CMC CH001A Fall15 x C Chegg, The Student Hub x Li Thermodynamics Ch x x C fi WWW

Question

CMC CH001A Fall15 x C Chegg, The Student Hub x Li Thermodynamics Ch x x C fi WWW sapling learning.com /ibiscms/mod/ibis/view.php?id-1947117 Question 32 of 36 100 22 Map University Science Books 4th Edition General Chemistry 23 00 nted by Sapling L Donald McQ Peter A. Rock .Ethan Gallogly 24 90 Calculate the pressures, P, in atmospheres at which the mean free path, l, of a hydrogen molecule will be 3.50 um, 3.50 mm, and 3.50 m at 20.0 °C. The diameter of a H2 molecule is 270 pm. 25 95 3.50 mm 3.50 m 3.50 Hm 26 100 Number Number Number 27 99 P- atm atm atm 28 00 100 29 30 90 00 32 33 34 00 35 00 Previous 3 Give Up & View Solution Check Answer Next Exit Hin 36 00 HE Search the web and Windows

Explanation / Answer

To calculate the pressure we require following equation:
mean free path; l = k T/(sq root 2)(pi)d^2 P

k = Boltzmann constant , 1.38X10^-23 J/K
T = Kelvin temperature ,20+273=293 K
d = diameter; 270 pm or 2.7*10^-10 m
P = pressure in Pa

3.5 X 10^-6 m = 1.38X10^-23 J/K (293 K)/1.414(3.14)(3.50X10^-10)^2*P
P = 3.5 X 10^-6 / 7.434*10^-3 Pa

P = 4.70*10^-4 Pa

P =4.64 × 10-9 atm

3.5 X 10^-6 m = 1.38X10^-23 J/K (293 K)/1.414(3.14)(3.50X10^-9)^2*P
P = 3.5 X 10^-6 / 7.434*10^-5 Pa

P=0.0470 Pa

P =4.64 × 10-7 atm

3.5 X 10^-6 m = 1.38X10^-23 J/K (293 K)/1.414(3.14)(3.50)^2*P
P = 3.5 X 10^-6 / 7.434*10^-23 Pa

P = 4.70X 10^16 Pa

P = 4.64 × 10+11 atm

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