Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
Question
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Chapter 3, Problem 3A.4AE

(i)

Interpretation Introduction

Interpretation: The change in entropies of the system, surroundings and the total change in entropy, when a sample of nitrogen gas of mass 14g at 298K doubles its volume in an reversible isothermal expansion has to be calculated.

Concept introduction: Entropy is a thermodynamic quantity which represents the disorder or randomness of a system.  It is the thermal energy per unit temperature which is unavailable to do the useful work.  Entropy change depends only on the initial and final states.  In an isothermal process, the change in temperature is zero while in an adiabatic process the change in heat absorbed or given out is zero.

(i)

Expert Solution
Check Mark

Answer to Problem 3A.4AE

The entropy change of the given the given system is 2.88JK-1_, the entropy change of the surroundings is 2.88JK-1_, and the total entropy change is 0JK-1_.

Explanation of Solution

The relation between entropy change of the system, surroundings and the total entropy change is given by the equation

    ΔStotal=ΔSsys+ΔSsurr                                                                                     (1)

Where,

  • ΔStotal is the total change in entropy.
  • ΔSsys is the change in entropy for system.
  • ΔSsurr is the change in entropy for surroundings.

For a reversible isothermal expansion, the total entropy change is zero.  Therefore  Therefore,

    ΔStotal=0

Substitute the value of  ΔStotal in equation (1).Substitute the value of ΔStotal in equation (1).

  ΔStotal=ΔSsys+ΔSsurr0=ΔSsys+ΔSsurr

Hence,

  ΔSsurr=ΔSsys                                                                                               (2)

The mass of nitrogen gas is 14g.

The molar mass of nitrogen gas is 28gmol-1

The number of moles of nitrogen gas is calculated by the formula,

      Numberofmoles=GivenmassMolarmass

Substitute the values of given mass and molar mass in the above equation.

  Numberofmoles=GivenmassMolarmass=14g28gmol-1

The expression for entropy change is given as

  ΔSsys=2.303nRlogVfVi                                                                                    (3)

Where,

  • R is the gas constant (8.314 J K1 mol1).
  • n is the number of moles.
  • Vf is the final volume.
  • Vi is the initial volume.

The gas undergoes isothermal reversible expansion and doubles its volume. Hence,

    Vf=2Vi

Substitute the values of n, RVi and Vf in equation (3) to calculate the change in entropy of system of nitrogen gas as

  ΔSsys=2.303×14g28gmol1×8.314JK-1mol-1×log(2ViVi)=2.88JK-1_

Substitute the value of ΔSsys in equation (2).

  ΔSsurr=ΔSsys=2.88JK-1_

Hence, the change in entropy of system for nitrogen gas is 2.88JK-1_, the change in entropy for surroundings is 2.88JK-1_ and the total entropy change is 0JK-1_

(ii)

Interpretation Introduction

Interpretation: The change in entropies of the system, surroundings and the total change in entropy when a sample of nitrogen of mass 14g at 298K doubles its volume in an isothermal irreversible expansion against pex=0 has to be calculated.

Concept introduction: Entropy is a thermodynamic quantity which represents the disorder or randomness of a system.  It is the thermal energy per unit temperature which is unavailable to do the useful work.  Entropy change depends only on the initial and final states.  In an isothermal process, the change in temperature is zero while in an adiabatic process the change in heat absorbed or given out is zero.

(ii)

Expert Solution
Check Mark

Answer to Problem 3A.4AE

The entropy change of the given system is 2.88JK-1_, the entropy change of the surroundings is 0JK-1_ and the total entropy change is 2.88JK-1_.

Explanation of Solution

The relation between entropy change of the system, surroundings and the total entropy change is given by the equation

    ΔStotal=ΔSsys+ΔSsurr                                                                                    (1)

Where,

  • ΔStotal is the total change in entropy
  • ΔSsys is the change in entropy for system
  • ΔSsurr is the change in entropy for surroundings

For an irreversible isothermal expansion against pex=0  the entropy change of the surroundings is zero.  Therefore

  ΔSsurr=0

Substitute the value of ΔSsurr=0 in equation (1)

    ΔStotal=ΔSsys+ΔSsurrΔStotal=ΔSsys+0

Hence,

    ΔStotal=ΔSsys                                                                                                 (2)

The mass of nitrogen gas is 14g.

The molar mass of nitrogen gas is 28gmol-1.

The number of moles of nitrogen gas is calculated by the formula,

      Numberofmoles=GivenmassMolarmass

Substitute the values of given mass and molar mass in the above equation

  Numberofmoles=GivenmassMolarmass=14g28gmol-1

The expression for entropy change is given as

  ΔSsys=2.303nRlogVfVi                                                                                    (3)

Where,

  • R is the gas constant (8.314 J K1 mol1).
  • n is the number of moles.
  • Vf is the final volume.
  • Vi is the initial volume.

The gas undergoes isothermal reversible expansion and doubles its volume Hence

    Vf=2Vi

Substitute the values of n, RVi and Vf  in equation (3) to calculate the change in entropy of system of nitrogen gas as,

  ΔSsys=2.303×14g28gmol1×8.314JK-1mol-1×log(2ViVi)=2.88JK-1_

Substitute the value of ΔSsys in equation (2).

  ΔStotal=ΔSsys=2.88JK-1_

Hence, the change in entropy of system for nitrogen gas is 2.88JK-1_, the change in entropy for surroundings is 0JK-1_ and the total entropy change is 2.88JK-1_.

(iii)

Interpretation Introduction

Interpretation: The change in entropies of the system, surroundings and the total change in entropy when a sample of nitrogen of mass 14g at 298K doubles its volume in an adiabatic reversible expansion has to be calculated.

Concept introduction: Entropy is a thermodynamic quantity which represents the disorder or randomness of a system.  It is the thermal energy per unit temperature which is unavailable to do the useful work.  Entropy change depends only on the initial and final states.  In an isothermal process, the change in temperature is zero while in an adiabatic process, the change in heat absorbed or given out is zero.

(iii)

Expert Solution
Check Mark

Answer to Problem 3A.4AE

The entropy change of the given system is 2.88JK-1_, the entropy change of the surroundings is 0_ and the total entropy change is 2.88JK-1_.

Explanation of Solution

The relation between entropy change of the system, surroundings and the total entropy change is given by the equation,

    ΔStotal=ΔSsys+ΔSsurr                                                                                     (1)

Where,

  • ΔStotal is the total change in entropy
  • ΔSsys is the change in entropy for system
  • ΔSsurr is the change in entropy for surroundings

For adiabatic reversible expansion the total entropy change is zero.  Therefore

    ΔStotal=0

Substitute the value of ΔStotal in equation (1).

    ΔStotal=ΔSsys+ΔSsurr0=ΔSsys+ΔSsurr

Hence,

    ΔSsurr=ΔSsys                                                                                                (2)

The expression for entropy change of surroundings is given as, as,

    ΔSsurr=qT                                                                                                       (3)

For a reversible adiabatic process,,

  q=0

Substitute the value of q in equation (3).

    ΔSsurr=qT=0T=0

Therefore

  ΔSsurr=0

Substitute the value of ΔSsurr in equation (2).

ΔSsys=ΔSsurr=0JK-1_

Hence, the change in entropy of system for nitrogen gas is 0JK-1_, the change in entropy for surroundings is 0JK-1_ and the total entropy change is 0JK-1_.

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Chapter 3 Solutions

Atkins' Physical Chemistry

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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