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At room temperature, show that KBT≈1/ 40 eV. It is useful to memorize this result, because it tells a lot about what phenomena are likely to occur at room temperature.

Short Answer

Expert verified

The result can be proved using the energy equation.

Step by step solution

01

Identification of given data

  • Boltzmann’s constant iskb=8.62×10-5eVK .
  • The room temperature isT=20°C=293K .
02

Concept of energy

The ability to do tasks is known as energy. Depending on the form it takes, it may be either potential or kinetic or thermal or electrical or chemical, or nuclear.

The energy equation is given by,

E=kbT…… (i)

Here kbis Boltzmann’s constant.

Tis the room temperature.

03

Evaluation of energy equation

The energy equation can be proved using equation (i)

E=8.62×105eVK×293KE=8.62×10-5×293.1eV1K×1KE=0.025eVE140eV

Thus, the temperature equation is proved.

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Most popular questions from this chapter

A block of copper at a temperature of 50°C is placed in contact with a block of aluminium at a temperature of45°C in an insulated container. As a result of a transfer of 2500 J of energy from the copper to the aluminium, the final equilibrium temperature of the two blocks is 48°C. (a) What is the approximate change in the entropy of the aluminium block? (b) What is the approximate change in the entropy of the copper block? (c) What is the approximate change in the entropy of the Universe? (d) What is the change in the energy of the Universe?

The interatomic spring stiffness for tungsten is determined from Young’s modulus measurements to be 90 N. The mass of one mole of tungsten is 0185 kg . If we model a block of tungsten as a collection of atomic “oscillators” (masses on springs), note that since each oscillator is attached to two “springs,” and each “spring” is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above.Use these precise values for the constants: h̶=1.05×10-34J.s(Planck’s constant divided by 2π), Avogadro’s number = 6.0221×1023molecule/mole, kB=1.3807×10-23J/K(the Boltzmann constant). (a) What is one quantum of energy for one of these atomic oscillators? (b) Figure 12.56 contains the number of ways to arrange a given number of quanta of energy in a particular block of tungsten. Fill in the blanks to complete the table, including calculating the temperature of the block. The energy E is measured from the ground state. Nothing goes in the shaded boxes. Be sure to give the temperature to the nearest 0.1 kelvin. (c) There are about 60 atoms in this object. What is the heat capacity on a per-atom basis? (Note that at high temperatures the heat capacity on a per-atom basis approaches the classical limit of 3kB = 4.2×10−23 J/K/atom.)

Verify that this equation gives the correct number of ways to arrange 0, 1, 2, 3, or 4 quanta among 3 one-dimensional oscillators, given in earlier tables (1, 3, 6, 10, 15).






Q1

Q2

#ways1

#ways2

#ways1

#ways2

0

4

1

15

15

1

3

3

10

30

2

2

6

6

36

3

1

10

3

30

4

0

15

1

15






Young’s modulus for copper is measured by stretching a copper wire to be about 1.2×1011N/m2. The density of copper is about 9g/cm3, and the mass of a mole is .Starting from a very low temperature, use these data to estimate roughly the temperature T at which we expect the specific heat for copper to approach 3 kB . Compare your estimate with the data shown on a graph in this chapter.

A carbon nanoparticle (very small particle) contains 6000 carbon atoms. According to the Einstein model of a solid, how many oscillators are in this block?

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