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SOLUTION TO EXERCISE 9
SOLUTION TO EXERCISE 9 (a) The surface area of a sphere is given by S = 4 * pi * r^2. If the diameter of the container is reduced to one-fifth, then the surface area goes down by a factor of 5^2, or 25. Since P = F/A, and the force of the gas remains the same, the pressure will change accordingly. P (before reduction) = 0.007 atm P (after reduction) = 0.007 atm / (1/25) = 0.175 atm. (b) According to Charles' Law, P (before re-entry) P (during re-entry) -------------------- = ---------------------. So, T (before re-entry) T (during re-entry) 0.175 atm P --------- = ------- and thus P = (0.175 * 700) / 260 = 0.47 atm 260 K 700 K The designers of the spacecraft had to make sure that the container could withstand at least 0.47 atm of pressure without rupturing. (Most likely, they designed the pressure tolerance of that container to be MUCH higher.) |