Question

Part A According to the equipartition principle, what is the overall energy in J of one mole of water at 1400 K if translational, rotational, and vibrational degrees of freedom are considered? The values of the vibrational constants of water are roughly 1600 cm-and 3700 cm ). kg = 1.381 . 10-23 JK-1 R - 8.314 JK-1 mol-1 Calculate the higher limit for the overall energy in J of one mole of water at 1400 K assuming that all vibrations contribute completely to the number of equipartition energy contributions Nep Express your answer using two significant figures. 190 A¢ * O O ? Ehigh = Submit Request Answer Part B Calculate the lower limit for the overall energy in J of one mole of water at 1400 K by omitting the vibrations entirely. Express your answer using two significant figures. AEM O 2 ? Erow

Answer 1

Answer " The total no of degrees of freedom of water molecule - 3N (No no of atom) 2 3x3 No of translational motion - 3 No of rotational motion 3 (because H₂O. is nonlinear) No of Vibrobional motion (30-6) - (0-6) 23 The energy allotted to each translational motion 2 £ KGT KB- Bolte man const = 1.38181023 JKT Ta temperature ink The enegy allotted to each rotational motion Ź ke T The energy allotted to each vibrational motion z Kis T Total energy of water and lecale E todel? 3x1/2 KgT + 3x į KBT + 3 KBT - 9 kg7

1 At Talbook Eeter = 2 x 1.381X16*34" x 1400k -21 = 87.003 X 10 ] This is the higher limit of overall energy Parit B By Omiting vibrational Centerbudion entirely We get be get Etetal = 3 3 KST + 2 KBT 23KgT At Tz 1400k Staten = 3x11381x10 JK' x14006 258.002X102J This is the lower limit of the overall energy -23 -21