Question

Name CHEM 108 Section KINETIC MOLECULAR THEORY HOMEWORK Part 2 Directions: Answer the following questions, then check your work by completing the kinetic Molecular Theory Homework (Part 1) found in Canvas. You can either accept your grade for the Canvas version or turn this in to your TA at CHEM 108 class on Tuesday. In Activities IV and V. you spent some time learning how to interpret distribution functions and to use them to determine the rms speed of gas molecules as well as to compare the rms speed and thus average kinetic energy of a gas at different temperatures and several gases at one temperature. Here we are going to help you take these concepts one step further - - % of molecules - - 1000 Molecular speed (m/s) Five gases are in a container at a temperature of 25°C with a pressure of 1.0 atm. There are 0.10 moles of each HCI, CCL, CO,. Ne, and Oz. B a) Which of these would have the highest average kinetic energy? A. Ne CO, CCL D. They all have the same average kinetic energy. There is not enough information to answer this question. b) Calculate the volume that each of these gases occupies. Show your work PV = nRT c) Match the five gases with the curves in the plot above d) What is the rms speed of O.? Show your work or explain how you determined it. 2019-2020

Answer 1

Q1. Ne

According to the formula of average K.E. = 3 /2 nRT

= 3 / 2 w RT/M

As Aver. K.E. $\alpha$ 1 / Molar mass

If molar mass will increase then the Average K.E. will decreases and vice-versa.

Mall - Boltzmann carrer % meleb cel - Molecular speed (m/ree) mano hop highest peak of low got molar molecular speed

Tz25c, pot atm noo-lo moles oma ZRT. -8.59 kg - 222) Kole UM Makmal vol of youus PV-ART VwxRxT нур we mans of gas, M. Molar Mass. - 0.10X B. 314x298 (T-273+25 298K - 247.751

As all the gant have equal presure, equel no. of molep, same temperature so volume will be o equal. Vom ince) = 34 3.314 298 neda )+ 35.5. 36.5] V 365 x 10-3 M +36-5x 10 kg/mol 203.63 Hom /se... 451.6 msec [Cela e 12 + 443815 Vrme (cl) = 3x8.314 7 298 C F 15413 =1643m M = 154x18 kg/mol 10.154 220.03 free

Verme (coo) | 3RT c o2 = 44 g/mol +44 x 10 kg/mol 3x 3.31498 44 x 10-3 7455.96 411-64 / ne 0.04 Very (No) Swee lo amot] wlox 10-18ml 13x8.314x298 10 x 10-3 17455.96 v0.olo - 863.47 m/rec

10 • 32gfool 38 31498 32 > 10-3 ✓ 7455.96 0.032 482.69 mere - Average kinetic energy - K= f'm things X = 3 MRT 3 WARXT Aver Kinetic energy & New I maket