Question 38: Thermodynamics & KTG

Question

The total number of degrees of freedom associated with $2 \mathrm{~cm}^3$ of Nitrogen gas at normal temperature and pressure is:
[Given Avogadro number as ' N ' ]

Accepted Answer

First, find how many NÃ¢ÂÂ molecules are in 2 cmÃÂ³ at NTP:
Volume per mole at NTP: 22.4 L = 22400 cmÃÂ³
Number of moles in 2 cmÃÂ³:
$$n = \frac{2\ {m cm}^3}{22400\ {m cm}^3/m mol} = \frac1{11200}\ {m mol}$$
Number of molecules:
$$N_{m mol} = N 	imes n = \frac{N}{11200}$$
At Ã¢ÂÂnormalÃ¢ÂÂ (room) temperature the vibrational mode of NÃ¢ÂÂ is essentially frozen, so each molecule has
Ã¢ÂÂ¢ 3 translational + 2 rotational = 5 degrees of freedom.
Total degrees of freedom:
$$ 	ext{dof}_{m total} = 5 	imes \frac{N}{11200} = \frac{5N}{11200} = \frac{N}{2240}. $$
Answer: Option C, $\displaystyle \frac{N}{2240}$.

Answer

$\frac{N}{44800}$

Answer

$\frac{N}{4480}$

Answer

$\frac{N}{2240}$

Answer

$\frac{N}{22400}$

Thermodynamics & KTG

The total number of degrees of freedom associated with 2 \mathrm{~cm}^3 of Nitrogen gas at normal temperature and pressure is: [Given Avogadro number as ' N ' ]

Choose the correct answer: