This set of Aerodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Vibrational Rate Equations”.

1. What is the value of transition probability?

a) 0

b) More than 1

c) Less than 1

d) 1

View Answer

Explanation: Transition probability is the probability that a molecule will jump to another i + 1 level after the molecular collision. The transition of the molecule moving to a higher energy level requires several number of collisions. This probability value is always less than 1.

2. What does the product of collision frequency and transition probability yield?

a) Number of transitions per particle per second

b) Number of collisions per second

c) Number of collisions per second per particle

d) Transitions per collision

View Answer

Explanation: The collision frequency (Z) when multiplied with the transition probability (P

_{i, i + 1}) yield the number of transitions per particle per second. Since the collision frequency is the number of collisions taking place per particle and the transition probability gives the number of transitions taking place per collision per particle.

3. What is the formula to compute net rate change of the population of the i^{th} level?

a) \(\frac {dN_i}{dt}\) = P_{i + 1, i} ZN_{i + 1} + P_{i – 1, i} ZN_{i – 1} – P_{i, i + 1} ZN_{i} – P_{i, i – 1} ZN_{i}

b) \(\frac {dN_i}{dt}\) = – P_{i + 1, i} ZN_{i + 1} + P_{i – 1, i} ZN_{i – 1} + P_{i, i + 1} ZN_{i} + P_{i, i – 1} ZN_{i}

c) \(\frac {dN_i}{dt}\) = P_{i + 1, i} ZN_{i + 1} – P_{i, i + 1} ZN_{i}

d) \(\frac {dN_i}{dt}\) = + P_{i – 1, i} ZN_{i – 1} – P_{i, i – 1} ZN_{i}

View Answer

Explanation: The rate of change of population of the molecules in i

^{th}level is computed by adding rate of increase of N

_{i}(population of i

^{th}level) and rate of decrease of N

_{i}. Thus the formula is:

\(\frac {dN_i}{dt}\) = P

_{i + 1, i}ZN

_{i + 1}+ P

_{i – 1, i}ZN

_{i – 1}– P

_{i, i + 1}ZN

_{i}– P

_{i, i – 1}ZN

_{i}

Where, P

_{i + 1, i}ZN

_{i + 1}+ P

_{i – 1, i}ZN

_{i – 1}is the rate of increase of population in i

^{th}level due to the molecules jumping up from i – 1 and down from i + 1 levels.

P

_{i, i + 1}ZN

_{i}– P

_{i, i – 1}ZN

_{i}is the rate of decrease of population in the i

^{th}level due to the molecules jumping from i

^{th}level to i + 1 and i – 1 levels.

4. What is the master equation for vibrational relaxation?

a) \(\frac {dN_i}{dt}\) = k_{i + 1, i} ZN_{i + 1} + k_{i – 1, i} ZN_{i – 1} – k_{i, i + 1} ZN_{i} – k_{i, i – 1} ZN_{i}

b) \(\frac {dN_i}{dt}\) = k_{i + 1, i} N_{i + 1} + k_{i – 1, i} N_{i – 1} – k_{i, i + 1}N_{i} – k_{i, i – 1} N_{i}

c) \(\frac {dN_i}{dt}\) = ZN_{i + 1} + ZN_{i – 1} – ZN_{i, i – 1} – ZN_{i}

d) \(\frac {dN_i}{dt}\) = P_{i + 1, i} ZN_{i + 1} + P_{i – 1, i} ZN_{i – 1}

View Answer

Explanation: In the formula derived to obtain then net rate of change of population of the i

^{th}level, the product of transition probability and collision frequency is expressed in the form of a new variable known as vibrational rate constant k

_{i + 1, i}= P

_{i + 1, i}Z (this is an example of one of the transitions). Thus the formula is reduced from \(\frac {dN_i}{dt}\) = P

_{i + 1, i}ZN

_{i + 1}+ P

_{i – 1, i}ZN

_{i – 1}– P

_{i, i + 1}ZN

_{i}– P

_{i, i – 1}ZN

_{i}to:

\(\frac {dN_i}{dt}\) = k

_{i + 1, i}N

_{i + 1}+ k

_{i – 1, i}N

_{i – 1}– k

_{i, i + 1}N

_{i}– k

_{i, i – 1}N

_{i}

5. Which of these is the vibrational rate equation?

a) \(\frac {de_{vib}}{dt} = \frac {1}{τ}\)(e\(_{vib}^{eq}\) – e_{vib})

b) τ = \(\frac {1}{k_{1, 0} (1 – e^{- hv/kT} )}\)

c) e_{vib} = τ(e\(_{vib}^{eq}\) – e_{vib})

d) \(\frac {de_{vib}}{dt}\) = \(\frac {1}{τ}\)(e\(_{vib}^{eq}\) – e_{vib})

View Answer

Explanation: The vibrational rate equation gives a relation between e

_{vib}which is the time rate change with the difference between the equilibrium and the local instantaneous non – equilibrium value (e\(_{vib}^{eq}\) – e

_{vib}). It is given by the following relation which is a differential equation:

\(\frac {de_{vib}}{dt} = \frac {1}{τ}\)(e\(_{vib}^{eq}\) – e

_{vib})

6. In case of translation-vibration transfers, there can be an increase or decrease in kinetic energy.

a) True

b) False

View Answer

Explanation: When a molecule undergoes a translation-vibration transfer also known as T – V transfer, molecules tend to lose or gain vibrational energy. This energy is reflected in the form of an increase or decrease in the kinetic energy of the molecule undergoing collision.

**Sanfoundry Global Education & Learning Series – Aerodynamics.**

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