Explanation Given: Height of the cylinder is 22 .6 m , temperature is 22 o C , and new height is 23 .8 m . Explanation: Assume the pressure and amount of the gas is constant. Assuming the radius to be r , calculate the volume of the cylinder on first day ( V 1 ) as shown below: V 1 = π r 2 h …… (1) Substitute 3.14 for π and 22.6 m for h . Thus V 1 = 3 .14× r 2 ×22 .6 m = ( 70 .96 m ) r 2 Use equation (1) to calculate the volume of the cylinder on the second day ( V 2 ) with height 23.8 m : V 2 = 3 .14× r 2 ×23 .8 m = ( 74 .73 m ) r 2 The temperature is given in degree Celsius ( o C )

Chemistry: A Molecular Approach, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (4th Edition)

4th Edition

ISBN: 9780134162454

Author: Nivaldo J. Tro

Publisher: PEARSON

See similar textbooks

Concept explainers

Ideal and Real Gases

Ideal gases obey conditions of the general gas laws under all states of pressure and temperature. Ideal gases are also named perfect gases. The attributes of ideal gases are as follows,

Gas Laws

Gas laws describe the ways in which volume, temperature, pressure, and other conditions correlate when matter is in a gaseous state. The very first observations about the physical properties of gases was made by Robert Boyle in 1662. Later discoveries we…

Gaseous State

It is well known that matter exists in different forms in our surroundings. There are five known states of matter, such as solids, gases, liquids, plasma and Bose-Einstein condensate. The last two are known newly in the recent days. Thus, the detailed fo…

Question

Chapter 5, Problem 121E

Interpretation Introduction

To calculate: The temperature on the second day.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5, Problem 120E

Chapter 5, Problem 122E

Students have asked these similar questions

4. Predict the major product(s) for each of the following reactions. HBr (1 equiv.) peroxide, A a. b. NBS, peroxide, A

In addition to the separation techniques used in this lab (magnetism, evaporation, and filtering), there are other commonly used separation techniques. Some of these techniques are:Distillation – this process is used to separate components that have significantly different boiling points. The solution is heated and the lower boiling point substance is vaporized first. The vapor can be collected and condensed and the component recovered as a pure liquid. If the temperature of the mixture is then raised, the next higher boiling component will come off and be collected. Eventually only non-volatile components will be left in the original solution.Centrifugation – a centrifuge will separate mixtures based on their mass. The mixture is placed in a centrifuge tube which is then spun at a high speed. Heavier components will settle at the bottom of the tube while lighter components will be at the top. This is the technique used to separate red blood cells from blood plasma.Sieving – this is…

Briefly describe a eutectic system.

Chapter 5 Solutions

Chemistry: A Molecular Approach, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (4th Edition)

Ch. 5 - Prob. 1SAQ Ch. 5 - Prob. 2SAQ Ch. 5 - Prob. 3SAQ Ch. 5 - Prob. 4SAQ Ch. 5 - Prob. 5SAQ Ch. 5 - Prob. 6SAQ Ch. 5 - Prob. 7SAQ Ch. 5 - Prob. 8SAQ Ch. 5 - Q9. A gas sample at STP contains 1.15 g oxygen and...Ch. 5 - Prob. 10SAQ

