Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 5, Problem 5.95P

(a)

Interpretation Introduction

Interpretation:

The volume of a gas increase, decrease or remain unchanged if the pressure is decreased from 2 atm to 1 atm while the temperature is decreased from 200 °C to 100 °C is to be identified.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

PV=constant

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

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

V α T

Here, T is the temperature and V is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

α n

Here, n is the mole of the gas and V is the volume.

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

(a)

Expert Solution
Check Mark

Answer to Problem 5.95P

The volume of the gas is increased if the pressure is decreased from 2 atm to 1 atm and the temperature is decreased from 200 °C to 100 °C.

Explanation of Solution

The formula to convert °C to Kelvin is:

TK=T°C+273.15 (1)

Substitute 200 °C for T°C to calculate T1 in equation (1).

TK=200+273.15=473.15 K

Substitute 100 °C for T°C to calculate T2 in equation (1).

TK=100+273.15=373.15 K

The expression to calculate the final volume is as follows:

P1V1T1=P2V2T2 (2)

Here, P1 is the initial pressure, P2 is the final pressure, V1 is the initial volume, V2 is the final volume, T1 is the initial temperature and T2 is the final temperature.

Rearrange the equation (2) to calculate V2.

V2=(P1V1T2T1P2) (3)

Substitute the value 2 atm for P1, 1 atm for P2, 373.15 K for T2 and 473.15 K for T1 in the equation (3)

V2=((V1)(373.15 K)(2 atm)(473.15 K)(1 atm))=1.577 V1 

Hence, the volume of the gas is increased if the pressure is decreased from 2 atm to 1 atm and the temperature is decreased from 200 °C to 100 °C.

Conclusion

The volume of the gas is increased if the pressure is decreased from 2 atm to 1 atm and the temperature is decreased from 200 °C to 100 °C.

(b)

Interpretation Introduction

Interpretation:

The volume of a gas increase, decrease or remain unchanged if the pressure is decreased from 1 atm to 3 atm while the temperature is decreased from 100 °C to 300 °C is to be identified.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

PV=constant

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

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

V α T

Here, T is the temperature and V is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

α n

Here, n is the mole of the gas and V is the volume.

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

(b)

Expert Solution
Check Mark

Answer to Problem 5.95P

The volume of the gas is decreased if the pressure is decreased from 1 atm to 3 atm and the temperature is decreased from 100 °C to 300 °C.

Explanation of Solution

Substitute 100 °C for T°C to calculate T1 in equation (1).

TK=200+273.15=373.15 K

Substitute 300 °C for T°C to calculate T2 in equation (1).

TK=300+273.15=573.15 K

Substitute the value 1 atm for P1, 3 atm for P2, 573.15 K for T2 and 373.15 K for T1 in the equation (3)

V2=((V1)(573.15 K)(1 atm)(373.15 K)(3 atm))=0.5120 V1 

Hence, the volume of the gas is decreased if the pressure is decreased from 1 atm to 3 atm and the temperature is decreased from 100 °C to 300 °C.

Conclusion

The volume of the gas is decreased if the pressure is decreased from 1 atm to 3 atm and the temperature is decreased from 100 °C to 300 °C.

(c)

Interpretation Introduction

Interpretation:

The volume of a gas increase, decrease or remain unchanged if the pressure is decreased from 3 atm to 6 atm while the temperature is decreased from 73 °C to 127 °C is to be identified.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

PV=constant

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

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

V α T

Here, T is the temperature and V is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

α n

Here, n is the mole of the gas and V is the volume.

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

(c)

Expert Solution
Check Mark

Answer to Problem 5.95P

The volume of the gas is unchanged if the pressure is decreased from 3 atm to 6 atm and the temperature is decreased from 73 °C to 127 °C.

Explanation of Solution

Substitute 73 °C for T°C to calculate T1 in equation (1).

TK=73+273.15=200.15 K

Substitute 127 °C for T°C to calculate T2 in equation (1).

TK=127+273.15=400.15 K

Substitute the value 3 atm for P1, 6 atm for P2, 400.15 K for T2 and 200.15 K for T1 in the equation (3)

V2=((V1)(400.15 K)(3 atm)(200.15 K)(6 atm))=0.9996 V1 

Hence, the volume of the gas is unchanged if the pressure is decreased from 3 atm to 6 atm and the temperature is decreased from 73 °C to 127 °C.

Conclusion

The volume of the gas is unchanged if the pressure is decreased from 3 atm to 6 atm and the temperature is decreased from 73 °C to 127 °C.

(d)

Interpretation Introduction

Interpretation:

The volume of a gas increase, decrease or remain unchanged if the pressure is decreased from 0.2 atm to 0.4 atm while the temperature is decreased from 300 °C to 150 °C is to be identified.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

PV=constant

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

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

V α T

Here, T is the temperature and V is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

α n

Here, n is the mole of the gas and V is the volume.

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

(d)

Expert Solution
Check Mark

Answer to Problem 5.95P

The volume of the gas is decreased if the pressure is decreased from 0.2 atm to 0.4 atm and the temperature is decreased from 300 °C to 150 °C.

Explanation of Solution

Substitute 300 °C for T°C to calculate T1 in equation (1).

TK=300+273.15=573.15 K

Substitute 150 °C for T°C to calculate T2 in equation (1).

TK=150+273.15=423.15 K

Substitute the value 0.2 atm for P1, 0.4 atm for P2, 423.15 K for T2 and 573.15 K for T1 in the equation (3)

V2=((V1)(423.15 K)(0.2 atm)(573.15 K)(0.4 atm))=0.369 V1 

Hence, the volume of the gas is decreased if the pressure is decreased from 0.2 atm to 0.4 atm and the temperature is decreased from 300 °C to 150 °C.

Conclusion

The volume of the gas is decreased if the pressure is decreased from 0.2 atm to 0.4 atm and the temperature is decreased from 300 °C to 150 °C.

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Chapter 5 Solutions

Chemistry: The Molecular Nature of Matter and Change

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