Bundle: Physical Chemistry, 2nd + Student Solutions Manual
Bundle: Physical Chemistry, 2nd + Student Solutions Manual
2nd Edition
ISBN: 9781285257594
Author: David W. Ball
Publisher: Cengage Learning
Question
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Chapter 17, Problem 17.57E
Interpretation Introduction

Interpretation:

The values of logarithm of N!, N=1 to 100, explicitly and by using Stirling’s approximation is to be calculated and the values are to be compared. The approximate value at which the Stirling’s approximation agrees with the true value to within 1% is to be stated.

Concept introduction:

Combination formula is a possible way of grouping of distinguishable objects into various subsystems. The equation for combination formula is as follows:

C=N!ni!i=1m

Where,

N is the object.

If the value of N is large, then the Stirling’s approximation method is used to estimate factorials. The Stirling’s approximation for large value of N is,

lnN!NlnNN

Expert Solution & Answer
Check Mark

Answer to Problem 17.57E

For N90 Stirling’s approximation agrees with the true value to within 1%.

Explanation of Solution

The Stirling’s approximation for large value of N is,

lnN!NlnNN …(1)

NlnN!NlnNN

10120.6931470.6137131.7917590.29583743.1780541.545177

54.7874923.0471966.5792514.75055778.5251616.621371810.60468.635532

912.8018310.775021015.1044113.025851117.5023115.376851219.9872117.81888

1322.5521620.344341425.1912222.94681527.8992725.620751630.6718628.36142

1733.5050731.164631836.3954534.026691939.3398836.944342042.3356239.91465

2145.3801442.934972248.4711846.002932351.6066849.116372454.7847352.27329

2558.0036155.47192661.261758.710512764.5575461.98762867.8897465.30173

2971.2570468.651583074.6582472.035923178.0922275.45363281.5579678.90355

3385.0544782.384753488.5808385.896263592.1361889.437183695.7196993.00668

3799.3306196.6039638102.9682100.228339106.6318103.878940110.3206107.5552

41114.0342111.256542117.7719114.982143121.5331118.731644125.3173122.5043

45129.1239126.299846132.9526130.117547136.8027133.956948140.6739137.8176

49144.5657141.699250148.4778145.601251152.4096149.523152156.3608153.4647

53160.3311157.425554164.3201161.405155168.3274165.403356172.3528169.4197

57176.3958173.453958180.4563177.505759184.5338181.574760188.6282185.6607

61192.739189.763362196.8662193.882363201.0093198.017564205.1682202.1685

65209.3426206.335266213.5322210.517267217.7369214.714468221.9564218.9265

69226.1905223.153370230.439227.394771234.7017231.650372238.9784235.92

73243.2688240.203574247.5729244.500875251.8904248.811676256.2211253.1357

77260.5649257.47378264.9216261.823379269.2911266.186480273.6731270.5621

81278.0676274.950482282.4743279.35183286.8931283.763884291.324288.1886

85295.7666292.625486300.2209297.073987304.6869301.53488309.1642306.0056

89313.6528310.488690318.1526314.982991322.6635319.488292327.1853324.0045

93331.7179328.531894336.2612333.069795340.8151337.618396345.3794342.1774

97349.9541346.74798354.5391351.326899359.1342355.9169100363.7394360.517

For N90 Stirling’s approximation agrees with the true value to within 1%. These values are calculated using the percentage difference between the lnN! and NlnNN for values N=1 to 100.

Conclusion

For N90 Stirling’s approximation agrees with the true value to within 1%.

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Chapter 17 Solutions

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