Physics Solution Manual for 1100 and 2101

We also know that p1v1 p2v2 then v1 p2 p gh gh 1

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Unformatted text preview: 1 kg 3 m = 2.55 ×105 mol ( 4.002 60 g/mol ) = 1.0 × 10 kg 1000 g ______________________________________________________________________________ ( ) 17. REASONING The maximum number of balloons that can be filled is the volume of helium available at the pressure in the balloons divided by the volume per balloon. The volume of helium available at the pressure in the balloons can be determined using the ideal gas law. Since the temperature remains constant, the ideal gas law indicates that PV = nRT = constant, and we can apply it in the form of Boyle’s law, PiVi = PfVf. In this expression Vf is the final volume at the pressure in the balloons, Vi is the volume of the cylinder, Pi is the initial pressure in the cylinder, and Pf is the pressure in the balloons. However, we need to remember that a volume of helium equal to the volume of the cylinder will remain in the cylinder when its pressure is reduced to atmospheric pressure at the point when balloons can no longer be filled. Chapter 14 Problems 733 SOLUTION Using Boyle’s law we find that Vf = PVi i Pf The volume of helium available for filling balloons is Vf − 0.0031 m3 = PVi i − 0.0031 m3 Pf The maximum number of balloons that can be filled is N Balloons = PVi i − 0.0031 m3 Pf VBalloon = (1.6 ×107 Pa ) ( 0.0031 m3 ) − 0.0031 m3 1.2 × 105 Pa 0.034 m3 = 12 18. REASONING According to the ideal gas law, PV = nRT , the absolute pressure P is directly proportional to the temperature T when the volume is held constant, provided that the temperature is measured on the Kelvin scale. The pressure is not proportional to the Celsius temperature. SOLUTION a. According to Equation 14.1, the pressures at the two temperatures are P= 1 nRT1 V and P2 = nRT2 V Taking the ratio P2/P1 of the final pressure to the initial pressure gives nRT2 P2 T 70.0 K = V = 2= = 2.00 nRT1 P T1 35.0 K 1 V b. Before determining the ratio of the pressures, we must convert the Celsius temperature scale to the Kelvin temperature scale. Thus, the ratio of the pressures at the temperatures of 35.0 °C and 70.0 °C is P2 T ( 273.15 + 70.0 ) K = 1.11 = 2= P T1 ( 273.15 + 35.0 ) K 1 ______________________________________________________________________________ 734 THE IDEAL GAS LAW AND KINETIC THEORY 19. SSM REASONING According to the ideal gas law (Equation 14.1), PV = nRT . Since n, the number of moles, is constant, n1R = n2 R . Thus, according to Equation 14.1, we have PV1 1 T1 = P2V2 T2 SOLUTION Solving for T2 , we have P V 48.5 P V1 /16 1 T2 = 2 2 T1 = P V P V (305 K)= 925 K 1 1 1 1 ______________________________________________________________________________ 20. REASONING The pressure P of the water vapor in the container can be found from the ideal gas law, Equation 14.1, as P = nRT / V , where n is the number of moles of water, R is the universal gas constant, T is the Kelvin temperature, and V is the volume. The variables R, T, and V are known, and the number of moles can be obtained by noting that it is equal to the mass m of the water divided by its mass per mole. SOLUTION Substituting n = m/(Mass per mole) into the ideal gas law, we have m Mass per mole RT nRT P= = V V The mass per mole (in g/mol) of water (H2O) has the same numerical value as its molecular mass. The molecular mass of water is 2 (1.00794 u) + 15.9994 u = 18.0153 u. The mass per mole of water is 18.0153 g/mol. Thus, the pressure of the water vapor is 4.0 g m [8.31 J/ ( mol ⋅ K )] ( 388 K ) RT Mass per mole 18.0153 g /mol P= = = 2.4 × 104 Pa 3 V 0.030 m ____________________________________________________________________________________________ 21. REASONING According to Equation 14.2, PV = NkT , where P is the pressure, V is the volume, N is the number of molecules in the sample, k is Boltzmann's constant, and T is the Kelvin temperature. The number of gas molecules per unit volume in the atmosphere is N / V = P /( kT ) . This can be used to find the desired ratio for the two planets. Chapter 14 Problems 735 SOLUTION We have ( N / V ) Venus ( N / V ) Earth = PVenus / (kTVenus ) PVenus TEarth 9.0 × 10 6 Pa 320 K = 39 = = PEarth / (kTEarth ) PEarth T Venus 1.0 × 10 5 Pa 740 K Thus, we can conclude that the atmosphere of Venus is 39 times "thicker" than that of Earth. ______________________________________________________________________________ 22. REASONING Pushing down on the pump handle lowers the piston from its initial height hi = 0.55 m above the bottom of the pump cylinder to a final height hf. The distance d the biker must push the handle down is the difference between these two heights: d = hi − hf . The initial and final heights of the piston determine the initial and final volumes of air inside the pump cylinder, both of which are the product of the piston height h and its crosssectional area A: Vi = Ahi Vf = Ahf and (1) We will assume that the air in the cylinder may be treated as an ideal gas, and that compressing it produces no change in temperature. Since none of the air escapes from the cylinder up to the point where the pressure inside equals the pressure in the i...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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