551_PartUniversity Physics Solution

551_PartUniversity Physics Solution - Thermal Properties of...

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Thermal Properties of Matter 18-9 18.40. IDENTIFY: rms 3 kT v m = . SET UP: 23 1.381 10 J/molecule K. k EXECUTE: (a) 23 3 rms 16 3(1.381 10 J/molecule K)(300 K) 6.44 10 m/s 6.44 mm/s 3.00 10 kg v ×⋅ == × = × EVALUATE: (b) No. The rms speed depends on the average kinetic energy of the particles. At this T , H 2 molecules would have larger v rms than the typical air molecules but would have the same average kinetic energy and the average kinetic energy of the smoke particles would be the same. 18.41. IDENTIFY: Use Eq.(18.24), applied to a finite temperature change. SET UP: 5/ 2 V CR = for a diatomic ideal gas and 3/ 2 V = for a monatomic ideal gas. EXECUTE: (a) () 5 2 V Qn CTnR T = Δ 5 2 (2.5 mol) (8.3145 J/mol K)(30.0 K) 1560 J Q =⋅ = (b) 3 2 V = Δ 3 2 (2.5 mol) (8.3145 J/mol K)(30.0 K) 935 J Q = EVALUATE: More heat is required for the diatomic gas; not all the heat that goes into the gas appears as translational kinetic energy, some goes into energy of the internal motion of the molecules (rotations). 18.42. IDENTIFY: The heat Q added is related to the temperature increase T Δ by . V CT = Δ SET UP: For 2 H , 2 ,H 20.42 J/mol K V C and for Ne (a monatomic gas), ,Ne 12.47 J/mol K. V C = EXECUTE: constant V Q n Δ= = , so 22 H Ne . VV Δ= Δ 2 2 Ne H 20.42 J/mol K (2.50 C ) 4.09 C 12.47 J/mol K V V C TT C ⎛⎞ Δ= = ⎜⎟ ⎝⎠ °° . EVALUATE: The same amount of heat causes a smaller temperature increase for 2 H since some of the energy input goes into the internal degrees of freedom. 18.43. IDENTIFY: CM c = , where C is the molar heat capacity and c is the specific heat capacity. . m pV nRT RT M SET UP: 2 3 N 2(14.007 g/mol) 28.014 10 kg/mol M × . For water, w 4190 J/kg K c = . For 2 N, 20.76 J/mol K V C . EXECUTE: (a) 2 N 3 20.76 J/mol K 741 J/kg K C c M = × . 2 w N 5.65 c c = ; w c is over five time larger. (b) To warm the water, 4 w (1.00 kg)(4190 J/mol K)(10.0 K) 4.19 10 J Qm cT = = × . For air, 2 4 N 5.65 kg (741 J/kg K)(10.0 K) Q m × = Δ⋅ . 3 35 (5.65 kg)(8.314 J/mol K)(293 K) 4.85 m (28.014 10 kg/mol)(1.013 10 Pa) mRT V Mp = ×× . EVALUATE: c is smaller for 2 N , so less heat is needed for 1.0 kg of 2 N than for 1.0 kg of water. 18.44. (a) IDENTIFY and SET UP: 1 2 R contribution to V C for each degree of freedom. The molar heat capacity C is related to the specific heat capacity c by . c = EXECUTE: 1 2 6 3 3(8.3145 J/mol K) 24.9 J/mol K. V R = = The specific heat capacity is 3 / (24.9 J/mol K)/(18.0 10 kg/mol) 1380 J/kg K. cCM × = (b) For water vapor the specific heat capacity is 2000 J/kg K. c = The molar heat capacity is 3 (18.0 10 kg/mol)(2000 J/kg K) 36.0 J/mol K. c × = EVALUATE: The difference is 36.0 J/mol K 24.9 J/mol K 11.1 J/mol K, −⋅ which is about ( ) 1 2 2.7 ; R the vibrational degrees of freedom make a significant contribution. 18.45. IDENTIFY: 3 V = gives V C in units of J/mol K . The atomic mass M gives the mass of one mole. SET UP: For aluminum, 3 26.982 10 kg/mol. M EXECUTE: (a) 3 24.9 J/mol K V . 3 24.9 J/mol K 923 J/kg K 26.982 10 kg/mol V c = × .
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551_PartUniversity Physics Solution - Thermal Properties of...

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