Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
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Question
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Chapter U3.11, Problem 4E
Interpretation Introduction

(a)

Interpretation:

Interpret the factors which remain constant and variable when a balloon with a volume of 240 mL at 25°C and 1.0 atm is carried up a mountain at a pressure of 0.75 atm and 25°C.

Concept introduction:

Due to the random movement of gas particles, they collide with other gas particles and also collide with the wall of the container. The collision between gas particles of air and the wall of the container exerts pressure on the wall of the container.

Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of the gas.

Pressure α 1volume  

Expert Solution
Check Mark

Answer to Problem 4E

Temperature remains constant for balloon. Pressure decreases whereas volume increases. Therefore, pressure and volume are variable factors for balloon.

Explanation of Solution

  • Initial volume = 240 mL
  • Initial temperature = 25°C
  • Initial pressure = 1.0 atm
  • Final pressure = 0.75 atm
  • Final temperature = 25°C.

The temperature of balloon is constant and the pressure at mountain decreases from 1.0 atm to 0.75 atm. Therefore, the volume of balloon will increase as with decrease in pressure, the volume of gas (filled in balloon) increases.

Interpretation Introduction

(b)

Interpretation:

Interpret the pressure outside and inside the balloon when a balloon with a volume of 240 mL at 25°C and 1.0 atm is carried up a mountain at the pressure of 0.75 atm and 25°C.

Concept introduction:

Due to the random movement of gas particles, they collide with other gas particles and also collide with the wall of the container. The collision between gas particles of air and the wall of the container exerts pressure on the wall of the container.

Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of the gas.

Pressure α 1volume  

Expert Solution
Check Mark

Answer to Problem 4E

The outside pressure decreases for balloon therefore inside higher pressure will decrease with the increase in volume of the balloon, so that the inside and outside pressure are in equilibrium with each other.

Explanation of Solution

  • Initial volume = 240 mL
  • Initial temperature = 25°C
  • Initial pressure = 1.0 atm
  • Final pressure = 0.75 atm
  • Final temperature = 25°C.

The outside pressure decreases for the balloon, therefore, inside higher pressure will decrease with the increase in the volume of the balloon. Thus, the inside and outside pressure are in equilibrium with each other.

Interpretation Introduction

(c)

Interpretation:

Interpret the volume occupied by the air inside the balloon if a balloon with a volume of 240 mL at 25°C and 1.0 atm is carried up a mountain at a pressure of 0.75 atm and 25°C.

Concept introduction:

Due to the random movement of gas particles, they collide with other gas particles and also collide with the wall of the container. The collision between gas particles of air and the wall of the container exerts pressure on the wall of the container.

Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of the gas.

Pressure α 1volume  

Expert Solution
Check Mark

Answer to Problem 4E

The volume of air occupied inside the balloon will increase.

Explanation of Solution

The outside pressure decreases for the balloon therefore the higher pressure will decrease with the increase in the volume of the balloon so that the inside and outside pressure are in equilibrium with each other. Thus, the volume of air occupied inside the balloon will increase.

Interpretation Introduction

(d)

Interpretation:

Interpret the new volume of the balloon if a balloon with a volume of 240 mL at 25°C and 1.0 atm is carried up a mountain at a pressure of 0.75 atm and 25°C.

Concept introduction:

Due to the random movement of gas particles, they collide with other gas particles and also collide with the wall of the container. The collision between gas particles of air and the wall of the container exerts pressure on the wall of the container.

Boyle’s law states that at constant temperature and amount of gas molecules, the volume is inversely proportional to the pressure of the gas.

Pressure α 1volume  

Expert Solution
Check Mark

Answer to Problem 4E

New volume = 320 mL

Explanation of Solution

  • Initial volume = 240 mL
  • Initial temperature = 25°C
  • Initial pressure = 1.0 atm
  • Final pressure = 0.75 atm
  • Final temperature = 25°C.

At constant temperature, the pressure of the gas is inversely proportional to the volume of gas.

P1V1 = P2V21.0 atm ×0.240 L = 0.75 atm  ×V2V2=1.0 atm ×0.240 L 0.75 atm V2=0.32L=320mL

