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Question: What is the average translational kinetic energy of nitrogen molecules at 1600 K?

Short Answer

Expert verified

Answer

The average translational kinetic energy of nitrogen moleculesis3.31×10-20J.

Step by step solution

01

Given data  

The temperature of the nitrogen gas molecules is 1600 K.

02

Understanding the concept

The expression for the average kinetic energy is given by,

Kavg=32KT

Here K is the Boltzmann constant, T is the temperature, Kavg is the average kinetic energy.

The value of k is equal to the1.38×10-23 J/molecule ·K .

03

Calculate the average translational kinetic energy

Using the expression of the kinetic average energy,

Kavg=32KT

Substitute 1.38×10-23 J/molecule ·Kfor k , 1600 k for T into the above equation,

Kavg=32×1.38×10-23×1600Kavg=3.31×10-20J

Therefore the average translational kinetic energy of nitrogen moleculesis3.31×10-20J.

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Most popular questions from this chapter

The temperature of3.00 molof a gas withCv=6.00 cal/mol.K is to be raised50.0 K . If the process is at constant volume, what are (a) the energy transferred as heat Q, (b) the work W done by the gas, (c) the changeΔ·¡int in internal energy of the gas, and (d) the changeΔ°­ in the total translational kinetic energy? If the process is at constant pressure, what are (e) Q, (f) W, (g) Δ·¡int, and (h) Δ°­? If the process is adiabatic, what are (i) Q, (j) W, (k)Δ·¡int , and (l)Δ°­?

The temperature of 2.00of an ideal monatomic gas is raised 15.0 Kin an adiabatic process. What are

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(b) the energy transferred as heat Q,

(c) the changeΔ·¡int in internal energy of the gas, and

(d) the changeΔ°­in the average kinetic energy per atom?

Suppose 4.00 molof an ideal diatomic gas, with molecular rotation but not oscillation, experienced a temperature increase of 60.0Kunder constant pressure conditions. What are

  1. The energy transferred as heat Q .
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  3. The work W done by the gas.
  4. The change ΔKin total translational kinetic energy of the gas?

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