PHYSICAL CHEMISTRY-WEBASSIGN
PHYSICAL CHEMISTRY-WEBASSIGN
11th Edition
ISBN: 9780357087411
Author: ATKINS
Publisher: CENGAGE L
Question
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Chapter 13, Problem 13A.1ST
Interpretation Introduction

Interpretation:

The ratio of the population of the levels with J=2 and J=1 of HCl at 25οC has to be calculated.

Concept introduction:

Statistical thermodynamics is used to describe all the possible configurations in a system at given physical quantities such as pressure, temperature, and a number of particles in the system.  The important quantity in the thermodynamic is partition function that is represented as shown below.

  qigieβi

It is also known as canonical ensemble partition function.

Expert Solution & Answer
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Answer to Problem 13A.1ST

The ratio of the population of the levels with J=2 and J=1 of HCl at 25οC is 1.359_.

Explanation of Solution

The expression for the energy of the state with quantum number J is shown below.

  εJ=hcB˜J(J+1)        (1)

Where,

  • h is Plank’s constant.
  • c is the velocity of light.
  • B˜ is rotational constant.
  • J is the rotational quantum number.

The energy of the state with quantum number J=1 is calculated by substituting J=1 in equation (1) as shown below.

  ε1=hcB˜×1(1+1)=2hcB˜

The energy of the state with quantum number J=2 is calculated by substituting J=2 in equation (1) as shown below.

  ε2=hcB˜×2(2+1)=6hcB˜

The expression for the Boltzmann population ratio is shown below.

  NiNj=eβ(εiεj)        (2)

Where,

  • NiNj is the ratio of the population of the upper state and lower state.
  • εi is the energy of the state with a quantum number i.
  • εj the energy of the state with a quantum number j.
  • β=1/kT, here k is Boltzmann’s constant and T is the temperature.

The energy separation of the states with J=2 and J=1 is shown below.

  ε2ε1=6hcB˜2hcB˜=4hcB˜

Substitute the values of i=2 and j=1 in equation (2) as shown below.

  N2N1=eβ(ε2ε1)

Substitute the value of ε2ε1 in the above equation.

  N2N1=e4hcB˜β

The upper level with rotational quantum number has five-fold degeneracy and the lower level has three-fold degeneracy.  So the relative population of the levels is calculated as shown below.

  N2N1=53e4hcB˜β

Substitute β=1/kT in the above equation as shown below.

  N2N1=53e4hcB˜kT        (3)

The value of kT/hc is 207.22cm1 at 25οC and the value of B˜ is 10.591cm1.  Substitute the value of kT/hc and B˜ in equation (3) as shown below.

  N2N1=53e4hcB˜kT=53e4×10.591cm1207.22cm1=53e42.364207.22=53e0.204

The above equation is further solved as shown below.

  N2N1=53e0.204=53×0.8154=1.359_

Therefore, the ratio of the population of the levels with J=2 and J=1 of HCl at 25οC is 1.359_.

