A box contains a mixture of {H_2} and He gase | vedclass

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(D) The average translational kinetic energy $(KE_{avg})$ of any gas molecule depends only on the absolute temperature $T$ and is given by the formula

Vedclass · Accepted Answer

Both $(A)$ and $(C)$.

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(D) The average translational kinetic energy $(KE_{avg})$ of any gas molecule depends only on the absolute temperature $T$ and is given by the formula $KE_{avg} = \frac{3}{2} k_B T$. Since both gases are in the same box,they are at the same temperature. Therefore,the average translational kinetic energies of ${H_2}$ molecules and $He$ atoms are equal.The total average energy $(E_{avg})$ of a gas molecule depends on its degrees of freedom $(f)$,given by $E_{avg} = \frac{f}{2} k_B T$. For a monoatomic gas like $He$,the degrees of freedom $f = 3$. For a diatomic gas like ${H_2}$ at room temperature,the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational). Since $f_{H_2} > f_{He}$,the average energy of ${H_2}$ molecules is greater than that of $He$ atoms.Thus,both statements $(A)$ and $(C)$ are correct.

$A$ box contains a mixture of ${H_2}$ and $He$ gases. Which of the following statements are correct?

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