In a certain gas $\frac{2}{5}$ th of the energy of molecules is associated with the rotation of molecules and the rest of it is associated with the motion of the centre of mass. The average translation energy of one such molecule, when the temperature is $27^{\circ} C$ is given by $x \times 10^{- 23} J$ ,then find $x$ ?

621

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623

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6.21

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62.1

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Solution

The correct option is A 621
Since two fifth energy is associated with the rotation, thus there are five degrees of freedom out of which two are rotational and three translational.

Hence, $E_{t r a n s .} = \frac{3}{2} k T = \frac{3}{2} (1.38 \times 10^{- 23}) (300) = 6.21 \times 10^{- 21} J$

Answer is $621 \times 10^{- 23} J$

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Similar questions

Q. In a certain gas

\frac{2}{5}

th of the energy of molecules is associated with the rotation of molecules and the rest of it is associated with the motion of the centre of mass. How much energy must be supplied to one mole of this gas at constant volume to raise the temperature by

1^{\circ} C

The oxygen molecule has a mass of $5.30 \times 10^{- 26} k g$ and a moment of inertia of $1.94 \times 10^{- 46} k g m^{2}$ about an axis through its centre perpendicular to the lines joining the two atoms.
Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Q. The oxygen molecule has a mass of 5.30 × 10¯²⁶ kg and a moment of inertia of 1.94×10-¯⁴⁶ kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Q. Given Avagadro number =

6.02 \times 10^{23}

and Boltzmann's constant =

1.38 \times 10^{- 23}

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In a certain gas 25th of the energy of molecules is associated with the rotation of molecules and the rest of it is associated with the motion of the centre of mass. The average translation energy of one such molecule, when the temperature is 27∘C is given by x×10−23 J,then find x?

The correct option is A 621Since two fifth energy is associated with the rotation, thus there are five degrees of freedom out of which two are rotational and three translational.Hence, Etrans.=32kT=32(1.38×10−23)(300)=6.21×10−21JAnswer is 621×10−23J