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
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Chapter 2, Problem 2B.3BE

(i)

Interpretation Introduction

Interpretation: The values of qwΔU and ΔH have to be calculated at constant pressure when the temperature is raised from 25 οC to 100 οC.

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state.  The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter.  Work is defined as the form of mechanical energy that is transferred from system to another.

(i)

Expert Solution
Check Mark

Answer to Problem 2B.3BE

The value of value of q and ΔH at constant pressure is 11.5802 kJ_.

The value of w at constant pressure is 0.62355 kJ_.

The value of ΔU at constant pressure is 10.95 kJ_.

Explanation of Solution

It is given that the constant-pressure heat capacity of a sample of a perfect gas varies with temperature according to the expression shown below.

    Cp/(J K1)=20.17+0.4001(T/K)                                                             (1)

The temperature is raised from 25 οC to 100 οC.

The conversion of Celsius to Kelvin is done as,

    οC=273 K

Therefore, the conversion of 25 οC to Kelvin is done as,

    25 οC=25+273 K=298 K

Similarly, the conversion of 100 οC to Kelvin is done as,

    100 οC=100+273 K=373 K

At constant pressure,

    q=ΔH=T1T2CpdT                                                                                          (2)

Substitute the value of heat capacity at constant pressure from equation (1) to equation (2).

  ΔH=T1T2CpdT=T1T220.17dT+T1T20.4001 TdT=20.17(T2T1)+0.4001(T222T122)

Now, substitute the values of initial and final temperature to calculate value of q and ΔH.

    ΔH=20.17(373298)+0.4001((373)22(298)22)=20.17×75+0.20005(13912988804)=11580.2 J=11.5802 kJ_

Therefore, the value of value of q and ΔH at constant pressure is 11.5802 kJ_ respectively.

The work done for a perfect gas is calculated by the formula,

    w=pdV=nRdT=(1)RdT=RdT                                                                                                 (3)

Substitute the values of universal gas constant and temperature change in the above equation to calculate the work done.

    w=8.314 J mol1 K1×(373 K298 K)=8.314 J mol1 K1×75 K=623.55 J=0.62355 kJ_

Therefore, the work done at constant pressure is 0.62355 kJ_.

The first law of thermodynamics is given by the expression,

    ΔU=q+w                                                                                                    (4)

Substitute the values of q and w in the above formula to calculate the internal energy.

    ΔU=11.5802 kJ+(0.62355 kJ)=11.5802 kJ0.62355 kJ=10.95 kJ_

Hence, the value of ΔU is 10.95 kJ_.

(ii)

Interpretation Introduction

Interpretation: The values of qwΔU and ΔH have to be calculated at constant volume when the temperature is raised from 25 οC to 100 οC.

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state.  The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter.  Work is defined as the form of mechanical energy that is transferred from system to another.

(ii)

Expert Solution
Check Mark

Answer to Problem 2B.3BE

The value of ΔU at constant volume is 10.9567 kJ_.

The value of value of ΔH at constant volume is 11.58025 kJ_.

At constant volume, w=0.

The value of q at constant volume is 10.9567 kJ_.

Explanation of Solution

At constant volume,

    CpCv=nR                                                                                                 (5)

For a perfect gas,

    n=1

Therefore,

    CpCv=RCv=CpR

Substitute the value of heat capacity at constant pressure in the above formula to calculate the heat capacity at constant volume.

    Cv=CpRCv=20.17+0.4001×T8.314Cv=11.856+0.4001×T

Now,

    Cv=(UT)vU=CvdT                                                                                                 (6)

Substitute the values of heat capacity at constant volume and integrate it.

    U=T1T2(11.856+0.4001×T)dT=T1T211.856dT+T1T20.4001×TdT=11.856(T2T1)+0.4001(T222T122)=11.856(373 K298 K)+0.4001((373)22(298)22)

Simplify the above equation,

    ΔU=11.856×75+0.20005(13912988804)=889.2+10067.5=10956.7 J=10.9567 kJ_

Hence, the value of ΔU is 10.9567 kJ_.

The relation between change in enthalpy, change in internal energy, and change in volume is,

    ΔH=ΔU+Δ(pV)                                                                                       (7)

Where,

  • Ø  ΔH is the change in enthalpy.
  • Ø  ΔU is the change in internal energy.
  • Ø  ΔV is the change in volume.
  • Ø  p is the pressure.

The equation (7) is also written as,

    ΔH=ΔU+Δ(pV)ΔH=ΔU+ΔpV+pΔV

At constant volume, pΔV=0. Therefore,

    ΔH=ΔU+ΔpVΔH=ΔU+V(nRT2VnRT1V)ΔH=ΔU+nR(T2T1)

Substitute the values of ΔU, universal gas constant and temperatures in the above formula to calculate the change in enthalpy.

    ΔH=10956.7 J+1×8.314 J mol1 K1(373 K298 K)=10956.7 J+8.314 J mol1 K1×75 K=11580.25 J=11.58025 kJ_

Hence, the value of value of ΔH is 11.58025 kJ_.

