Mastering Chemistry With Pearson Etext -- Standalone Access Card -- For General Chemistry: Principles And Modern Applications (11th Edition)
Mastering Chemistry With Pearson Etext -- Standalone Access Card -- For General Chemistry: Principles And Modern Applications (11th Edition)
11th Edition
ISBN: 9780133387803
Author: Ralph H. Petrucci; F. Geoffrey Herring; Jeffry D. Madura; Carey Bissonnette
Publisher: PEARSON
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Chapter 6, Problem 1E

Convert each pressure to an equivalent pressure in atmospheres. (a) 736 mmHg; (b) 0.776 bar; (C) 892 Torr; (d) 225 kPa.

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The pressure value 736 mmHg needs to be converted into atmosphere.

Concept introduction:

The pressure of gas can be expressed in different units. The equality factors can be used to convert the given unit of pressure to the desired unit of pressure. The SI unit of pressure is Pascal, but pressure is also commonly reported in atmosphere unit.

Some of the equality of pressure units are as follows:

1 atm = 760 mm Hg

1 atm = 1.01325 bar

1 atm = 760 torr

1 atm= 101325 Pa

Or, 1 kPa = 1000 Pa

Answer to Problem 1E

736 mm Hg is equivalent to 0.968 atm pressure.

Explanation of Solution

The given pressure is 736 mm Hg.

The following equality needs to be used to convert it into atm.

Equality: 1 atm = 760 mm Hg

Thus,

P=736 mm Hg ( 1atm760 mm Hg) = 0.968 atm

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The pressure value 0.776 bar needs to be converted into atmosphere.

Concept introduction:

The pressure of gas can be expressed in different units. The equality factors can be used to convert the given unit of pressure to the desired unit of pressure. The SI unit of pressure is Pascal, but pressure is also commonly reported in atmosphere unit.

Some of the equality of pressure units are as follows:

1 atm = 760 mm Hg

1 atm = 1.01325 bar

1 atm = 760 torr

1 atm= 101325 Pa

Or, 1 kPa = 1000 Pa

Answer to Problem 1E

0.776 bar is equivalent to 0.766 atm pressure.

Explanation of Solution

The given pressure is 0.776 bar.

The following equality is used to convert it into atm.

Equality: 1 atm = 1.01325 bar

Thus,

P=0.776 bar ×1atm1.01325 bar = 0.766 atm

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The pressure value 892 torr needs to be converted into atmosphere.

Concept introduction:

The pressure of gas can be expressed in different units. The equality factors can be used to convert the given unit of pressure to the desired unit of pressure. The SI unit of pressure is Pascal, but pressure is also commonly reported in atmosphere unit.

Some of the equality of pressure units are as follows:

1 atm = 760 mm Hg

1 atm = 1.01325 bar

1 atm = 760 torr

1 atm= 101325 Pa

Or, 1 kPa = 1000 Pa

Answer to Problem 1E

892 torr is equivalent to 1.17 atm pressure.

Explanation of Solution

The given pressure value is 892 torr. The following equality can be used to convert it into atm.

Equality: 1 atm = 760 torr

Thus,

P=892 torr ×1atm760 torr = 1.17 atm

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

The pressure value 225 kPa needs to be converted into atmosphere.

Concept introduction:

The pressure of gas can be expressed in different units. The equality factors can be used to convert the given unit of pressure to the desired unit of pressure. The SI unit of pressure is Pascal, but pressure is also commonly reported in atmosphere unit.

Some of the equality of pressure units are as follows:

1 atm = 760 mm Hg

1 atm = 1.01325 bar

1 atm = 760 torr

1 atm= 101325 Pa

Or, 1 kPa = 1000 Pa

Answer to Problem 1E

225 kPa is equivalent to 2.22 atm pressure.

Explanation of Solution

The given pressure is 225 kPa.

The following equality can be used to convert the given pressure into atm.

Equality:

1 atm= 101325 Pa

1 kPa = 1000 Pa

Thus,

P=225 kPa×1000 Pa 1 kPa×1 atm 101325 Pa = 2.22 atm

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07:37

Chapter 6 Solutions

Mastering Chemistry With Pearson Etext -- Standalone Access Card -- For General Chemistry: Principles And Modern Applications (11th Edition)

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