546_PartUniversity Physics Solution

546_PartUniversity - 18-4 Chapter 18 18.16 IDENTIFY F = pA and pV = nRT SET UP For a cube V A = L EXECUTE(a The force of any side of the cube is F

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18-4 Chapter 18 18.16. IDENTIFY: Fp A = and p Vn R T = SET UP: For a cube, / . VA L = EXECUTE: (a) The force of any side of the cube is () ( ) , An R T V R TL == = since the ratio of area to volume is / 1/ . AV L = For 20.0 C 293.15 K, ° T 4 (3 mol) (8.3145 J mol K) (293.15 K) 3.66 10 N. 0.200 m nRT F L (b) For 100.00 C 373.15 K, T = 4 (3 mol)(8.3145 J mol K)(373.15 K) 4.65 10 N. 0.200 m nRT F L = × EVALUATE: When the temperature increases while the volume is kept constant, the pressure increases and therefore the force increases. The force increases by the factor 21 /. TT 18.17. IDENTIFY: Example 18.4 assumes a temperature of 0 C ° at all altitudes and neglects the variation of g with elevation. With these approximations, / 0 . Mgy RT pp e = SET UP: ln( ) . x ex =− For air, 3 28.8 10 kg/mol. M EXECUTE: We want y for 0 0.90 p p = so / 0.90 Mgy RT e = and ln(0.90) 850 m. RT y Mg = EVALUATE: This is a commonly occurring elevation, so our calculation shows that 10% variations in atmospheric pressure occur at many locations. 18.18. IDENTIFY: From Example 18.4, the pressure at elevation y above sea level is / 0 . Mgy RT e = SET UP: The average molar mass of air is 3 M EXECUTE: At an altitude of 100 m, 32 1 (28.8 10 kg mol)(9.80 m s )(100 m) 0.01243, (8.3145 J mol K)(273.15 K) Mgy RT × and the percent decrease in pressure is 0.01243 0 1 1 0.0124 1.24%. e −= = = At an altitude of 1000 m, 2 0.1243 Mgy RT = and the percent decrease in pressure is 0.1243 1 0.117 11.7%. e −== EVALUATE: These answers differ by a factor of (11.7%) (1.24%) 9.44, = which is less than 10 because the variation of pressure with altitude is exponential rather than linear. 18.19. IDENTIFY: 0 Myg RT e = from Example 18.4. Eq.(18.5) says p ρ MRT . = Example 18.4 assumes a constant 273 K, so and Tp ρ = are directly proportional and we can write 0 . Mgy RT ρρ e = SET UP: From Example 18.4, 1.10 Mgy RT = when 8863 m. y = EXECUTE: For 100 m, y = 0.0124, Mgy RT = so 0.0124 00 0.988 . e ρ The density at sea level is 1.2% larger than the density at 100 m. EVALUATE: The pressure decreases with altitude. tot , m p VR T M = so when the pressure decreases and T is constant the volume of a given mass of gas increases and the density decreases. 18.20. IDENTIFY: / 0 Mgy RT e = from Example 18.4 gives the variation of air pressure with altitude. The density of the air is , pM RT = so is proportional to the pressure p . Let 0 be the density at the surface, where the pressure is 0 . p SET UP: From Example 18.4, 41 (28.8 10 kg/mol)(9.80 m/s ) 1.244 10 m . (8.314 J/mol K)(273 K) Mg RT × × EXECUTE: 3 (1.244 10 m )(1.00 10 m) 0.883 . p pe p −− −× × constant, M pR T so 0 0 p p = and 0 0.883 . p p ⎛⎞ ⎜⎟ ⎝⎠ The density at an altitude of 1.00 km is 88.3% of its value at the surface. EVALUATE: If the temperature is assumed to be constant, then the decrease in pressure with increase in altitude corresponds to a decrease in density.
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Thermal Properties of Matter 18-5 18.21. IDENTIFY: Use Eq.(18.5) and solve for p . SET UP: / pMR T ρ = and / p RT M = ( 56.5 273.15) K 216.6 K T =− + = For air 3 28.8 10 kg/mol M (Example 18.3) EXECUTE: 3 4 3 (8.3145 J/mol K)(216.6 K)(0.364 kg/m ) 2.28 10 Pa p == × × EVALUATE: The pressure is about one-fifth the pressure at sea-level.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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546_PartUniversity - 18-4 Chapter 18 18.16 IDENTIFY F = pA and pV = nRT SET UP For a cube V A = L EXECUTE(a The force of any side of the cube is F

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