At constant volume, w=0. Now, the value of q is calculated as,

    ΔU=q+0ΔU=qq=10.9567 kJ_

Hence, the value of q is 10.9567 kJ_.

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Chapter 2 Solutions

Atkins' Physical Chemistry

Ch. 2 - Prob. 2A.4DQCh. 2 - Prob. 2A.5DQCh. 2 - Prob. 2A.1AECh. 2 - Prob. 2A.1BECh. 2 - Prob. 2A.2AECh. 2 - Prob. 2A.2BECh. 2 - Prob. 2A.3AECh. 2 - Prob. 2A.3BECh. 2 - Prob. 2A.4AECh. 2 - Prob. 2A.4BECh. 2 - Prob. 2A.5AECh. 2 - Prob. 2A.5BECh. 2 - Prob. 2A.6AECh. 2 - Prob. 2A.6BECh. 2 - Prob. 2A.1PCh. 2 - Prob. 2A.2PCh. 2 - Prob. 2A.3PCh. 2 - Prob. 2A.4PCh. 2 - Prob. 2A.5PCh. 2 - Prob. 2A.6PCh. 2 - Prob. 2A.7PCh. 2 - Prob. 2A.8PCh. 2 - Prob. 2A.9PCh. 2 - Prob. 2A.10PCh. 2 - Prob. 2B.1DQCh. 2 - Prob. 2B.2DQCh. 2 - Prob. 2B.1AECh. 2 - Prob. 2B.1BECh. 2 - Prob. 2B.2AECh. 2 - Prob. 2B.2BECh. 2 - Prob. 2B.3AECh. 2 - Prob. 2B.3BECh. 2 - Prob. 2B.4AECh. 2 - Prob. 2B.4BECh. 2 - Prob. 2B.1PCh. 2 - Prob. 2B.2PCh. 2 - Prob. 2B.3PCh. 2 - Prob. 2B.4PCh. 2 - Prob. 2B.5PCh. 2 - Prob. 2C.1DQCh. 2 - Prob. 2C.2DQCh. 2 - Prob. 2C.3DQCh. 2 - Prob. 2C.4DQCh. 2 - Prob. 2C.1AECh. 2 - Prob. 2C.1BECh. 2 - Prob. 2C.2AECh. 2 - Prob. 2C.2BECh. 2 - Prob. 2C.3AECh. 2 - Prob. 2C.3BECh. 2 - Prob. 2C.4AECh. 2 - Prob. 2C.4BECh. 2 - Prob. 2C.5AECh. 2 - Prob. 2C.5BECh. 2 - Prob. 2C.6AECh. 2 - Prob. 2C.6BECh. 2 - Prob. 2C.7AECh. 2 - Prob. 2C.7BECh. 2 - Prob. 2C.8AECh. 2 - Prob. 2C.8BECh. 2 - Prob. 2C.1PCh. 2 - Prob. 2C.2PCh. 2 - Prob. 2C.3PCh. 2 - Prob. 2C.4PCh. 2 - Prob. 2C.5PCh. 2 - Prob. 2C.6PCh. 2 - Prob. 2C.7PCh. 2 - Prob. 2C.8PCh. 2 - Prob. 2C.9PCh. 2 - Prob. 2C.10PCh. 2 - Prob. 2C.11PCh. 2 - Prob. 2D.1DQCh. 2 - Prob. 2D.2DQCh. 2 - Prob. 2D.1AECh. 2 - Prob. 2D.1BECh. 2 - Prob. 2D.2AECh. 2 - Prob. 2D.2BECh. 2 - Prob. 2D.3AECh. 2 - Prob. 2D.3BECh. 2 - Prob. 2D.4AECh. 2 - Prob. 2D.4BECh. 2 - Prob. 2D.5AECh. 2 - Prob. 2D.5BECh. 2 - Prob. 2D.1PCh. 2 - Prob. 2D.2PCh. 2 - Prob. 2D.3PCh. 2 - Prob. 2D.4PCh. 2 - Prob. 2D.5PCh. 2 - Prob. 2D.6PCh. 2 - Prob. 2D.7PCh. 2 - Prob. 2D.8PCh. 2 - Prob. 2D.9PCh. 2 - Prob. 2E.1DQCh. 2 - Prob. 2E.2DQCh. 2 - Prob. 2E.1AECh. 2 - Prob. 2E.1BECh. 2 - Prob. 2E.2AECh. 2 - Prob. 2E.2BECh. 2 - Prob. 2E.3AECh. 2 - Prob. 2E.3BECh. 2 - Prob. 2E.4AECh. 2 - Prob. 2E.4BECh. 2 - Prob. 2E.5AECh. 2 - Prob. 2E.5BECh. 2 - Prob. 2E.1PCh. 2 - Prob. 2E.2PCh. 2 - Prob. 2.1IACh. 2 - Prob. 2.3IACh. 2 - Prob. 2.4IACh. 2 - Prob. 2.5IACh. 2 - Prob. 2.6IACh. 2 - Prob. 2.7IA
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