Physics Solution Manual for 1100 and 2101

# This can be found from the ideal gas law rt pv nrt p

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Unformatted text preview: __________________________________________ 40. REASONING The average kinetic energy KE of each molecule in the gas is directly proportional to the Kelvin temperature T, according to KE = 3 kT (Equation 14.6), where k 2 is the Boltzmann constant. Solving for the temperature, we obtain T= () 2 KE 3k (1) The average kinetic energy KE of one molecule is the total average kinetic energy of all the molecules divided by the total number N of molecules: KE ( )total KE = ( KE )total N (2) We will determine the number N of molecules in the gas by multiplying the number n of moles by Avogadro’s number NA: N = nN A (3) ( )total of all the molecules is equal to the kinetic energy The total average kinetic energy KE KEbullet of the bullet, which has a mass m and a speed v. Thus, according to Equation 6.2, 748 THE IDEAL GAS LAW AND KINETIC THEORY ( KE )total = KE bullet = 1 mv 2 2 (6.2) SOLUTION Substituting Equation (2) into Equation (1) yields () KE total 2 2 KE N 2 KE = total T= = 3k 3k 3kN () () (4) Substituting Equation (3) and Equation 6.2 into Equation (4), we obtain T= ( )total = 2 ( 1 mv2 ) = 2 2 KE 3kN 3k ( nN A ) mv 2 3knN A (5) The Boltzmann constant k is equal to the ratio of the universal gas constant R to Avogadro’s number: k = R N A . Making this substitution into Equation (5) yields the Kelvin temperature of the gas: T= 41. ( 3 SSM WWW ) 8.0 ×10−3 kg ( 770 m/s ) mv 2 mv 2 = = = 95 K 3Rn 3 8.31 J/ ( mol ⋅ K ) ( 2.0 mol ) R n NA NA 2 REASONING The rms-speed vrms of the sulfur dioxide molecules is 2 3 related to the Kelvin temperature T by 1 mvrms = 2 kT (Equation 14.6), where m is the mass 2 of a SO2 molecule and k is Boltzmann’s constant. Solving this equation for the rms-speed gives vrms = 3kT m (1) The temperature can be found from the ideal gas law, Equation 14.1, as T = PV / ( nR ) , where P is the pressure, V is the volume, n is the number of moles, and R is the universal gas constant. All the variables in this relation are known. Substituting this expression for T into Equation (1) yields Chapter 14 Problems vrms 749 PV 3k nR = 3 k PV = m nmR The mass m of a single SO2 molecule will be calculated in the Solutions section. SOLUTION Using the periodic table on the inside of the text’s back cover, we find the molecular mass of a sulfur dioxide molecule (SO2) to be 32.07 u 3 u) 14 244 + 2 (15.9994 3 = 64.07 u 4 14 244 4 Mass of a single sulfur atom Mass of two oxygen atoms Since 1 u = 1.6605 × 10−27 kg (see Section 14.1), the mass of a sulfur dioxide molecule is 1.6605 ×10−27 kg −25 m = ( 64.07 u ) kg = 1.064 × 10 1u The translational rms-speed of the sulfur dioxide molecules is vrms = = 3 k PV nm R 3 (1.38 × 10−23 J/K )( 2.12 ×104 Pa )( 50.0 m3 ) = 343 m/s ( 421 mol ) (1.064 × 10−25 kg ) 8.31 J/ ( mol ⋅ K ) ______________________________________________________________________________ 42. REASONING Power is the rate at which energy is produced, according to Energy (Equation 6.10b), so the time required for the engine to produce a certain Power = t amount of energy is given by t= Energy Power (1) Assuming that helium (a monatomic gas) behaves as an ideal gas, its internal energy U can be found from U = 3 nRT (Equation 14.7), where n is the number of moles of helium in the 2 container, R = 8.31 J/ ( mol ⋅ K ) is the universal gas constant, and T is the Kelvin temperature. We will determine the quantity nRT in Equation 14.7 from the pressure P and volume V of the helium via the ideal gas law P V = nRT (Equation 14.1). 750 THE IDEAL GAS LAW AND KINETIC THEORY SOLUTION The engine must produce an amount of energy equal to the internal energy U = 3 nRT (Equation 14.7) of the helium, so from Equation (1) we have that 2 3 nRT Energy U t= = =2 Power Power Power (2) Substituting P V = nRT (Equation 14.1) into Equation (2) yields t= 3 nRT 2 Power = 3PV ( Power ) 2 (3) Using the equivalence 1 hp = 746 W in Equation (3), we obtain t= 43. 3PV 3 ( 6.2 ×105 Pa ) ( 0.010 m3 ) = = 5.0 × 101 s ( Power ) 746 W 2 2 0.25 hp 1 hp ( ) SSM WWW REASONING AND SOLUTION a. Assuming that the direction of travel of the bullets is positive, the average change in momentum per second is ∆p/∆t = m∆v/∆t = (200)(0.0050 kg)[(0 m/s) – 1200 m/s)]/(10.0 s) = –120 N b. The average force exerted on the bullets is F = ∆p/∆t. According to Newton’s third law, the average force exerted on the wall is − F = 120 N . c. The pressure P is the magnitude of the force on the wall per unit area, so P = (120 N)/(3.0 × 10–4 m2) = 4.0 × 10 5 Pa ______________________________________________________________________________ 44. REASONING Because the container holds 1.000 mol of neon, we know that the number of neon atoms inside is equal to Avogadro’s number: N = NA. These NA atoms are continually bouncing back and forth between the walls of the cubical container, with an rms speed vrms 2 3 found from 1 mvrms = 2 kT (Equation 14.6), where m is the mass of a single neon atom, k is 2 the Boltzmann constant, and T is the Kelvin temperature. Following the development of kinetic theory given in Section 1...
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