ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
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Chapter 3, Problem 3C.9P

(a)

Interpretation Introduction

Interpretation: The reason of the plot of Cp,m(T)T against T2 is expected to be a straight line with a slope a and intercept b has to be stated.

Concept introduction: Cp,m represents the molar heat capacity.  At low temperatures, the value of Cp,m for solids can be calculated by the Debye’s law.  At low temperatures, the value of Cp,m for metals is calculated by the Debye contribution as well as the electronic contributions.

(a)

Expert Solution
Check Mark

Answer to Problem 3C.9P

The plot between Cp,m(T)T against T2 follows the straight line equation with a slope a and intercept with y axis as b.

Explanation of Solution

The heat capacity at low temperature is calculated by the following formula.

    Cp,m(T)=aT3+bT

Divide the above equation by T.

    Cp,m(T)T=aT2+b                                                                                        (1)

Where,

  • Cp,m(T)T is the molar heat capacity with respect to the temperature T.
  • a and b are the Debye’s constant.

Compare the above equation with the straight line equation, y=mx+c.

The vertical axis, y represents Cp,m(T)T

The horizontal axis, x represents T2.

The slope of the plot, m represents the Debye’s conatant a.

The intercept with y axis c, represents the Debye’s constant b.

The plot between Cp,m(T)T against T2 follows the straight line equation with a slope a and intercept with y axis as b.  Therefore, the plot of Cp,m(T)T against T2 is expected to be a straight line with a slope a and intercept b.

(b)

Interpretation Introduction

Interpretation: The values of the constants a and b has to be calculated.

Concept introduction: Cp,m represents the molar heat capacity.  At low temperatures, the value of Cp,m for solids can be calculated by the Debye’s law.  At low temperatures, the value of Cp,m for metals is calculated by the Debye contribution as well as the electronic contributions.

(b)

Expert Solution
Check Mark

Answer to Problem 3C.9P

The value of a and b are 2.44Jmol-1K-4_ and 2.087Jmol-1K-2_ respectively.

Explanation of Solution

The given data for the variation of molar heat capacity at very low temperature is shown below.

T/K0.200.250.300.350.400.450.500.55
Cp,m/(JK-1mol-1)0.4370.5600.6930.8380.9961.1701.3611.572

The table for the value of Cp,m(T)T against T2 is shown below.

T20.040.06250.090.12250.160.20250.250.3025
Cp,m(T)T2.1852.242.312.3942.492.62.7222.85

The plot of Cp,m(T)T against T2 is shown below.

ATKINS' PHYSICAL CHEMISTRY-ACCESS, Chapter 3, Problem 3C.9P

From the above table, values of T2 are 0.04 and 0.0625 and their corresponding values of Cp,m(T)T are 2.185 and 2.24 respectively.

Substitute the values of corresponding values of Cp,m(T)T and T2 in equation (1).

    2.185=0.04a+b                                                                                          (2)

    2.24=0.0625a+b                                                                                        (3)

Solve the linear equation (2) and (3).

    a=2.44Jmol-1K-4_ and b=2.087Jmol-1K-2_

Hence, the value of a and b are 2.44Jmol-1K-4_ and 2.087Jmol-1K-2_ respectively.

(c)

Interpretation Introduction

Interpretation: The expression for the molar entropy has to be predicted.

Concept introduction: Cp,m represents the molar heat capacity.  At low temperatures, the value of Cp,m for solids can be calculated by the Debye’s law.  At low temperatures, the value of Cp,m for metals is calculated by the Debye contribution as well as the electronic contributions.

(c)

Expert Solution
Check Mark

Answer to Problem 3C.9P

The expression for the molar entropy is Sm=a[T233T133]+b[T2T1].

Explanation of Solution

The molar entropy is calculated by the following formula.

    Sm=Cp,m(T)ln(T2T1)                                                                                    (4)

The value of Cp,mln(T2T1) is calculated by integrating the equation (1) in between the limits T2 and T1 with respect to the temperature.

Integrate the equation (1).

    T1T2Cp,m(T)TdT=T1T2aT2dT+T1T2bdT[Cp,m(T)ln(T)]T1T2=a[T33]T1T2+b[T]T1T2Cp,m(T)ln(T2T1)=a[T233T133]+b[T2T1]

Substitute the value of Cp,m(T)ln(T2T1) in equation (4).

    Sm=a[T233T133]+b[T2T1]

Hence the expression for molar entropy is Sm=a[T233T133]+b[T2T1].

(d)

Interpretation Introduction

Interpretation: The molar entropy of potassium at 2.0K has to be calculated.

Concept introduction: Standard entropy of reaction predicts the feasibility of the reaction.  For a reaction to be feasible, the change in the standard entropy of the reaction must be positive. Standard entropy of the reaction increases as the numbers of gaseous constituents in the reaction increases.

(d)

Expert Solution
Check Mark

Answer to Problem 3C.9P

The molar entropy of potassium at 2.0K is 21.8106JK-1mol-1_.

Explanation of Solution

The molar entropy at 2.0K is calculated by the following formula.

    ST=Cp,mΔTT                                                                                                 (5)

Where,

  • ST is the entropy at TK.
  • Cp,m is the molar heat capacity at constant pressure.
  • ΔT is the change in temperature.

The change in temperature is calculated by the following formula.

    ΔT=T2T1                                                                                                   (6)

It is given that the temperature T1 is 0.20K and temperature T2 is 2.0K.

