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?

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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