Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078129865
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 5, Problem 5.82P

(a)

Interpretation Introduction

Interpretation:

The root-mean-square speed of He at 0 °C is to be calculated. Also, the root-mean-square speed of He at 30 °C is to be calculated.

Concept introduction:

The expression to calculate the root-mean-square speed is as follows:

urms=3RTM

Here, M is the molar mass, T is the temperature and R is the gas constant.

The kinetic energy is directly proportional to the temperature and inversely proportional to the molar mass.

(a)

Expert Solution
Check Mark

Answer to Problem 5.82P

The root-mean-square speed of He at 0 °C is 1.30×103 m/s and the root-mean-square speed of He at 30 °C is 1.37×103 m/s.

Explanation of Solution

The formula to convert °C to Kelvin is as follows:

TK=T°C+273.15 (1)

Substitute °C for T°C in equation (1).

TK=0+273.15=273.15 K

The expression to calculate the root-mean-square velocity of He is as follows:

urms=3RTM (2)

Substitute the value 273.15 K for T, 8.314 JmolK for R and 0.004003 kg/mol for M in the equation (2).

urms=3(8.314 JmolK)(273.15 K)(0.004003 kg/mol)(kgm2/s2J)=1.3042×103 m/s1.30×103 m/s

Substitute 30 °C for T°C in equation (1).

TK=30+273.15=303.15 K

Substitute the value 303.15 K for T, 8.314 JmolK for R and 0.004003 kg/mol for M in the equation (2).

urms=3(8.314 JmolK)(303.15 K)(0.004003 kg/mol)(kgm2/s2J)=1.374×103 m/s1.37×103 m/s

Conclusion

The root-mean-square speed of He at 0 °C is 1.30×103 m/s. And the root-mean-square speed of He at 30 °C is 1.37×103 m/s.

(b)

Interpretation Introduction

Interpretation:

The root-mean-square speed of He and Xe at 30 °C are to be compared.

Concept introduction:

The expression to calculate the root-mean-square speed is as follows:

urms=3RTM

Here, M is the molar mass, T is the temperature and R is the gas constant.

The kinetic energy is directly proportional to the temperature and inversely proportional to the molar mass.

The mathematical expression of Graham’s law of effusion is as follows:

Rate ARate B=urms of Aurms of B=MBMA=time Btime A

Here, MB is the molar mass of gas B.

MA is the molar mass of gas A.

(b)

Expert Solution
Check Mark

Answer to Problem 5.82P

The root-mean-square speed of He and Xe at 30 °C is 5.73

Explanation of Solution

The formula to convert °C to Kelvin is:

TK=T°C+273.15 (1)

Substitute 30 °C for T°C in equation (1)

TK=30+273.15=303.15 K

The expression to calculate the root-mean-square velocity of Xe is as follows:

urms=3RTM (2)

Substitute the value 303.15 K for T, 8.314 JmolK for R and 0.004003 kg/mol for M in the equation (2).

urms=3(8.314 JmolK)(303.15 K)(0.13113 kg/mol)(kgm2/s2J)=239.913×103 m/s

The expression to compare the root-mean-square speed of He with that of Xe is as follows:

Rate of HeRate of Xe=urms of Heurmsof Xe (3)

Substitute the value 1.3740×103 m/s for urms of He and 239.913 m/s for urms of Xe in the equation (3).

Rate HeRate Xe=1.3740×103 m/s239.913 m/s=5.7270765.73

Conclusion

The root-mean-square speed He and Xe at 30 °C is 5.73.

(c)

Interpretation Introduction

Interpretation:

The average kinetic energy of He at 30 °C is to be calculated. Also, the average kinetic energy of Xe at 30 °C is to be calculated.

Concept introduction:

The expression to calculate the average kinetic energy is as follows:

Ek=12(m)(u)2

Here, Ek is the average kinetic energy, m is the mass and u is the speed.

The kinetic energy is directly proportional to the temperature.

(c)

Expert Solution
Check Mark

Answer to Problem 5.82P

The average kinetic energy of He at 30 °C is 3.78×103. And also the average kinetic energy of Xe at 30 °C is 3.78×103 J/mol.

Explanation of Solution

The expression to calculate the average kinetic energy of He is as follows:

EHe=12(m)(u)2 (4)

Substitute the value 1.3740×103 m/s for the speed of He and 0.004003 g/mol for the mass of He in the equation (4).

EHe=12[(0.004003 kg/mol)(1.3740 m/s)2(kgm2/s2J)]=3778.58 J/mol=3.78×103 J/mol

The expression to calculate the average kinetic energy of Xe is as follows:

EXe=12(m)(u)2 (5)

Substitute the value 239.913×103 m/s for the speed of Xe and 0.1313 g/mol for the mass of Xe in the equation (5).

EXe=12[(0.1313 Kg/mol)(239.913 m/s)2(kgm2/s2J)]=3778.70 J/mol3.78×103 J/mol

Conclusion

The average kinetic energy of He at 30 °C is 3.78×103. And also the average kinetic energy of Xe at 30 °C is 3.78×103 J/mol.

(d)

Interpretation Introduction

Interpretation:

The average kinetic energy per molecule of He at 30 °C is to be calculated.

Concept introduction:

The expression to calculate the average kinetic energy is as follows:

Ek=12(m)(u)2

Here, Ek is the average kinetic energy, m is the mass and u is the speed.

The kinetic energy is directly proportional to the temperature.

(d)

Expert Solution
Check Mark

Answer to Problem 5.82P

The average kinetic energy per molecule of He at 30 °C is 6.27×1021 J/atom.

Explanation of Solution

The expression to calculate the average kinetic energy of Xe is as follows:

EXe=12(m)(u)2 (6)

Rearrange the equation (6) in per molecule by dividing NA.

EXe=12(m)(u)2NA

Substitute the value 239.913×103 m/s for the speed of Xe and 0.1313 g/mol for the mass of Xe in the equation (5).

EXe=12[(0.1313 Kg/mol)(239.913 m/s)2(kgm2/s2J)](6.022×1023 atoms)=6.2746×1021 J/atom 6.27×1021 J/atom

Conclusion

The average kinetic energy per molecule of He at 30 °C is 6.27×1021 J/atom.

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Chapter 5 Solutions

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change

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