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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