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Chapter 13 Solutions

PHYSICAL CHEMISTRY-WEBASSIGN

Ch. 13 - Prob. 13A.1AECh. 13 - Prob. 13A.1BECh. 13 - Prob. 13A.2AECh. 13 - Prob. 13A.2BECh. 13 - Prob. 13A.3AECh. 13 - Prob. 13A.3BECh. 13 - Prob. 13A.4AECh. 13 - Prob. 13A.4BECh. 13 - Prob. 13A.5AECh. 13 - Prob. 13A.5BECh. 13 - Prob. 13A.6AECh. 13 - Prob. 13A.6BECh. 13 - Prob. 13A.1PCh. 13 - Prob. 13A.2PCh. 13 - Prob. 13A.4PCh. 13 - Prob. 13A.5PCh. 13 - Prob. 13A.6PCh. 13 - Prob. 13A.7PCh. 13 - Prob. 13B.1DQCh. 13 - Prob. 13B.2DQCh. 13 - Prob. 13B.3DQCh. 13 - Prob. 13B.1AECh. 13 - Prob. 13B.1BECh. 13 - Prob. 13B.2AECh. 13 - Prob. 13B.2BECh. 13 - Prob. 13B.3AECh. 13 - Prob. 13B.3BECh. 13 - Prob. 13B.4AECh. 13 - Prob. 13B.4BECh. 13 - Prob. 13B.7AECh. 13 - Prob. 13B.7BECh. 13 - Prob. 13B.8AECh. 13 - Prob. 13B.8BECh. 13 - Prob. 13B.9AECh. 13 - Prob. 13B.9BECh. 13 - Prob. 13B.10AECh. 13 - Prob. 13B.10BECh. 13 - Prob. 13B.11AECh. 13 - Prob. 13B.11BECh. 13 - Prob. 13B.12AECh. 13 - Prob. 13B.12BECh. 13 - Prob. 13B.4PCh. 13 - Prob. 13B.5PCh. 13 - Prob. 13B.6PCh. 13 - Prob. 13B.7PCh. 13 - Prob. 13B.8PCh. 13 - Prob. 13B.10PCh. 13 - Prob. 13C.1DQCh. 13 - Prob. 13C.2DQCh. 13 - Prob. 13C.1AECh. 13 - Prob. 13C.1BECh. 13 - Prob. 13C.6AECh. 13 - Prob. 13C.6BECh. 13 - Prob. 13C.7AECh. 13 - Prob. 13C.7BECh. 13 - Prob. 13C.3PCh. 13 - Prob. 13C.7PCh. 13 - Prob. 13C.8PCh. 13 - Prob. 13C.9PCh. 13 - Prob. 13D.1DQCh. 13 - Prob. 13D.2DQCh. 13 - Prob. 13D.3DQCh. 13 - Prob. 13D.4DQCh. 13 - Prob. 13D.1AECh. 13 - Prob. 13D.1BECh. 13 - Prob. 13D.1PCh. 13 - Prob. 13D.2PCh. 13 - Prob. 13E.1DQCh. 13 - Prob. 13E.2DQCh. 13 - Prob. 13E.3DQCh. 13 - Prob. 13E.4DQCh. 13 - Prob. 13E.5DQCh. 13 - Prob. 13E.6DQCh. 13 - Prob. 13E.1AECh. 13 - Prob. 13E.1BECh. 13 - Prob. 13E.2AECh. 13 - Prob. 13E.2BECh. 13 - Prob. 13E.3AECh. 13 - Prob. 13E.3BECh. 13 - Prob. 13E.4AECh. 13 - Prob. 13E.4BECh. 13 - Prob. 13E.5AECh. 13 - Prob. 13E.5BECh. 13 - Prob. 13E.6AECh. 13 - Prob. 13E.6BECh. 13 - Prob. 13E.7AECh. 13 - Prob. 13E.7BECh. 13 - Prob. 13E.8AECh. 13 - Prob. 13E.8BECh. 13 - Prob. 13E.9AECh. 13 - Prob. 13E.9BECh. 13 - Prob. 13E.1PCh. 13 - Prob. 13E.2PCh. 13 - Prob. 13E.3PCh. 13 - Prob. 13E.4PCh. 13 - Prob. 13E.7PCh. 13 - Prob. 13E.9PCh. 13 - Prob. 13E.10PCh. 13 - Prob. 13E.11PCh. 13 - Prob. 13E.14PCh. 13 - Prob. 13E.15PCh. 13 - Prob. 13E.16PCh. 13 - Prob. 13E.17PCh. 13 - Prob. 13F.1DQCh. 13 - Prob. 13F.2DQCh. 13 - Prob. 13F.3DQCh. 13 - Prob. 13F.1AECh. 13 - Prob. 13F.1BECh. 13 - Prob. 13F.2AECh. 13 - Prob. 13F.2BECh. 13 - Prob. 13F.3AECh. 13 - Prob. 13F.3BECh. 13 - Prob. 13F.3PCh. 13 - Prob. 13F.4PCh. 13 - Prob. 13F.5PCh. 13 - Prob. 13F.6PCh. 13 - Prob. 13.1IA
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