Question 15 (1 point) Main Content Given is 1 mole of [(14N2)] molecules at atmospheric pressure and temperature of 277K. Calculate the logic of the number of molecules having their speed between 309m/s and (309+1) m/s. Round your answer to the nearest hundredth (i.e. if your answer is 32.567, round it to 32.57). Take Avogadro's Constant as 6.022x1023 and the Ideal Gas Constant R=8.31J/mol*K. Your Answer: Given that, Lino of moles of nitrogen 2 1 mol We know, Molan merss of nitrogen, M = 28 9/mol = 28x 10 Kq / mol. temperature , T = 277.K V12 309 m/s , V2 = 310 mys NA = 6. 02 2 X 10 23, R = 8. 31 J/ mol:4 Maxwell - moltzman distribution provides molecules velocity distribution. If f (v) is the velocity distribution function, f( v ) = 4 rcV 2 ( M 3/2 ( - MV 2 2 RT ) 2ART x e so, the number of molecules having speed between d(v) = ( v 2 - VI ) is, IN = f ( v ) d ( v ) N Now from -0 4 rev = 4x 3.14 x ( 309 . m. 5 )2 = 1 2x 106 mis - 2 M 13/2 28 x 10 - 3 2 RRT 7 312 2X 3. 14 x 8.31 J/ molike x 27 TK = [ 1. 94 x 10-6 7 3/2 34 2 - 207 x 10 m-35 3 MV 2 28 *10 kg/mol x ( 309 m. 37 ) 2 2 RT = = 0.58 2 x 8. 31 Jumat x 277 - MV 21 - 0.58 e 2 RT 2 e = 0.5599: From equ" -D f ( v ) = ( 1.2 *+06 m 3 - 2 ) x (2.7 x10 m's "m 35 3) x 0. 5599. 2 108 14 x 105 - 3 m's substituting all values in -1 we get- 2N = f( v ) dv x N 2 ( 1 . 814 x 10 3 m's ) XI mistx , mole + 23 2 10 814 x 10 x 6. 022 x 10 . molecules - 1.09 x 10 21 molecules. log,o ( no of particles 2 log,0 ( 1.09 x102 ) 2 21.037 N 21:04 The log go of the number of molecules having their youed between Fog ups anw. 310 mis 21.04 Ans