Chapter U3 Solutions

Living By Chemistry: First Edition Textbook

Ch. U3.2 - Prob. 4ECh. U3.2 - Prob. 5ECh. U3.2 - Prob. 6ECh. U3.2 - Prob. 7ECh. U3.3 - Prob. 1TAICh. U3.3 - Prob. 1ECh. U3.3 - Prob. 2ECh. U3.3 - Prob. 3ECh. U3.3 - Prob. 4ECh. U3.3 - Prob. 5ECh. U3.3 - Prob. 6ECh. U3.3 - Prob. 7ECh. U3.3 - Prob. 8ECh. U3.3 - Prob. 9ECh. U3.4 - Prob. 1TAICh. U3.4 - Prob. 1ECh. U3.4 - Prob. 2ECh. U3.4 - Prob. 3ECh. U3.4 - Prob. 4ECh. U3.4 - Prob. 5ECh. U3.4 - Prob. 6ECh. U3.4 - Prob. 7ECh. U3.4 - Prob. 8ECh. U3.4 - Prob. 9ECh. U3.5 - Prob. 1TAICh. U3.5 - Prob. 1ECh. U3.5 - Prob. 2ECh. U3.5 - Prob. 3ECh. U3.5 - Prob. 4ECh. U3.5 - Prob. 5ECh. U3.5 - Prob. 6ECh. U3.5 - Prob. 7ECh. U3.5 - Prob. 8ECh. U3.5 - Prob. 9ECh. U3.5 - Prob. 10ECh. U3.5 - Prob. 11ECh. U3.5 - Prob. 12ECh. U3.6 - Prob. 1TAICh. U3.6 - Prob. 1ECh. U3.6 - Prob. 2ECh. U3.6 - Prob. 3ECh. U3.6 - Prob. 4ECh. U3.6 - Prob. 5ECh. U3.6 - Prob. 6ECh. U3.6 - Prob. 7ECh. U3.6 - Prob. 8ECh. U3.7 - Prob. 1TAICh. U3.7 - Prob. 1ECh. U3.7 - Prob. 2ECh. U3.7 - Prob. 3ECh. U3.7 - Prob. 4ECh. U3.7 - Prob. 5ECh. U3.7 - Prob. 6ECh. U3.8 - Prob. 1TAICh. U3.8 - Prob. 1ECh. U3.8 - Prob. 2ECh. U3.8 - Prob. 3ECh. U3.8 - Prob. 4ECh. U3.8 - Prob. 5ECh. U3.8 - Prob. 6ECh. U3.8 - Prob. 7ECh. U3.8 - Prob. 8ECh. U3.8 - Prob. 9ECh. U3.8 - Prob. 10ECh. U3.9 - Prob. 1TAICh. U3.9 - Prob. 1ECh. U3.9 - Prob. 2ECh. U3.9 - Prob. 3ECh. U3.9 - Prob. 4ECh. U3.9 - Prob. 5ECh. U3.9 - Prob. 6ECh. U3.9 - Prob. 7ECh. U3.9 - Prob. 8ECh. U3.9 - Prob. 10ECh. U3.10 - Prob. 1TAICh. U3.10 - Prob. 1ECh. U3.10 - Prob. 2ECh. U3.10 - Prob. 4ECh. U3.10 - Prob. 5ECh. U3.10 - Prob. 6ECh. U3.10 - Prob. 7ECh. U3.11 - Prob. 1TAICh. U3.11 - Prob. 1ECh. U3.11 - Prob. 2ECh. U3.11 - Prob. 3ECh. U3.11 - Prob. 4ECh. U3.11 - Prob. 5ECh. U3.12 - Prob. 1TAICh. U3.12 - Prob. 1ECh. U3.12 - Prob. 2ECh. U3.12 - Prob. 3ECh. U3.12 - Prob. 4ECh. U3.12 - Prob. 5ECh. U3.12 - Prob. 6ECh. U3.12 - Prob. 7ECh. U3.12 - Prob. 8ECh. U3.13 - Prob. 1TAICh. U3.13 - Prob. 1ECh. U3.13 - Prob. 2ECh. U3.13 - Prob. 3ECh. U3.13 - Prob. 4ECh. U3.13 - Prob. 5ECh. U3.13 - Prob. 6ECh. U3.13 - Prob. 7ECh. U3.14 - Prob. 1TAICh. U3.14 - Prob. 1ECh. U3.14 - Prob. 2ECh. U3.15 - Prob. 1TAICh. U3.15 - Prob. 1ECh. U3.15 - Prob. 2ECh. U3.15 - Prob. 3ECh. U3.15 - Prob. 4ECh. U3.15 - Prob. 5ECh. U3.15 - Prob. 6ECh. U3.15 - Prob. 7ECh. U3.15 - Prob. 8ECh. U3.16 - Prob. 1TAICh. U3.16 - Prob. 1ECh. U3.16 - Prob. 2ECh. U3.16 - Prob. 3ECh. U3.16 - Prob. 4ECh. U3.16 - Prob. 5ECh. U3.16 - Prob. 6ECh. U3.16 - Prob. 7ECh. U3.17 - Prob. 1TAICh. U3.17 - Prob. 1ECh. U3.17 - Prob. 2ECh. U3.17 - Prob. 3ECh. U3.17 - Prob. 4ECh. U3.17 - Prob. 5ECh. U3.17 - Prob. 6ECh. U3.17 - Prob. 7ECh. U3.17 - Prob. 8ECh. U3.18 - Prob. 1TAICh. U3.18 - Prob. 1ECh. U3.18 - Prob. 2ECh. U3.18 - Prob. 3ECh. U3.18 - Prob. 4ECh. U3.18 - Prob. 5ECh. U3.18 - Prob. 6ECh. U3.18 - Prob. 7ECh. U3.18 - Prob. 8ECh. U3.18 - Prob. 9ECh. U3.18 - Prob. 10ECh. U3.18 - Prob. 11ECh. U3.18 - Prob. 12ECh. U3.19 - Prob. 1TAICh. U3.19 - Prob. 1ECh. U3.19 - Prob. 2ECh. U3.19 - Prob. 4ECh. U3 - Prob. SI1RECh. U3 - Prob. SI2RECh. U3 - Prob. SI3RECh. U3 - Prob. SI4RECh. U3 - Prob. SI5RECh. U3 - Prob. SII1RECh. U3 - Prob. SII2RECh. U3 - Prob. SII3RECh. U3 - Prob. SII4RECh. U3 - Prob. SII5RECh. U3 - Prob. SIII1RECh. U3 - Prob. SIII2RECh. U3 - Prob. SIII3RECh. U3 - Prob. SIII4RECh. U3 - Prob. SIII5RECh. U3 - Prob. 1RECh. U3 - Prob. 2RECh. U3 - Prob. 3RECh. U3 - Prob. 4RECh. U3 - Prob. 5RECh. U3 - Prob. 6RECh. U3 - Prob. 7RECh. U3 - Prob. 8RECh. U3 - Prob. 9RECh. U3 - Prob. 10RECh. U3 - Prob. 11RECh. U3 - Prob. 12RECh. U3 - Prob. 13RE
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