Question

Assume that you have 40000 molecules of carbon dioxide in a box at a temperature of 300 K.  Also, imagine if the speed distribution of the molecules is governed by the following function

N(v) = Ave^{(-Mv^2/2RT)} instead of a Boltzmann distribution N(v) = Av²e^{(-Mv^2/2RT)}

Normalize this function to solve for A.  Also, solve for the v_rms of the carbon dioxide gas.

[Use .044 kg/mol for the molar mass of carbon dioxide]

Group of answer choices

a) A = .706, Vrms = 348.25 m/s

b) A = .588, Vrms = 322.67 m/s

c) A = .822, Vrms = 348.25 m/s

d) A = .706, Vrms = 336.71 m/s

e) A = .588, Vrms = 336.71 m/s

Answer 1

, M.-40000 x0. Given No. of molecules 40000 molar mass of CO2 = 0.044 kg total mass in 60.044 Griven distoöbutim fenetin MV2/2RT diw) to normal se this funtion @ distribution "SN (o) do al IAS - Ave C6 e "Savem du = 1 Mv2 -92 A fré ZRT dre. Il dal ㅗ pris da 922 A у е ame) a=M ART o A = aa 1 =) A 2 a A = 2 M PRT RT A 40000 x 0.044 > A = 1760 8.314 X 300 8.314 X300

vivemo Here for whole distribution they have considered MEM=1760 1760 kg A=07056 A=0.706 Normalised value of a and Rmy velonty (104) of Ä v. emelt drol 1 VO ค -Mu2 <voy - 22 4,20 @ art idre a3 2 B -an a da al za 2 e M22 die <V² = .A. I i3 e 22T Sep Ivedi -902 are s 2 RT 2 ZRT ZRT Wrons =/<0? =

7 ZRT To calculate Ring valece Yms = Here Mis only moleculel mass they have considered according to option: 2 x8.314 x 300 0.044 Ninas = 336070 m/s V my 0 A M Actual concept tis RT only but there according to optim JA m. RT we need to take this this mly original Avalece is Az 0.044 8.314x3n A 1.764 X105