556_PartUniversity Physics Solution

# 556_PartUniversity Physics Solution - 18-14 Chapter 18...

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18-14 Chapter 18 18.63. IDENTIFY: Apply Bernoulli&s equation to relate the efflux speed of water out the hose to the height of water in the tank and the pressure of the air above the water in the tank. Use the ideal-gas equation to relate the volume of the air in the tank to the pressure of the air. (a) SET UP: Points 1 and 2 are shown in Figure 18.63. 5 1 4.20 10 Pa p 5 2a i r 1.00 10 Pa pp == × large tank implies 1 0 v Figure 18.63 EXECUTE: 22 11 111 2 2 2 pg y v y v ρ ρρ ++ = 2 1 212 12 2 () vpp g yy =−+ 21 2 1 2 (2/ )( ) 2 ( ) vp p g y y =− + 2 26.2 m/s v = (b) 3.00 m h = The volume of the air in the tank increases so its pressure decreases. constant, pV nRT = = so 00 p Vp V = 0 ( p is the pressure for 0 3.50 m h = and p is the pressure for 3.00 m) h = (4.00 m ) (4.00 m ) p hA p h A −= 55 0 0 4.00 m 4.00 m 3.50 m (4.20 10 Pa) 2.10 10 Pa 4.00 m 4.00 m 3.00 m h h −− ⎛⎞ × = × ⎜⎟ ⎝⎠ Repeat the calculation of part (a), but now 5 1 p and 1 3.00 m. y = 2 1 2 2/ ( ) 2 ( ) p g y y + 2 16.1 m/s v = 2.00 m h = 0 0 4.00 m 4.00 m 3.50 m 1.05 10 Pa 4.00 m 4.00 m 2.00 m h h × = × 2 1 2 ) 2 ( ) p g y y + 2 5.44 m/s v = (c) 2 0 v = means ) 2 ( ) 0 pp g −+ p y y 1.00 m yyh −=− 5 0 0.50 m 0.50 m . 4.00 m 4.00 m hh × This is 1 , p so 2 3 0.50 m (9.80 m/s )(1000 kg/m )(1.00 m ) 4.00 m h h ×− × = (210/(4.00 )) 100 9.80 9.80 , −− = − with h in meters. 210 (4.00 )(109.8 9.80 ) 2 9.80 149 229.2 0 = and 2 15.20 23.39 0 = quadratic formula: ( ) 2 1 2 15.20 (15.20) 4(23.39) (7.60 5.86) m h = ± h must be less than 4.00 m, so the only acceptable value is 7.60 m 5.86 m 1.74 m h = EVALUATE: The flow stops when p gy y +− equals air pressure. For 1.74 m, h = 4 9.3 10 Pa p and 4 0 . 7 1 0 P a , −=× so 5 1 . 0 1 0 P a , y y = × which is air pressure.

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Thermal Properties of Matter 18-15 18.64. IDENTIFY: Use the ideal gas law to find the number of moles of air taken in with each breath and from this calculate the number of oxygen molecules taken in. Then find the pressure at an elevation of 2000 m and repeat the calculation. SET UP: The number of molecules in a mole is 23 A 6.022 10 molecules/mol N . 0.08206 L atm/mol K R =⋅ . Example 18.4 shows that the pressure variation with altitude y , when constant temperature is assumed, is / 0 Mgy RT pp e = . For air, 3 28.8 10 kg/mol M . EXECUTE: (a) p Vn R T = gives (1.00 atm)(0.50 L) 0.0208 mol (0.08206 L atm/mol K)(293.15 K) pV n RT == = ⋅⋅ . 23 21 A (0.210) (0.210)(0.0208 mol)(6.022 10 molecules/mol) 2.63 10 molecules Nn N × = × . (b) 32 (28.8 10 kg/mol)(9.80 m/s )(2000 m) 0.2316 (8.314 J/mol K)(293.15 K) Mgy RT × . /0 . 2 3 1 6 0 (1.00 atm) 0.793 atm Mgy RT e e −− = . N is proportional to n , which is in turn proportional to p , so 21 21 0.793 atm (2.63 10 molecules) 2.09 10 molecules 1.00 atm N ⎛⎞ = × ⎜⎟ ⎝⎠ . (c) Less 2 O is taken in with each breath at the higher altitude, so the person must take more breaths per minute. EVALUATE: A given volume of gas contains fewer molecules when the pressure is lowered and the temperature is kept constant. 18.65. IDENTIFY and SET UP: Apply Eq.(18.2) to find n and then use Avogadro&s number to find the number of molecules. EXECUTE: Calculate the number of water molecules N .
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556_PartUniversity Physics Solution - 18-14 Chapter 18...

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