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 U3.10, Problem 7E

(a)

Interpretation Introduction

Interpretation: The relation between P and V as P = k1V  if 1atm = 14.7 lb/in2 needs to be explained.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

(a)

Expert Solution
Check Mark

Answer to Problem 7E

In the given data, the constant value of k proves that:

  P = k1VPV=k

Explanation of Solution

    Trial Volume (mL) Pressure (lb/in2) Pressure (atm) 1VP = k1V PV=k
    1 60 mL 10 lb/in2 10 lb/in214.7 lb/in2 = 0.680 atm10.060L=  16.70.060 L×0.680 atm = 0.0408 L.atm
    2 40 mL 15 lb/in2 15 lb/in214.7 lb/in2 =1.02 atm10.040L=  250.040 L×1.02 atm = 0.0408 L.atm
    3 30 mL 20 lb/in2 20 lb/in214.7 lb/in2 = 1.36 atm10.030L=  33.30.020 L×1.36 atm = 0.0408 L.atm
    4 15 mL 40 lb/in2 40 lb/in214.7 lb/in2 = 2.72 atm10.015 L=  66.70.015 L×2.72 atm = 0.0408 L.atm
    5 10 mL 60 lb/in2 60 lb/in214.7 lb/in2 = 4.08 atm10.010L=  1000.010 L×4.08 atm = 0.0408 L.atm

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

  Pressure α 1volume  

In the given data, the constant value of k proves that:

  P = k1VPV=k

(b)

Interpretation Introduction

Interpretation: The relation between P and V as PV = k if 1atm = 14.7 lb/in2 needs to be explained.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

(b)

Expert Solution
Check Mark

Answer to Problem 7E

In the given data, the constant value of k proves that:

  P = k1VPV=k

Explanation of Solution

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

  Pressure α 1volume  

    Trial Volume (mL) Pressure (lb/in2) Pressure (atm) PV = k
    1 60 mL 10 lb/in2 10 lb/in214.7 lb/in2 = 0.680 atm0.060 L×0.680 atm = 0.0408 L.atm
    2 40 mL 15 lb/in2 15 lb/in214.7 lb/in2 =1.02 atm0.040 L×1.02 atm = 0.0408 L.atm
    3 30 mL 20 lb/in2 20 lb/in214.7 lb/in2 = 1.36 atm0.020 L×1.36 atm = 0.0408 L.atm
    4 15 mL 40 lb/in2 40 lb/in214.7 lb/in2 = 2.72 atm0.015 L×2.72 atm = 0.0408 L.atm
    5 10 mL 60 lb/in2 60 lb/in214.7 lb/in2 = 4.08 atm0.010 L×4.08 atm = 0.0408 L.atm

(c)

Interpretation Introduction

Interpretation: The graph of P versus V needs to be drawn and the relation between gas pressure and volume needs to be explained.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

(c)

Expert Solution
Check Mark

Answer to Problem 7E

  Living By Chemistry: First Edition Textbook, Chapter U3.10, Problem 7E , additional homework tip  1

The curve between P and V interprets that with increase in pressure of gas, the volume of gas increases.

Explanation of Solution

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

  Pressure α 1volume  

    Trial Volume (mL) Pressure (lb/in2) Pressure (atm) PV = k
    1 60 mL 10 lb/in2 10 lb/in214.7 lb/in2 = 0.680 atm0.060 L×0.680 atm = 0.0408 L.atm
    2 40 mL 15 lb/in2 15 lb/in214.7 lb/in2 =1.02 atm0.040 L×1.02 atm = 0.0408 L.atm
    3 30 mL 20 lb/in2 20 lb/in214.7 lb/in2 = 1.36 atm0.030 L×1.36 atm = 0.0408 L.atm
    4 15 mL 40 lb/in2 40 lb/in214.7 lb/in2 = 2.72 atm0.015 L×2.72 atm = 0.0408 L.atm
    5 10 mL 60 lb/in2 60 lb/in214.7 lb/in2 = 4.08 atm0.010 L×4.08 atm = 0.0408 L.atm

The curve between P and V must be:

  Living By Chemistry: First Edition Textbook, Chapter U3.10, Problem 7E , additional homework tip  2

The curve between P and V interprets that with increase in pressure of gas, the volume of gas increases. Overall pressure is inversely proportional to the volume of gas.

