Let v denote the rms speed of the molecules in an ideal diatomic gas at absolute temperature T. The mass of a molecule is ‘m’ Neglecting vibrational energy terms, the false statement is :

Question

Let v denote the rms speed of the molecules in an ideal diatomic gas at absolute temperature T. The mass of a molecule is ‘m’ Neglecting vibrational energy terms, the false statement is : (A) A molecule can have a speed greater than √2 v (B) v is proportional to √T (C) The average rotational K.E. of a molecule is (1/4) m v2 (D) The average K.E. of a molecule is (5/6) m ν2

Tardigrade Admin · Accepted Answer

Total K.E. = (1/2) fkT = (1/2) (5)k ((mν2/3k)) = (5/6) mν2

Q. Let v denote the rms speed of the molecules in an ideal diatomic gas at absolute temperature T. The mass of a molecule is ‘m’ Neglecting vibrational energy terms, the false statement is :

A molecule can have a speed greater than $2 v$

$v$ is proportional to $T$

The average rotational $K . E .$ of a molecule is $\frac{1}{4} m v^{2}$

The average K.E. of a molecule is $\frac{5}{6} m ν^{2}$

Total $K . E . = \frac{1}{2} f k T = \frac{1}{2} (5) k (\frac{m ν ^{2}}{3 k}) = \frac{5}{6} m ν^{2}$

Q. Let v denote the rms speed of the molecules in an ideal diatomic gas at absolute temperature T. The mass of a molecule is ‘m’ Neglecting vibrational energy terms, the false statement is :

A molecule can have a speed greater than 2​v

v is proportional to T​

The average rotational K.E. of a molecule is 41​mv2

The average K.E. of a molecule is 65​mν2

Total K.E.=21​fkT=21​(5)k(3kmν2​)=65​mν2

A molecule can have a speed greater than $2 v$

$v$ is proportional to $T$

The average rotational $K . E .$ of a molecule is $\frac{1}{4} m v^{2}$

The average K.E. of a molecule is $\frac{5}{6} m ν^{2}$

Total $K . E . = \frac{1}{2} f k T = \frac{1}{2} (5) k (\frac{m ν ^{2}}{3 k}) = \frac{5}{6} m ν^{2}$