Substitute the values of T1 and T2 in equation (6).

    ΔT=2.0K-0.20K=1.8K

Therefore, the change in temperature is 1.8K.

The molar heat capacity is calculated by the given expression.

    Cp,m(T)=aT3+bT                                                                                        (7)

It is given that the temperature T is 2.0K.

The value of a is 2.44Jmol-1K-4.

The value of b is 2.087Jmol-1K-2.

Cp,m(2.0K)=2.44Jmol-1K-4(2K)3+2.087Jmol-1K-2(2K)=19.52Jmol-1K-1+4.174Jmol-1K-1=24.234Jmol-1K-1

Therefore the value of Cp,m(2.0K) is 24.234Jmol-1K-1.

Substitute the value of T,ΔT and Cp,m(2.0K) in equation (1).

    Sm,2.0K=24.234Jmol-1K-1×1.8K2.0K=21.8106JK-1mol-1_

Hence, the molar entropy at 2.0K is 21.8106JK-1mol-1_.

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Chapter 3 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 3 - Prob. 3A.1AECh. 3 - Prob. 3A.1BECh. 3 - Prob. 3A.2AECh. 3 - Prob. 3A.2BECh. 3 - Prob. 3A.3AECh. 3 - Prob. 3A.3BECh. 3 - Prob. 3A.4AECh. 3 - Prob. 3A.4BECh. 3 - Prob. 3A.5AECh. 3 - Prob. 3A.5BECh. 3 - Prob. 3A.6AECh. 3 - Prob. 3A.6BECh. 3 - Prob. 3A.1PCh. 3 - Prob. 3A.2PCh. 3 - Prob. 3A.3PCh. 3 - Prob. 3A.4PCh. 3 - Prob. 3A.5PCh. 3 - Prob. 3A.6PCh. 3 - Prob. 3A.7PCh. 3 - Prob. 3B.1DQCh. 3 - Prob. 3B.1AECh. 3 - Prob. 3B.1BECh. 3 - Prob. 3B.2AECh. 3 - Prob. 3B.2BECh. 3 - Prob. 3B.3AECh. 3 - Prob. 3B.3BECh. 3 - Prob. 3B.4AECh. 3 - Prob. 3B.4BECh. 3 - Prob. 3B.5AECh. 3 - Prob. 3B.5BECh. 3 - Prob. 3B.6AECh. 3 - Prob. 3B.6BECh. 3 - Prob. 3B.7AECh. 3 - Prob. 3B.7BECh. 3 - Prob. 3B.1PCh. 3 - Prob. 3B.2PCh. 3 - Prob. 3B.3PCh. 3 - Prob. 3B.4PCh. 3 - Prob. 3B.5PCh. 3 - Prob. 3B.6PCh. 3 - Prob. 3B.7PCh. 3 - Prob. 3B.8PCh. 3 - Prob. 3B.9PCh. 3 - Prob. 3B.10PCh. 3 - Prob. 3B.12PCh. 3 - Prob. 3C.1DQCh. 3 - Prob. 3C.1AECh. 3 - Prob. 3C.1BECh. 3 - Prob. 3C.2AECh. 3 - Prob. 3C.2BECh. 3 - Prob. 3C.3AECh. 3 - Prob. 3C.3BECh. 3 - Prob. 3C.1PCh. 3 - Prob. 3C.2PCh. 3 - Prob. 3C.4PCh. 3 - Prob. 3C.5PCh. 3 - Prob. 3C.6PCh. 3 - Prob. 3C.7PCh. 3 - Prob. 3C.8PCh. 3 - Prob. 3C.9PCh. 3 - Prob. 3C.10PCh. 3 - Prob. 3D.1DQCh. 3 - Prob. 3D.2DQCh. 3 - Prob. 3D.1AECh. 3 - Prob. 3D.1BECh. 3 - Prob. 3D.2AECh. 3 - Prob. 3D.2BECh. 3 - Prob. 3D.3AECh. 3 - Prob. 3D.3BECh. 3 - Prob. 3D.4AECh. 3 - Prob. 3D.4BECh. 3 - Prob. 3D.5AECh. 3 - Prob. 3D.5BECh. 3 - Prob. 3D.1PCh. 3 - Prob. 3D.2PCh. 3 - Prob. 3D.3PCh. 3 - Prob. 3D.4PCh. 3 - Prob. 3D.5PCh. 3 - Prob. 3D.6PCh. 3 - Prob. 3E.1DQCh. 3 - Prob. 3E.2DQCh. 3 - Prob. 3E.1AECh. 3 - Prob. 3E.1BECh. 3 - Prob. 3E.2AECh. 3 - Prob. 3E.2BECh. 3 - Prob. 3E.3AECh. 3 - Prob. 3E.3BECh. 3 - Prob. 3E.4AECh. 3 - Prob. 3E.4BECh. 3 - Prob. 3E.5AECh. 3 - Prob. 3E.5BECh. 3 - Prob. 3E.6AECh. 3 - Prob. 3E.6BECh. 3 - Prob. 3E.1PCh. 3 - Prob. 3E.2PCh. 3 - Prob. 3E.3PCh. 3 - Prob. 3E.4PCh. 3 - Prob. 3E.5PCh. 3 - Prob. 3E.6PCh. 3 - Prob. 3E.8PCh. 3 - Prob. 3.1IACh. 3 - Prob. 3.2IA
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