(d)

Interpretation Introduction

Interpretation: The graph of P versus 1/V needs to be drawn and the relation between gas pressure and volume needs to be explained.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

(d)

Expert Solution
Check Mark

Answer to Problem 7E

  Living By Chemistry: First Edition Textbook, Chapter U3.10, Problem 7E , additional homework tip  3

The curve between P and 1/V interprets that pressure is directly proportional to the inverse volume.

  Pressure α1Volume

Explanation of Solution

    Trial Volume (mL) Pressure (lb/in2) Pressure (atm) 1V
    1 60 mL 10 lb/in2 10 lb/in214.7 lb/in2 = 0.680 atm10.060L=  16.7
    2 40 mL 15 lb/in2 15 lb/in214.7 lb/in2 =1.02 atm10.040L=  25
    3 30 mL 20 lb/in2 20 lb/in214.7 lb/in2 = 1.36 atm10.030L=  33.3
    4 15 mL 40 lb/in2 40 lb/in214.7 lb/in2 = 2.72 atm10.015 L=  66.7
    5 10 mL 60 lb/in2 60 lb/in214.7 lb/in2 = 4.08 atm10.010L=  100

The curve between P and 1/V must be:

  Living By Chemistry: First Edition Textbook, Chapter U3.10, Problem 7E , additional homework tip  4

The curve between P and 1/V interprets that pressure is directly proportional to the inverse volume.

  Pressure α1Volume

(e)

Interpretation Introduction

Interpretation: The volume of gas at 30 lb/in2 pressure with the help of given graph needs to be determined.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

(e)

Expert Solution
Check Mark

Answer to Problem 7E

Volume at 30 lb/in2 = 0.020 L = 20.0 mL

Explanation of Solution

    Trial Volume (mL) Pressure (lb/in2) Pressure (atm)
    1 60 mL 10 lb/in2 10 lb/in214.7 lb/in2 = 0.680 atm
    2 40 mL 15 lb/in2 15 lb/in214.7 lb/in2 =1.02 atm
    3 30 mL 20 lb/in2 20 lb/in214.7 lb/in2 = 1.36 atm
    4 15 mL 40 lb/in2 40 lb/in214.7 lb/in2 = 2.72 atm
    5 10 mL 60 lb/in2 60 lb/in214.7 lb/in2 = 4.08 atm

Convert 30 lb/in2 to atm:

1 atm = 14.7 lb/in2

  1atm ×30 lb/in214.7 lb/in2 = 2.04 atm

  Living By Chemistry: First Edition Textbook, Chapter U3.10, Problem 7E , additional homework tip  5

Thus the volume of gas at 2.04 atm will be 20.0 mL.

(f)

Interpretation Introduction

Interpretation: The pressure of 50 mL gas with the help of given graph needs to be determined.

Concept Introduction:

Due to random movement of gas particles, they colloid with other gas particles and also colloid with wall of container. The collision between gas particles of air and wall of container exert pressure on the wall of container. The gas pressure is inversely proportional to the volume of gas. This is because as the gas pressure increases, the gas particles come close to each other. It decreases the intermolecular distance between particles and volume decreases.

(f)

Expert Solution
Check Mark

Answer to Problem 7E

The pressure of 0.050 L or 50.0 mL of gas must be 0.82 atm.

Explanation of Solution

    Trial Volume (mL) Pressure (lb/in2) Pressure (atm)
    1 60 mL 10 lb/in2 10 lb/in214.7 lb/in2 = 0.680 atm
    2 40 mL 15 lb/in2 15 lb/in214.7 lb/in2 =1.02 atm
    3 30 mL 20 lb/in2 20 lb/in214.7 lb/in2 = 1.36 atm
    4 15 mL 40 lb/in2 40 lb/in214.7 lb/in2 = 2.72 atm
    5 10 mL 60 lb/in2 60 lb/in214.7 lb/in2 = 4.08 atm

Volume = 50 mL

  1Volume=10.050L=20

  Living By Chemistry: First Edition Textbook, Chapter U3.10, Problem 7E , additional homework tip  6

Thus the pressure of 0.050 L or 50.0 mL of gas must be 0.82 atm.

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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