GENERAL ORGANIC & BIOCHEMISTRY >ACCESS<
GENERAL ORGANIC & BIOCHEMISTRY >ACCESS<
4th Edition
ISBN: 9781264096602
Author: SMITH
Publisher: MCG
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Question
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Chapter 7, Problem 31P
Interpretation Introduction

(a)

Interpretation:

The following table is to be completed by assuming the gas at a constant temperature.

P1V1P2V2
2.0atm3.0L8.0atm?

Concept Introduction:

The concept of Boyle's law will be used. This is represented as-

  P1V1=P2V2

Or,

  V2=P1V1P2

Here, P is pressure and V is the volume of the gas.

Expert Solution
Check Mark

Answer to Problem 31P

P1V1P2V2
2.0atm3.0L8.0atm0.75L

Explanation of Solution

According to Boyle's law, the pressure of the ideal gas is inversely proportional to the volume of the ideal gas at constant temperature. This is represented as

  PV

Or,

  P1V1=P2V2 ..................... (1)

Or,

  V2=P1V1P2

Given that −

P1, initial pressure = 2.0atm

V1, initial volume = 3.0L

P2, final pressure = 8.0atm

Put the above values in equation (1),

  V2=2.0 atm×3.0 L8.0 atm=0.75 L

Final volume = 0.75 L

Interpretation Introduction

(b)

Interpretation:

The following table is to be completed by assuming the gas at a constant temperature.

P1V1P2V2
55 mmHg0.35 L18 mmHg?

Concept Introduction:

The concept of Boyle's law will be used. This is represented as-

  P1V1=P2V2

Or,

  V2=P1V1P2

Here, P is pressure and V is the volume of the gas.

Expert Solution
Check Mark

Answer to Problem 31P

P1V1P2V2
55mmHg0.35L18mmHg1.07L

Explanation of Solution

According to Boyle's law, the pressure of the ideal gas is inversely proportional to the volume of the ideal gas at a constant temperature. This is represented as

  PV

Or,

  P1V1=P2V2 ..................... (1)

Or,

  V2=P1V1P2 .............. (2)

Given that −

P1, initial pressure = 55mmHg

V1, initial volume = 0.35mL

P2, final pressure = 18mmHg

Put the above values in equation (2)

  V2=55mmHg×0.35L18mmHg=1.07L

Final volume = 1.07L

Interpretation Introduction

(c)

Interpretation:

The following table is to be completed by assuming the gas at a constant temperature.

P1V1P2V2
705mmHg215mL?1.52L

Concept Introduction:

The concept of Boyle's law will be used. This is represented as-

  P1V1=P2V2

Or,

  V2=P1V1P2

Here, P is pressure and V is the volume of the gas.

Expert Solution
Check Mark

Answer to Problem 31P

P1V1P2V2
705mmHg215mL100mmHg1.52L

Explanation of Solution

According to Boyle's law, the pressure of the ideal gas is inversely proportional to the volume of the ideal gas at a constant temperature. This is represented as

  PV

Or,

  P1V1=P2V2 ..................... (1)

Or,

  V2=P1V1P2 .............. (2)

