How about a basic, straightforward problem?Compute the rms rate of an oxygen molecule at room temperature.Use the outcomes of parta. to recognize the rms speed of a hydrogen molecule at room temperature.Use the outcomes of partb. to identify the rms rate of a mercury atom at 1200K.

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Here come the solutions…

Use the formula. Respeak to that oxygen is a diatomic molecule in day-to-day situations. Let"s not go nuts below via precision. Just use the approximate molecular mass of oxygen (2×16u=32u) and the approximate worth of room temperature (300K). Let"s attempt calculating it both ways: first through the mass of a molecule…

 vO2=√ 3kT m
vO2=√
 3(1.38×10−23J/K)(300K) (32u)(1.66× 10−27kg/u)
vO2=483.5m/s

and then with the mass of a mole…

 vO2=√ 3RT M
vO2=√
 3(8.31J/molK)(300K) (0.032kg/mol)
vO2=483.4m/s

The two answers are slightly various in the fourth considerable digit, however I said to be reasonable with the precision. Let"s simply say the answer is…

vO2=480m/s

Exploit the straightforward proportion of the 2 molecular masses. Oxygen is 16 times heavier than hydrogen on a per atom or per molecule comparikid (considering that both gases are diatomic in our daily lives). RMS rate is inversely proportional to the square root of mass (molecular or molar). This implies the rms speed of hydrogen need to be √16=4 times much faster. If you would like to check out the mathematical reasoning presented formally, here it is…

 vH2 = √3kT/mH2 vO2 √3kT/mO2
vH2
 =√ mO2 vO2 mH2
vH2
 =√ 32u =4 vO2 2u
vH2
 =4 vO2
 vH2=4vO2vH2=4(480m/s)vH2=1,920m/s

Another question via rigged numbers. The atomic mass of mercury (200u) is 100 times that of molecular hydrogen (2u). This distinction reduces the speed by 1√100=110. In a comparable vein, the temperature of these mercury atoms is 4 times that of the hydrogen molecules in partb. This adjust raises the rms rate by a aspect of √4=2. Combining both transforms provides a new rms speed that"s 210 of the old one. Again, if you would certainly choose to watch the mathematical reasoning presented formally, here it is…

 vHg = √3kTHg/mHg vH2 √3kTH2/mH2
vHg
 =√ THgmH2 vH2 TH2mHg
vHg
 =√ (1200K)(2u) vH2 (300K)(200u)
vHg
 = 2 = 1 vH2 10 5
 vHg=⅕vH2vHg=⅕(1,920m/s)vHg=384m/s

### exercise difficulty 2

Write somepoint else.
solution

### exercise difficulty 3

Write somepoint various.
solution

### exercise trouble 4

Derive the legislation of Dulong and also Petit by applying the equipartition of power to the atoms in a solid.
solution

Atoms in a solid have six levels of liberty. (Why?) As such, the power per atom is

 ⟨K⟩= 6 kT=3kT 2

Multiply by Avogadro"s continuous to acquire the interior power in a mole of atoms.

U=⟨K⟩NA=3kTNA

Specific heat is the price of adjust in inner energy via respect to temperature. Molar particular warm is this derivative used to the inner power in one mole of atoms.

 CV= ∂ (3kTNA)=3kNA ∂T

This mirrors that molar certain warm is a constant for all materials given that Boltzmann"s constant (k) and Avogadro"s consistent (NA) are both continuous. Substitution making use of unusually specific worths for the two constants returns the typically declared value of this constant.

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 CV=3kNA CV=3(1.3806503×10−23J/K)(6.0221415×1023/mol) CV=24.94J/molK  The slrfc.org Hypertextbook©1998–2021 Glenn ElertAuthor, Illustrator, Webmaster

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