problem16_09

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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16.9: Solving Eq. (16.10) for the temperature, ( 29 ( 29 K, 191 K mol J 3145 . 8 40 . 1 hr km 6 . 3 s m 1 0.85 h km 850 mol) kg 10 8 . 28 ( 2 3 2 = × = = - R Mv T γ or C. 82 ° - b) See the results of Problem 18.88, the variation of atmospheric pressure with altitude, assuming a non-constant temperature. If we know the altitude we can use the result of Problem 18.88, . 1 0 0 α α - = R Mg T y p p Since , y T T o α - = m, C 10 6 . K, 191 for 2 ° × = α = - T and ). ft. 840 , 44 ( m 667 , 13 K, 273 0 = = y T Although a very high altitude for commercial aircraft, some military aircraft fly this high. This result assumes a uniform decrease in temperature that is solely due to the increasing altitude.
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Unformatted text preview: Then, if we use this altitude, the pressure can be found: , K 273 m) (13,667 ) m C 10 6 (. 1 p m) C 10 K)(.6 mol J 315 . 8 ( ) s m mol)(9.8 kg 10 8 . 28 ( 2 o 2 2 3 ° × ⋅ ×--- ° ×-= p and , p 13 . ) 70 (. p o 66 . 5 o = = p or about .13 atm. Using an altitude of 13,667 m in the equation derived in Example 18.4 gives , p 18 . o = p which overestimates the pressure due to the assumption of an isothermal atmosphere....
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