Ch. 5 - Prob. 11SAQ Ch. 5 - Prob. 12SAQ Ch. 5 - Prob. 13SAQ Ch. 5 - Prob. 14SAQ Ch. 5 - Prob. 15SAQ Ch. 5 - Prob. 1E Ch. 5 - Prob. 2E Ch. 5 - Prob. 3E Ch. 5 - Prob. 4E Ch. 5 - Prob. 5E Ch. 5 - Prob. 6E Ch. 5 - Prob. 7E Ch. 5 - Prob. 8E Ch. 5 - Prob. 9E Ch. 5 - Prob. 10E Ch. 5 - Prob. 11E Ch. 5 - Prob. 12E Ch. 5 - Prob. 13E Ch. 5 - Prob. 14E Ch. 5 - Prob. 15E Ch. 5 - Prob. 16E Ch. 5 - Prob. 17E Ch. 5 - Prob. 18E Ch. 5 - Prob. 19E Ch. 5 - Prob. 20E Ch. 5 - Prob. 21E Ch. 5 - Prob. 22E Ch. 5 - Prob. 23E Ch. 5 - Prob. 24E Ch. 5 - Prob. 25E Ch. 5 - Prob. 26E Ch. 5 - Prob. 27E Ch. 5 - Prob. 28E Ch. 5 - 29. Given a barometric pressure of 762.4 mmHg,...Ch. 5 - Prob. 30E Ch. 5 - Prob. 31E Ch. 5 - Prob. 32E Ch. 5 - Prob. 33E Ch. 5 - Prob. 34E Ch. 5 - Prob. 35E Ch. 5 - Prob. 36E Ch. 5 - Prob. 37E Ch. 5 - Prob. 38E Ch. 5 - Prob. 39E Ch. 5 - Prob. 40E Ch. 5 - Prob. 41E Ch. 5 - Prob. 42E Ch. 5 - Prob. 43E Ch. 5 - Prob. 44E Ch. 5 - Prob. 45E Ch. 5 - Prob. 46E Ch. 5 - Prob. 47E Ch. 5 - Prob. 48E Ch. 5 - Prob. 49E Ch. 5 - Prob. 50E Ch. 5 - Prob. 51E Ch. 5 - Prob. 52E Ch. 5 - Prob. 53E Ch. 5 - Prob. 54E Ch. 5 - Prob. 55E Ch. 5 - Prob. 56E Ch. 5 - Prob. 57E Ch. 5 - Prob. 58E Ch. 5 - Prob. 59E Ch. 5 - Prob. 60E Ch. 5 - Prob. 61E Ch. 5 - Prob. 62E Ch. 5 - Prob. 63E Ch. 5 - 64. A 275-mL flask contains pure helium at a...Ch. 5 - Prob. 65E Ch. 5 - Prob. 66E Ch. 5 - Prob. 67E Ch. 5 - Prob. 68E Ch. 5 - Prob. 69E Ch. 5 - Prob. 70E Ch. 5 - Prob. 71E Ch. 5 - Prob. 72E Ch. 5 - Prob. 73E Ch. 5 - Prob. 74E Ch. 5 - Prob. 75E Ch. 5 - Prob. 76E Ch. 5 - Prob. 77E Ch. 5 - 78. Ozone is depleted in the stratosphere by...Ch. 5 - Prob. 79E Ch. 5 - Prob. 80E Ch. 5 - Prob. 81E Ch. 5 - Prob. 82E Ch. 5 - Prob. 83E Ch. 5 - Prob. 84E Ch. 5 - Prob. 85E Ch. 5 - Prob. 86E Ch. 5 - Prob. 87E Ch. 5 - Prob. 88E Ch. 5 - Prob. 89E Ch. 5 - Prob. 90E Ch. 5 - Prob. 91E Ch. 5 - Prob. 92E Ch. 5 - Prob. 93E Ch. 5 - Prob. 94E Ch. 5 - 95. Modern pennies are composed of zinc coated...Ch. 5 - Prob. 96E Ch. 5 - Prob. 97E Ch. 5 - Prob. 98E Ch. 5 - Prob. 99E Ch. 5 - Prob. 100E Ch. 5 - Prob. 101E Ch. 5 - Prob. 102E Ch. 5 - Prob. 103E Ch. 5 - Prob. 104E Ch. 5 - Prob. 105E Ch. 5 - Prob. 106E Ch. 5 - Prob. 107E Ch. 5 - Prob. 108E Ch. 5 - Prob. 109E Ch. 5 - Prob. 110E Ch. 5 - Prob. 111E Ch. 5 - Prob. 112E Ch. 5 - Prob. 113E Ch. 5 - Prob. 114E Ch. 5 - Prob. 115E Ch. 5 - Prob. 116E Ch. 5 - Prob. 117E Ch. 5 - Prob. 118E Ch. 5 - Prob. 119E Ch. 5 - Prob. 120E Ch. 5 - Prob. 121E Ch. 5 - Prob. 122E Ch. 5 - Prob. 123E Ch. 5 - Prob. 124E Ch. 5 - Prob. 125E Ch. 5 - Prob. 126E Ch. 5 - Prob. 127E Ch. 5 - Prob. 128E Ch. 5 - Prob. 129E Ch. 5 - Prob. 130E Ch. 5 - Prob. 131E Ch. 5 - Prob. 132E Ch. 5 - Prob. 133E Ch. 5 - Prob. 134E Ch. 5 - Prob. 135E Ch. 5 - Prob. 136E Ch. 5 - Prob. 137E Ch. 5 - Prob. 138E Ch. 5 - Prob. 139E Ch. 5 - Prob. 140E Ch. 5 - Prob. 141E Ch. 5 - Prob. 142E Ch. 5 - Prob. 143E Ch. 5 - Prob. 144E Ch. 5 - Prob. 145E Ch. 5 - Prob. 146E Ch. 5 - Prob. 147E Ch. 5 - Prob. 148QGW Ch. 5 - Prob. 149QGW Ch. 5 - Prob. 150QGW Ch. 5 - Prob. 151QGW Ch. 5 - Prob. 152QGW Ch. 5 - 153. Air contains about 78.08% nitrogen gas,...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.