we need two pieces of information. First, how does pressure affect volume? Second, what is the pressure at a depth of 40 meters?

If we assume that the air in the balloon is an ideal gas, we can use the ideal gas law to relate temperature, pressure, and volume:
` P V = n R T = const`

At constant temperature `T` , for a fixed amount of air in the balloon` n` , the volume `V` will simply be inversely proportional to the pressure `P` . (Think about what happens if we put the balloon in vacuum: volume goes to infinity---in other words, the balloon pops!)

`P_1 V_1 = P_2 V_2`

To answer the second question, the formula for pressure within a given fluid is `d g h` , where `d` is the density of the fluid,` g` is the acceleration of gravity, and `h` is the height of fluid above us. Here we have to add both air and water pressure; the air pressure we can assume to be about 1 atmosphere, or 100 kPa. Then we need the water pressure (remember that the density of water is 1000 kg/m^3).

`P_{H_2O} = (1000 kg/m^3)(9.8 m/s^2)(40 m) = 392,000 Pa = 400 kPa`

Thus, the total pressure on the balloon at a depth of 40 meters is 500 kPa, the sum of the air and the water.

` P = P_{H2O} + P_{air} = 400 kPa + 500 kPa`

So, we have increased the pressure from about 100 kPa to about 500 kPa, a factor of 5. This means that we must decrease the volume by the same factor, so it will shrink from 7 cubic meters to 1.4 cubic meters.
`V_2 = V_1 P_1/P_2 = (7 m^3) (100 kPa)/(500 kPa) = 1.4 m^3`

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