Homework Problem Set 5 Solutions ( )2

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5. continued 14 c.) Calculate the most probable speed for atoms weighing 12.0 g/mol at 300 K and 1000 K. Express your answers in m/s. The most probable speed v mp is € v mp = ⎛⎛ ⎜⎜ ⎝⎝ 2k B T m 1/ 2 ⎞⎞ ⎟⎟ . ⎠⎠ At 300 K, the most probable speed is therefore € 2 1.38062 × 10 −23 JK −1 1/ 2 ⎡⎡ ( )( 300 K) ⎤⎤ v mp = ⎢⎢ ⎥⎥ ⎢⎢ ( 0.012 kg/mol) ( 1mol/6.02217 × 10 ⎣⎣ 23 ) ⎥⎥ ⎦⎦ v mp = 645 m/s. At 1000 K, the most probable speed is € 2 1.38062 × 10 −23 JK −1 1/ 2 ⎡⎡ ( )( 1000 K) ⎤⎤ v mp = ⎢⎢ ⎥⎥ ⎢⎢ ( 0.012 kg/mol) ( 1mol/6.02217 × 10 ⎣⎣ 23 ) ⎥⎥ ⎦⎦ v mp = 1180 m/s. d.) From the most € probable speeds calculated in part (c), determine the most probable kinetic energy of one mole of atoms at 300 K and 1000 K. Express your answers in J/mol. The most probable kinetic energy KE mp is given by KE mp = 1 2 m v 2 mp . At 300 K, the most probable kinetic energy is € KE mp = 1 ( 0.012 kg/mol )( 2 645m/s)2 KE mp = 2490 J/mol. At 1000 K, the most probable kinetic energy is € KE mp = 1 0.012 kg/mol 2 ( ) ( 1180 m/s )2 KE mp = 8310 J/mol. €
15 6. The equipartition theorem states that each degree of freedom contributes 1 2 k BT to the internal energy of a particle (or 1 2 RT to the internal energy of one mole of particles). a.) Use the equipartition theorem to calculate the internal energy of a mole of atoms at 300 K. Express your answer in J/mol. Since an atom has only three translational degrees of freedom (and no vibrational or rotational degrees of freedom), the internal energy U for one mole of atoms at 300 K is U = 3 2 RT ( )( 300 K) = 3 8.314 J/molK 2 U = 3740 J/mol. b.) Repeat the calculation € of the internal energy for a mole of atoms at 1000 K. your answer in J/mol. Express The internal energy U for one mole of atoms at 1000 K is € U = 3 2 RT ( )( 1000 K) = 3 8.314 J/molK 2 U = 12500 J/mol. c.) Compare your results from € parts (a) and (b) to the kinetic energy determined in part 5(d). Are the values similar The internal energy values are similar in magnitude to the most probably kinetic energy values, 3740 vs. 2490 J/mol at 300 K, and 12500 vs. 8310 J/mol at 1000 K. The internal energy values are a little lower than the most probable kinetic energy. This is because the distribution of speeds is skewed to higher speeds; therefore, the average speed is greater than the most probable speed. The equipartition theorem provides a measure of the average kinetic energy of the system rather than the most probable kinetic energy; thus, the internal energy computed from the equipartition theorem is larger than the most probable kinetic energy.
Page 1 and 2: Chemistry 380.37 Dr. Jean M. Standa
Page 3 and 4: 1 d.) continued 3 To complete the d
Page 5 and 6: 2. continued 5 b.) What is the kine
Page 7 and 8: 3. continued 7 b.) Using the analyt
Page 9 and 10: 4. Again consider the simple one-di
Page 11 and 12: 4 a.) continued 11 Averaging the se
Page 13: 5a, b.) continued 13 The velocity d

Homework Problem Set 5 Solutions ( )2

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