Physics Solution Manual for 1100 and 2101

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Unformatted text preview: ygen ) PV RT (1.01×105 Pa ) ( 2.5 m )( 4.0 m )(5.0 m ) = 0.79 ( 28.0 g/mol ) + 0.21( 32.0 g/mol ) 8.31 J/ ( mol ⋅ K ) ( 295 K ) = 5.9 ×104 g 26. REASONING The ideal gas law is PV = nRT. Since the number of moles is constant, this PV PV equation can be written as = nR = constant . Thus, the value of is the same T T initially and finally, and we can write P0V0 T0 = Pf Vf Tf This expression can be solved for the final temperature Tf. (1) Chapter 14 Problems 739 We have no direct data for the initial and final pressures. However, we can deal with this by realizing that pressure is the magnitude of the force applied perpendicularly to a surface divided by the area of the surface. Thus, the magnitudes of the forces that the initial and final pressures apply to the piston (and, therefore, to the spring) are given by Equation 11.3 as (2) F0 = P0 A and Ff = Pf A 14 244 4 3 14 244 4 3 Force applied by final pressure Force applied by initial pressure The forces in Equations (2) are applied to the spring, and the force FxApplied that must be applied to stretch an ideal spring by an amount x with respect to its unstrained length is given by Equation 10.1 as FxApplied = kx (3) where k is the spring constant. Lastly, we note that the final volume is the initial volume plus the amount by which the volume increases as the spring stretches. The increased volume due to the additional stretching is A ( xf − x0 ) . Therefore, we have Vf = V0 + A ( xf − x0 ) (4) SOLUTION The final temperature can be obtained by rearranging Equation (1) to show that PV (5) Tf = f f T0 PV 0 0 Into this result we can now substitute expressions for P0 and Pf. These expressions can be obtained by using Equations (2) in Equation (3) as follows (and recognizing that, for the initial and final forces, P0A = FxApplied and Pf A = FxApplied): P0 A = kx0 14243 and Pf A = kxf 14243 (6) Force applied to spring by final pressure Force applied to spring by initial pressure Thus, we find that P0 = kx0 A and Pf = kxf A (7) Finally, we substitute Equations (7) for the initial and final pressure and Equation (4) for the final volume into Equation (5). With these substitutions Equation (5) becomes 740 THE IDEAL GAS LAW AND KINETIC THEORY kxf A PV Tf = f f T0 = PV 0 0 = V0 + A ( xf − x0 ) T0 x V + A ( x − x ) T f 0 f 00 = kx0 x0V0 A V0 ( 0.1000 m ) 6.00 ×10−4 m3 + ( 2.50 ×10−3 m 2 ) ( 0.1000 m − 0.0800 m ) ( 273 K ) ( 0.0800 m ) ( 6.00 ×10−4 m3 ) = 3.70 × 102 K 27. SSM REASONING The graph that accompanies Problem 75 in Chapter 12 can be used to determine the equilibrium vapor pressure of water in the air when the temperature is 30.0 °C (303 K). Equation 12.6 can then be used to find the partial pressure of water in the air at this temperature. Using this pressure, the ideal gas law can then be used to find the number of moles of water vapor per cubic meter. SOLUTION According to the graph that accompanies Problem 75 in Chapter 12, the equilibrium vapor pressure of water vapor at 30.0 °C is approximately 4250 Pa. According to Equation 12.6, ( )( ) Partial 1 pressure of = Percent relative Equilibrium vapor pressure of × water at the existing temperature 100 humidity water vapor = ( 55 )( 4250 Pa ) = 2.34 × 103 Pa 100 The ideal gas law then gives the number of moles of water vapor per cubic meter of air as n P (2.34 × 10 3 Pa) 3 (14.1) = = = 0.93 mol/m V RT [8.31 J/(mol ⋅ K)](303 K) ______________________________________________________________________________ 28. REASONING AND SOLUTION If the pressure at the surface is P1 and the pressure at a depth h is P2, we have that P2 = P1 + ρ gh. We also know that P1V1 = P2V2. Then, V1 P2 P + ρ gh ρ gh = =1 = 1+ V2 P P P 1 1 1 Chapter 14 Problems 741 Therefore, V1 (1.000 × 103 kg/m3 )(9.80 m/s 2 )(0.200 m) = 1+ = 1.02 V2 1.01× 105 Pa ______________________________________________________________________________ 29. REASONING The desired percentage is the volume the atoms themselves occupy divided by the total volume that the gas occupies, multiplied by the usual factor of 100. The volume VAtoms that the atoms themselves occupy is the volume of an atomic sphere ( 4 π r 3 , where r 3 is the atomic radius), times the number of atoms present, which is the number n of moles times Avogadro’s number NA. The total volume VGas that the gas occupies can be taken to be that calculated from the ideal gas law, because the atoms themselves occupy such a small volume. SOLUTION The total volume VGas that the gas occupies is given by the ideal gas law as VGas = nRT/P, where the temperature and pressure at STP conditions are 273 K and 1.01 × 105 Pa. Thus, we can write the desired percentage as Percentage = = VAtoms VGas 4π 3 ( 4 π r 3 ) nN A ×100 = ( 4 π r 3 ) NA P ×100 3 3 ×100 = nRT P RT ( 2.0 ×10−10 m ) ( 6.022 ×1023 mol−1 )(1.01×105 Pa ) ×100 = 0.090 % 3 8.31 J/ ( mol ⋅ K ) ( 273 K ) 30. REASONING AND SOLUTION We need to determine the amount of He inside the balloon. We begin by using Archimedes’ principle; the balloon is being buoyed up by a force equal to the weight of the air displa...
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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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