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Get Full Access to University Physics - 13 Edition - Chapter 18 - Problem 37e
Get Full Access to University Physics - 13 Edition - Chapter 18 - Problem 37e

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# Oxygen (O2) has a molar mass of 32.0 g/mol. What is (a)

ISBN: 9780321675460 31

## Solution for problem 37E Chapter 18

University Physics | 13th Edition

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Problem 37E

Oxygen (O2) has a molar mass of 32.0 g/mol. What is (a) the average translational kinetic energy of an oxygen molecule at a temperature of 300 K; (b) the average value of the square of its speed; (c) the root-mean-square speed; (d) the momentum of an oxygen molecule traveling at this speed? (e) Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.10 m on a side. What is the average force the molecule exerts on one of the walls of the container? (Assume that the molecule’s velocity is perpendicular to the two sides that it strikes.) (f) What is the average force per unit area? (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm? (h) Compute the number of oxygen molecules that are contained in a vessel of this size at 300 K and atmospheric pressure. (i) Your answer for part (h) should be three times as large as the answer for part (g). Where does this discrepancy arise?

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Solution 37E Given, molar mass of oxygen M = 32.0 g/mol = 0.032 kg/mol Temperature T = 300 K (a) Average translational kinetic energy K tr= 2 × kBT , k iB Boltzmann’s constant K = 3 × 1.38 × 10 23J/K × 300 K tr 2 23 K tr= 450 × 1.38 × 10 J 23 K tr= 621 × 10 J K tr= 6.21 × 10 21J (b) The average value of the square of its speed, 2 3RT v av = M 2 3×8.314 J/mol.K×300 K v av = 0.032 kg/mol v 2 = 233831 m /s2 2 av (c) The root-mean -square speed, v rms= 233831 m/s v rms= 483.5 m/s 0.032 kg/mol (d) Momentum = m O 2 = 6.02×10/mol× 483.56 m/s Momentum = 2.57 × 10 23kg.m/s (e) otal distance moves by the oxygen molecule = 2 × 0.10 m = 0.20 m 4 Time taken by the molecule t = 0.20 m/(483.5 m/s) = 4.1 × 10 s Therefore, the average force = Momentum change/time 23 4 = 2 × (2.57 × 10 kg.m/s)/(4.1 × 10 s) 19 = 1.25 × 10 N 19 2 2 17 2 (f) Average force per unit area = F/A = 1.25 × 10 N/0.1 m = 1.25 × 10 N/m (g) Pressure P = 1 atm = 1.01 × 10 Pa 5 3 3 Volume V = (0.1) m 23 23 m = M/(6.02 × 10 ) = (0.032 kg/mol)/(6.02 × 10 /mol) v rms = 483.5 m/s Nmv 2 From the formula P = Vrms , we can calculate N, the number of molecules. Substituting the values from above in this equation, we get 2 1.01 × 10 =5 N×0.032×(483.5) 6.02×103×(0.1) N = 8.12 × 10 21 molecules 5 (h) P = 1 atm = 1.01 × 10 Pa T = 300 K Volume V = (0.1) m 3 3 R = 8.314 J/mol.K From the ideal gas equation, PV = nRT 5 3 3 1.01 × 10 Pa × (0.1) m = n × 8.314 J/K.mol × 300 K n = 0.04 mol n = 0.04 × 6.02 × 10 23 atoms 22 n = 2.40 × 10 atoms The number of molecules computed in part h is almost three times as large as that done in part g. Part g considers the root-mean-square speed of the molecules. Not all the molecules have the exact rms value of speed. But the part h considers all the molecules. This is the reason for the observed discrepancy.

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