Masteringchemistry with Pearson Etext -- Standalone Access Card -- For Chemistry
Masteringchemistry with Pearson Etext -- Standalone Access Card -- For Chemistry
3rd Edition
ISBN: 9780321806383
Author: Nivaldo J. Tro
Publisher: PEARSON
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Chapter 5, Problem 1SAQ
Interpretation Introduction

To determine: The pressure (in atm) when volume of the sample is decreased to 225 mL.

Expert Solution & Answer
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Answer to Problem 1SAQ

Solution: (b) 1.60 atm.

Explanation of Solution

Given:

Initial pressure, P1 = 547 mmHg

Initial volume, V1 = 0.500 L

Final volume, V2 =225mL=0.225L(1L=1000mL)

According to Boyle’s law,

P1V1=P2V2

Substitute all the given values in the above equation to get

547mmHg×0.500L=P2×0.225L

 P2=547mmHg×0.500L0.225LP2=273.50.225mmHgP2=1215.55mmHg

To convert 1215.5mmHg into atmospheres, use the following relation:

760 mmHg=1 atm

or 1 mmHg=1760atm

1215.5 mmHg=1215.5760 atm 1.60 atm

Therefore, the final pressure is 1.60atm.

Conclusion

The pressure at which the volume of the sample is decreased to 225mL is 1.60atm.

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Chapter 5 Solutions

Masteringchemistry with Pearson Etext -- Standalone Access Card -- For Chemistry

Ch. 5 - Prob. 11SAQCh. 5 - Prob. 12SAQCh. 5 - Prob. 13SAQCh. 5 - Prob. 14SAQCh. 5 - Prob. 15SAQCh. 5 - Prob. 1ECh. 5 - Prob. 2ECh. 5 - Prob. 3ECh. 5 - Prob. 4ECh. 5 - Prob. 5ECh. 5 - Prob. 6ECh. 5 - Prob. 7ECh. 5 - Prob. 8ECh. 5 - Prob. 9ECh. 5 - Prob. 10ECh. 5 - Prob. 11ECh. 5 - Prob. 12ECh. 5 - Prob. 13ECh. 5 - Prob. 14ECh. 5 - Prob. 15ECh. 5 - Prob. 16ECh. 5 - Prob. 17ECh. 5 - Prob. 18ECh. 5 - Prob. 19ECh. 5 - Prob. 20ECh. 5 - Prob. 21ECh. 5 - Prob. 22ECh. 5 - Prob. 23ECh. 5 - Prob. 24ECh. 5 - Prob. 25ECh. 5 - Prob. 26ECh. 5 - Prob. 27ECh. 5 - Prob. 28ECh. 5 - 29. Given a barometric pressure of 762.4 mmHg,...Ch. 5 - Prob. 30ECh. 5 - Prob. 31ECh. 5 - Prob. 32ECh. 5 - Prob. 33ECh. 5 - Prob. 34ECh. 5 - Prob. 35ECh. 5 - Prob. 36ECh. 5 - Prob. 37ECh. 5 - Prob. 38ECh. 5 - Prob. 39ECh. 5 - Prob. 40ECh. 5 - Prob. 41ECh. 5 - Prob. 42ECh. 5 - Prob. 43ECh. 5 - Prob. 44ECh. 5 - Prob. 45ECh. 5 - Prob. 46ECh. 5 - Prob. 47ECh. 5 - Prob. 48ECh. 5 - Prob. 49ECh. 5 - Prob. 50ECh. 5 - Prob. 51ECh. 5 - Prob. 52ECh. 5 - Prob. 53ECh. 5 - Prob. 54ECh. 5 - Prob. 55ECh. 5 - Prob. 56ECh. 5 - Prob. 57ECh. 5 - Prob. 58ECh. 5 - Prob. 59ECh. 5 - Prob. 60ECh. 5 - Prob. 61ECh. 5 - Prob. 62ECh. 5 - Prob. 63ECh. 5 - 64. A 275-mL flask contains pure helium at a...Ch. 5 - Prob. 65ECh. 5 - Prob. 66ECh. 5 - Prob. 67ECh. 5 - Prob. 68ECh. 5 - Prob. 69ECh. 5 - Prob. 70ECh. 5 - Prob. 71ECh. 5 - Prob. 72ECh. 5 - Prob. 73ECh. 5 - Prob. 74ECh. 5 - Prob. 75ECh. 5 - Prob. 76ECh. 5 - Prob. 77ECh. 5 - 78. Ozone is depleted in the stratosphere by...Ch. 5 - Prob. 79ECh. 5 - Prob. 80ECh. 5 - Prob. 81ECh. 5 - Prob. 82ECh. 5 - Prob. 83ECh. 5 - Prob. 84ECh. 5 - Prob. 85ECh. 5 - Prob. 86ECh. 5 - Prob. 87ECh. 5 - Prob. 88ECh. 5 - Prob. 89ECh. 5 - Prob. 90ECh. 5 - Prob. 91ECh. 5 - Prob. 92ECh. 5 - Prob. 93ECh. 5 - Prob. 94ECh. 5 - 95. Modern pennies are composed of zinc coated...Ch. 5 - Prob. 96ECh. 5 - Prob. 97ECh. 5 - Prob. 98ECh. 5 - Prob. 99ECh. 5 - Prob. 100ECh. 5 - Prob. 101ECh. 5 - Prob. 102ECh. 5 - Prob. 103ECh. 5 - Prob. 104ECh. 5 - Prob. 105ECh. 5 - Prob. 106ECh. 5 - Prob. 107ECh. 5 - Prob. 108ECh. 5 - Prob. 109ECh. 5 - Prob. 110ECh. 5 - Prob. 111ECh. 5 - Prob. 112ECh. 5 - Prob. 113ECh. 5 - Prob. 114ECh. 5 - Prob. 115ECh. 5 - Prob. 116ECh. 5 - Prob. 117ECh. 5 - Prob. 118ECh. 5 - Prob. 119ECh. 5 - Prob. 120ECh. 5 - Prob. 121ECh. 5 - Prob. 122ECh. 5 - Prob. 123ECh. 5 - Prob. 124ECh. 5 - Prob. 125ECh. 5 - Prob. 126ECh. 5 - Prob. 127ECh. 5 - Prob. 128ECh. 5 - Prob. 129ECh. 5 - Prob. 130ECh. 5 - Prob. 131ECh. 5 - Prob. 132ECh. 5 - Prob. 133ECh. 5 - Prob. 134ECh. 5 - Prob. 135ECh. 5 - Prob. 136ECh. 5 - Prob. 137ECh. 5 - Prob. 138ECh. 5 - Prob. 139ECh. 5 - Prob. 140ECh. 5 - Prob. 141ECh. 5 - Prob. 142ECh. 5 - Prob. 143ECh. 5 - Prob. 144ECh. 5 - Prob. 145ECh. 5 - Prob. 146ECh. 5 - Prob. 147E
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