621_Physics ProblemsTechnical Physics

621_Physics ProblemsTechnical Physics - Chapter 21 *P21.56...

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Chapter 21 623 *P21.56 The work done by the gas on the bullet becomes its kinetic energy: 1 2 1 2 11 10 792 23 2 mv = .. kg 120 m s J bg . The work on the gas is 1 1 γ −= PV ff i i di . J . Also f f i i γγ = PP V V fi i f = F H G I K J . So F H G I K J L N M M O Q P P 1 040 . . J V V V if i f i . And V f =+ = 12 50 13 5 cm cm 0.03 cm cm 32 3 . . Then P i = L N M O Q P = 10 13 5 12 574 10 566 6 12 13 5 1.40 6 . . . J 0.40 cm m cm cm Pa atm 33 a f ch . P21.57 The pressure of the gas in the lungs of the diver must be the same as the absolute pressure of the water at this depth of 50.0 meters. This is: PP g h =+ = + × 0 3 1 00 1 03 10 9 80 50 0 ρ . . atm kg m m s m ej e j a f or P × × F H G I K J = 1 00 5 05 10 5 98 5 . atm Pa 1.00 atm 1.013 10 Pa atm 5 If the partial pressure due to the oxygen in the gas mixture is to be 1.00 atmosphere (or the fraction 1 598 . of the total pressure) oxygen molecules should make up only 1 . of the total number of molecules. This will be true if 1.00 mole of oxygen is used for every 4.98 mole of helium. The ratio by
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