Given that −

P1, initial pressure = 705mmHg

V1, initial volume = 215mL

Or,

  =215mL×1L1000mL=0.215L

V2, final pressure = 1.52 L

  P2=P1V1V2

Put the values in the above equation,

  P2=705 mmHg×0.215 L1.52 L=99.7 mmHg100 mmHg

Final pressure = 100mmHg

The final table is −

P1V1P2V2
705mmHg215mL100mmHg1.52L

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Chapter 7 Solutions

GENERAL ORGANIC & BIOCHEMISTRY >ACCESS<

Ch. 7.4 - Prob. 7.7PPCh. 7.4 - Prob. 7.5PCh. 7.5 - Prob. 7.8PPCh. 7.5 - Prob. 7.6PCh. 7.6 - CO2 was added to a cylinder containing 2.5 atm of...Ch. 7.6 - Prob. 7.10PPCh. 7.6 - Prob. 7.7PCh. 7.7 - Prob. 7.8PCh. 7.7 - Prob. 7.9PCh. 7.7 - Prob. 7.11PPCh. 7.7 - Which species in each pair has stronger...Ch. 7.7 - Prob. 7.12PPCh. 7.7 - Prob. 7.11PCh. 7.8 - Prob. 7.13PPCh. 7.8 - Would you predict the surface tension of gasoline,...Ch. 7.9 - Prob. 7.13PCh. 7.10 - Prob. 7.14PPCh. 7.10 - The human body is composed of about 70% water. How...Ch. 7.10 - How much energy is required to heat 28.0 g of iron...Ch. 7.10 - Prob. 7.15PCh. 7.10 - Prob. 7.16PPCh. 7.10 - If the initial temperature of 120. g of ethanol is...Ch. 7.11 - Use the heat of fusion of water from Sample...Ch. 7.11 - Answer the following questions about water, which...Ch. 7.11 - Prob. 7.19PPCh. 7.12 - Answer the following questions about the graph...Ch. 7.12 - How much energy (in calories) is released when...Ch. 7.12 - How much energy (in calories) is required to melt...Ch. 7 - Prob. 19PCh. 7 - Prob. 20PCh. 7 - Prob. 21PCh. 7 - The compressed air tank of a scuba diver reads...Ch. 7 - Assume that each of the following samples is at...Ch. 7 - Use the diagrams in problem 7.23 to answer the...Ch. 7 - Prob. 25PCh. 7 - Prob. 26PCh. 7 - Prob. 27PCh. 7 - Prob. 28PCh. 7 - Prob. 29PCh. 7 - Prob. 30PCh. 7 - Prob. 31PCh. 7 - Prob. 32PCh. 7 - Prob. 33PCh. 7 - If you pack a bag of potato chips for a snack on a...Ch. 7 - Prob. 35PCh. 7 - Prob. 36PCh. 7 - Prob. 37PCh. 7 - Prob. 38PCh. 7 - Prob. 39PCh. 7 - Prob. 40PCh. 7 - Prob. 41PCh. 7 - Prob. 42PCh. 7 - Prob. 43PCh. 7 - Prob. 44PCh. 7 - Prob. 45PCh. 7 - Prob. 46PCh. 7 - Prob. 47PCh. 7 - Consider balloons A and B, which contain CH4 and...Ch. 7 - Prob. 49PCh. 7 - Prob. 50PCh. 7 - Prob. 51PCh. 7 - Prob. 52PCh. 7 - Prob. 53PCh. 7 - Prob. 54PCh. 7 - Prob. 55PCh. 7 - Prob. 56PCh. 7 - Prob. 57PCh. 7 - Prob. 58PCh. 7 - Prob. 59PCh. 7 - Prob. 60PCh. 7 - Prob. 61PCh. 7 - Prob. 62PCh. 7 - Prob. 63PCh. 7 - Prob. 64PCh. 7 - Which molecules are capable of intermolecular...Ch. 7 - Prob. 66PCh. 7 - Prob. 67PCh. 7 - Explain why the boiling point of A is higher than...Ch. 7 - Prob. 69PCh. 7 - Prob. 70PCh. 7 - Prob. 71PCh. 7 - Prob. 72PCh. 7 - Prob. 73PCh. 7 - Prob. 74PCh. 7 - Prob. 75PCh. 7 - Prob. 76PCh. 7 - Prob. 77PCh. 7 - Prob. 78PCh. 7 - Prob. 79PCh. 7 - Prob. 80PCh. 7 - Prob. 81PCh. 7 - How many calories of heat are needed to increase...Ch. 7 - Prob. 83PCh. 7 - If it takes 37.0 cal of heat to raise the...Ch. 7 - Prob. 85PCh. 7 - What phase change is shown in the accompanying...Ch. 7 - Prob. 87PCh. 7 - Which process requires more energy, melting 250 g...Ch. 7 - Consider the cooling curve drawn below a. Which...Ch. 7 - Prob. 90PCh. 7 - Draw the heating curve that is observed when...Ch. 7 - Prob. 92PCh. 7 - Use the following values to answer each part. The...Ch. 7 - Prob. 94PCh. 7 - Prob. 95PCh. 7 - Prob. 96PCh. 7 - Prob. 97PCh. 7 - Explain why you feel cool when you get out of a...Ch. 7 - Prob. 99CPCh. 7 - As we learned in Chapter 5, an automobile airbag...
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