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Chapter 19: Q26P (page 578)

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

Question: A container encloses 2 mol of an ideal gas that has molar mass M1 and 0.5 mol of a second ideal gas that has molar mass m2 = 3 .m1 What fraction of the total pressure on the container wall is attributable to the second gas? (The kinetic theory explanation of pressure leads to the experimentally discovered law of partial pressures for a mixture of gases that do not react chemically: The total pressure exerted by the mixture is equal to the sum of the pressures that the several gases would exert separately if each were to occupy the vessel alone.)

Figure shows a cycle undergone by 1.00molof an ideal monoatomic gas. The temperatures are T1=300K, T2=600K and T3=455K.For 1→2, what are- (a)Heat Q (b)The change in internal energyΔEint (c) The work done W? For3→1, (d) What is Q? (e) What is Q? (f) What is W? For3→1, (g)What is Q?(h) What isΔEint? (i) What is W? For the full cycle, (j) What is Q? (k)What is ΔEint? (l) What is W? The initial pressure at point 1 is 1.00atm.What are the (m) volume and (n) pressure at point 2? What are the (o) volume and (p) pressure at point 3?

Question: Oxygen (O2) gas at 273 K and 1 atm is confined to a cubical container 10 cm on a side. CalculateΔUg/Kavg , whereΔUg is the change in the gravitational potential energy of an oxygen molecule falling the height of the box and Kavg is the molecule’s average translational kinetic energy.

Suppose 12.0gof oxygen (O2) gas is heated at constant atmospheric pressure from25.0∘Cto125.0∘Cto.

  1. How many moles of oxygen are present?
  2. How much energy is transferred to the oxygen as heat? (The molecules rotate but do not oscillate).
  3. What fraction of heat is used to raise the internal energy of the oxygen?

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 changeΔEint in internal energy of the gas, and (d) the changeΔK in the total translational kinetic energy? If the process is at constant pressure, what are (e) Q, (f) W, (g) ΔEint, and (h) ΔK? If the process is adiabatic, what are (i) Q, (j) W, (k)ΔEint , and (l)ΔK?
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