I need help with B E and F HW 20Exercise 18.33 Enhanced with Solution Resources

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I need help with B E and F

HW 20Exercise 18.33 Enhanced with Solution Resources previous| 1 of 11 | next » Exercise 18.33 - Enhanced with Solution You may want to review ( pages 593-599) For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Molecular kinetic energy and vrms Part A Oxygen 02 has a molar mass of 32.0 g/mol. What is the average translational kinetic energy of an oxygen molecule at a temperature of 299 K? | K= 6.19x10-21 Submit My Answers Give Up Correct IDENTIFY and SET UP: Apply the analysis of Section 18.3 EXECUTE: m(v2),,-3 k7-3 (1.38 × 10-23 J/molecule . K) (299K)-619 x 10-21 J MasteringPhysics: HW20-Google Chrome Part B Secure l https://session.masteringphysics.com/myct/item.. What is the average value of the square of its speed? Submitted Answers ANSWER 1: Deduction:-3% (H)av = 480 (H),v= 1.07.105 m2/s2 Submit My Answers G ANSWER 2: Deduction:-3% Incorrect, Iry Again, 3 attempts remaining =444 m Part C ANSWER 3: Deduction:-3% What is the root-mean-square speed? - 480 m urns= 483 m/s Submit My Answers Give Up Correct

Explanation / Answer

(b)

the mass of m of one atom is

m = M/ NA = 32 * 10^-3/6.022 * 10^23 = 5.314 * 10^-26 kg

1/2 * mv ( v avg)^2 = 6.19 * 10^-21 J

vavg^2 = 2 ( 6.19 * 10^-21 J)/5.314 * 10^-26 kg

=2.329 * 10^5 m^2/s^2

(e)

the time betwen collisions with one wall is

t= d/v rms = 0.30 m/483 = 6.21 * 10^-4 s

in a collision v chnages direction

del p = 2 m v rms = 2( 2.57 * 10^-23) = 5.14 * 10^-23 kg m/s

F = dP/ dt = 5.14 * 10^-23 kg m/s/ 6.21 * 10^-4 s = 0.828 * 10^-19 N

(f)

)pressure P = F/A = 0.828 * 10^-19 N/(0.15)^2 = 36.8 * 10^-19 Pa

rounding to two significant digits P = 37 * 10^-19 Pa

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