Physics Solution Manual for 1100 and 2101

1 into equation 2 yields t 3 nrt 2 power 3pv power

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Unformatted text preview: ced. The buoyant force, Fb, therefore, is equal to 3 Fb = mg = ρ Vg = (1.19 kg/m3 ) 4 π (1.50 m ) ( 9.80 m/s 2 ) = 164.9 N 3 Since the material from which the balloon is made has a mass of 3.00 kg (weight = 29.4 N), the He inside the balloon weighs 164.9 N – 29.4 N = 135.5 N. Hence, the mass of the ( ) helium present in the balloon is m = (135.5 N ) / 9.80 m/s2 = 13.8 kg . Now we can determine the number of moles of He present in the balloon: n= m 13.8 kg = = 3450 mol M 4.0026 ×10−3 kg/mol 742 THE IDEAL GAS LAW AND KINETIC THEORY Using the ideal gas law to find the pressure, we have P= nRT ( 3450 mol ) [8.31 J/(mol ⋅ K) ] ( 305 K ) = = 6.19 × 105 Pa 4 π (1.50 m )3 V 3 ______________________________________________________________________________ 31. SSM REASONING According to the ideal gas law (Equation 14.1), PV = nRT . Since n, the number of moles of the gas, is constant, n1R = n2 R . Therefore, PV1 / T1 = P2V2 / T2 , 1 where T1 = 273 K and T2 is the temperature we seek. Since the beaker is cylindrical, the volume V of the gas is equal to Ad, where A is the cross-sectional area of the cylindrical volume and d is the height of the region occupied by the gas, as measured from the bottom of the beaker. With this substitution for the volume, the expression obtained from the ideal gas law becomes P d1 P2 d 2 1 = (1) T1 T2 where the pressures P and P2 are equal to the sum of the atmospheric pressure and the 1 pressure caused by the mercury in each case. These pressures can be determined using Equation 11.4. Once the pressures are known, Equation (1) can be solved for T2 . SOLUTION Using Equation 11.4, we obtain the following values for the pressures P and 1 P2 . Note that the initial height of the mercury is h1 = 1 (1.520 m) = 0.760 m , while the final 2 height of the mercury is h2 = 1 (1.520 4 m) = 0.380 m . P = P0 + ρ gh1 = (1.01× 105 Pa ) + (1.36 ×104 kg/m3 )(9.80 m/s 2 )(0.760 m) = 2.02 ×105 Pa 1 P2 = P0 + ρ gh2 = (1.01×105 Pa ) + (1.36 ×104 kg/m3 ) ( 9.80 m/s 2 ) (0.380 m) = 1.52 × 105 Pa In these pressure calculations, the density of mercury is ρ = 1.36 × 104 kg/m3 . In Equation (1) we note that d1 = 0.760 m and d 2 = 1.14 m . Solving Equation (1) for T2 and substituting values, we obtain (1.52 × 105 Pa ) (1.14 m ) Pd T2 = 2 2 T1 = (273 K) = 308 K Pd 5 ( 2.02 ×10 Pa ) ( 0.760 m ) 11 ______________________________________________________________________________ Chapter 14 Problems 743 32. REASONING AND SOLUTION The volume of the cylinder is V = AL where A is the cross-sectional area of the piston and L is the length. We know P1V1 = P2V2 so that the new pressure P2 can be found. We have V AL L P2 = P 1 = P 1 1 = P 1 1 1 1 V2 A2 L2 L2 (since A1 = A2 ) L 4 = (1.01×105 Pa ) = 5.05 ×10 Pa 2L The force on the piston and spring is, therefore, F = P2A = (5.05 × 104 Pa)π (0.0500 m)2 = 397 N The spring constant is k = F/x (Equation 10.1), so F 397 N = = 1.98 ×103 N/m x 0.200 m ______________________________________________________________________________ k= 33. REASONING The average kinetic energy per molecule is proportional to the Kelvin temperature of the carbon dioxide gas. This relation is expressed by Equation 14.6 as 2 3 1 2 mvrms = 2 k T , where m is the mass of a carbon dioxide molecule. The mass m is equal to the molecular mass of carbon dioxide (44.0 u), expressed in kilograms. SOLUTION Solving Equation 14.6 for the temperature of the gas, we have 1.66 × 10−27 kg 2 44.0 u ( 650 m /s ) 2 1u mv T = rms = = 750 K −23 3k 3 1.38 × 10 J/K ( ) ( ) ______________________________________________________________________________ 34. REASONING According to the kinetic theory of gases, the average kinetic energy of an 2 3 atom is related to the temperature of the gas by 1 mvrms = 2 kT (Equation 14.6). We see that 2 the temperature is proportional to the product of the mass and the square of the rms-speed. 2 Therefore, the tank with the greatest value of mvrms has the greatest temperature. Using the information from the table given with the problem statement, we see that the values of 2 mvrms for each tank are: 744 THE IDEAL GAS LAW AND KINETIC THEORY Product of the mass and the square of the rms-speed 2 mvrms Tank A m ( 2vrms ) = 4mvrms 2 B 2 C 2 2 ( 2m ) vrms = 2mvrms D 2m ( 2vrms ) = 8mvrms 2 2 Thus, tank D has the greatest temperature, followed by tanks B, C, and A. SOLUTION The temperature of the gas in each tank can be determined from Equation 14.6: TA = TB = TC 2 mvrms (3.32 × 10 kg ) (1223 m /s ) = 3 (1.38 × 10 J /K ) −26 −23 3k ( m 2vrms 3k TD = ( 2 2m 2vrms 3k = 1200 K ) = (3.32 × 10 kg ) ( 2 × 1223 m /s ) 3 (1.38 × 10 J /K ) 2 ( 3.32 × 10 kg ) (1223 m /s ) = = 3 (1.38 × 10 J /K ) −26 2 −23 −26 2 ( 2m ) vrms = 3k 2 2 −23 ) 2 = ( 2 3.32 × 10 ( = 4800 K −26 ) 2400 K kg ( 2 ×1223 m /s ) 3 1.38 × 10 −23 J /K ) 2 = 9600 K These results confirm the conclusion reached in the REASONING. ______________________________________________________________________________ 35. REASON...
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