Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
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Chapter 13, Problem 96A
Interpretation Introduction

Interpretation:

The pressure of the gas is to be calculated for a given new volume when the temperature and amount of gas are constant.

Concept introduction:

When temperature and amount of gas are constant, pressure of gas is inversely proportional to the volume of the gas. This law is known as Boyle’s Law.

According to this law,

P1V1=P2V2

Where

P1 is initial pressure of the gas

V1 is initial volume of the gas

P2 is the final pressure of the gas

V2 is the final volume of the gas

Expert Solution & Answer
Check Mark

Answer to Problem 96A

The new pressure of the gas is 1.17atm.

Explanation of Solution

The change in volume on changing the pressure of the gas keeping the temperature fixed is studied by a law named as Boyle’s Law. According to this law, Product of pressure and volume of a given mass of gas is constant at constant temperature.

If V1 is the volume of the gas at pressure P1 and volume is increased to V2 Then pressure is decreased to P2. The formula used to calculate P2 is

P1V1=P2V2

Or

P2=P1V1V2

Given data is:

V1=1LP1=3.50atmV2=1+2=3L

On substituting the values in the formula:

P2=1×3.503 =1.17atm

Therefore new pressure is 1.17atm.

Conclusion

The pressure of the gas on opening the valve is 1.17atm.

Chapter 13 Solutions

Chemistry: Matter and Change

Ch. 13.1 - Prob. 11PPCh. 13.1 - Prob. 12PPCh. 13.1 - Prob. 13PPCh. 13.1 - Prob. 14SSCCh. 13.1 - Prob. 15SSCCh. 13.1 - Prob. 16SSCCh. 13.1 - Prob. 17SSCCh. 13.1 - Prob. 18SSCCh. 13.1 - Prob. 19SSCCh. 13.2 - Prob. 20PPCh. 13.2 - Prob. 21PPCh. 13.2 - Prob. 22PPCh. 13.2 - Prob. 23PPCh. 13.2 - Prob. 24PPCh. 13.2 - Prob. 25PPCh. 13.2 - Prob. 26PPCh. 13.2 - Prob. 27PPCh. 13.2 - Prob. 28PPCh. 13.2 - Prob. 29PPCh. 13.2 - Prob. 30PPCh. 13.2 - Prob. 31SSCCh. 13.2 - Prob. 32SSCCh. 13.2 - Prob. 33SSCCh. 13.2 - Prob. 34SSCCh. 13.2 - Prob. 35SSCCh. 13.2 - Prob. 36SSCCh. 13.2 - Prob. 37SSCCh. 13.3 - Prob. 38PPCh. 13.3 - Prob. 39PPCh. 13.3 - Prob. 40PPCh. 13.3 - Prob. 41PPCh. 13.3 - Prob. 42PPCh. 13.3 - Prob. 43PPCh. 13.3 - Prob. 44PPCh. 13.3 - Prob. 45PPCh. 13.3 - Prob. 46SSCCh. 13.3 - Prob. 47SSCCh. 13.3 - Prob. 48SSCCh. 13.3 - Prob. 49SSCCh. 13 - Prob. 50ACh. 13 - Prob. 51ACh. 13 - Prob. 52ACh. 13 - Prob. 53ACh. 13 - Prob. 54ACh. 13 - Prob. 55ACh. 13 - Prob. 56ACh. 13 - Prob. 57ACh. 13 - Prob. 58ACh. 13 - Prob. 59ACh. 13 - Prob. 60ACh. 13 - Prob. 61ACh. 13 - Prob. 62ACh. 13 - Prob. 63ACh. 13 - Prob. 64ACh. 13 - Prob. 65ACh. 13 - Prob. 66ACh. 13 - Prob. 67ACh. 13 - Prob. 68ACh. 13 - Prob. 69ACh. 13 - Prob. 70ACh. 13 - Prob. 71ACh. 13 - Prob. 72ACh. 13 - Prob. 73ACh. 13 - Prob. 74ACh. 13 - Prob. 75ACh. 13 - Prob. 76ACh. 13 - Prob. 77ACh. 13 - Prob. 78ACh. 13 - Prob. 79ACh. 13 - Prob. 80ACh. 13 - Prob. 81ACh. 13 - Prob. 82ACh. 13 - Prob. 83ACh. 13 - Prob. 84ACh. 13 - Prob. 85ACh. 13 - Prob. 86ACh. 13 - Prob. 87ACh. 13 - Prob. 88ACh. 13 - Prob. 89ACh. 13 - Prob. 90ACh. 13 - Prob. 91ACh. 13 - Prob. 92ACh. 13 - Prob. 93ACh. 13 - Prob. 94ACh. 13 - Oxygen Consumption If 5.00 L of hydrogen...Ch. 13 - Prob. 96ACh. 13 - If 2.33 L of propane at 24°C and 67.2 kPa is...Ch. 13 - Prob. 98ACh. 13 - Prob. 99ACh. 13 - Prob. 100ACh. 13 - Prob. 101ACh. 13 - Apply Calculate the pressure of (4.671022)...Ch. 13 - Analyze When nitroglycerin (C3H5N3O9) explodes,...Ch. 13 - Prob. 104ACh. 13 - Prob. 105ACh. 13 - Prob. 106ACh. 13 - Prob. 107ACh. 13 - Prob. 108ACh. 13 - Prob. 109ACh. 13 - Prob. 110ACh. 13 - Prob. 111ACh. 13 - Prob. 112ACh. 13 - Prob. 113ACh. 13 - Prob. 114ACh. 13 - Prob. 115ACh. 13 - Prob. 116ACh. 13 - Prob. 117ACh. 13 - Prob. 118ACh. 13 - Prob. 119ACh. 13 - Prob. 1STPCh. 13 - Prob. 2STPCh. 13 - Prob. 3STPCh. 13 - Prob. 4STPCh. 13 - Prob. 5STPCh. 13 - Prob. 6STPCh. 13 - Prob. 7STPCh. 13 - Prob. 8STPCh. 13 - Prob. 9STPCh. 13 - Prob. 10STPCh. 13 - Prob. 11STPCh. 13 - Prob. 12STPCh. 13 - Prob. 13STPCh. 13 - Prob. 14STP
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