# Energy and root mean square velocity question

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## Homework Statement

I read the expression E=fRT/2 where E is internal energy of ideal gas and f is degrees of freedom, and $V_{rms} = \sqrt{\frac{3RT}{M}}$ Since internal energy for an ideal gas is purely kinetic (according to KTG) I can write 1/2 mv^2 = fRT/2. Now H2 is a diatomic molecule and has 5 degrees of freedom, yet I see the expression $V_{rms} = \sqrt{\frac{3RT}{M}}$ being used for it instead of $V_{rms} = \sqrt{\frac{5RT}{M}}$ What am I missing?

## Homework Equations

All related to KTG

## The Attempt at a Solution

for conceptual clarity, as mentioned

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Chandra Prayaga
The extra two degrees of freedom for the diatomic molecule are rotational degrees of freedom. They have nothing to do with the translational motion of the molecule as a whole. For a diatomic molecule, the motion of the molecule "as a whole" is the motion of the center of mass, which has three degrees of freedom The average energy of the gas is RT/2 per degree of freedom per mole, so it is equal to 5RT/2 per mole. The average energy of a molecule is no longer equal to 1/2 mVrms2, but has additional contribution from rotations.
You should be careful about the distinction between the average energy of a molecule, and the average energy of the gas, which has, as you know, a lot of molecules in it. For example, the average energy of a molecule is not given by fRT/2.
1/2 mv^2 = fRT/2
.
The right hand side would be the average energy of one mole of the gas. The average kinetic energy of a molecule is (3/2) kBT where kB is the Boltzmann constant.

Gold Member
The extra two degrees of freedom for the diatomic molecule are rotational degrees of freedom. They have nothing to do with the translational motion of the molecule as a whole. For a diatomic molecule, the motion of the molecule "as a whole" is the motion of the center of mass, which has three degrees of freedom The average energy of the gas is RT/2 per degree of freedom per mole, so it is equal to 5RT/2 per mole. The average energy of a molecule is no longer equal to 1/2 mVrms2, but has additional contribution from rotations.
You should be careful about the distinction between the average energy of a molecule, and the average energy of the gas, which has, as you know, a lot of molecules in it. For example, the average energy of a molecule is not given by fRT/2.
.
The right hand side would be the average energy of one mole of the gas. The average kinetic energy of a molecule is (3/2) kBT where kB is the Boltzmann constant.
Alright, I understand. Thank you very much for explaining this :D
there's a numerical in my book which seems to have a discrepancy though (from which my question on this post arised). Its as follows-
"calculate the rms speed of an ideal diatomic gas having molecular weight 32 gm/mol at 273 K If Cp=9.3 J/mol K and Cv=6.34 J/mol K"
Using Cp-Cv=R and putting it into the equation $V_{rms} = \sqrt{\frac{3RT}{M}}$ gives an incorrect answer while plugging it into $V_{rms} = \sqrt{\frac{5RT}{M}}$ gives the correct one. Is this a mistake on the part of the book?

I'm joining this discussion in the middle, but your numbers for $C_P$ and $C_V$ look incorrect. The gas constant $R=8.314$ J/mol K. The $C_p=9.3$ and $C_V=6.34$ look incorrect. $C_V=\frac{5}{2} R \approx 20.8$. That would make $C_P \approx 29.1$.
I'm joining this discussion in the middle, but your numbers for $C_P$ and $C_V$ look incorrect. The gas constant $R=8.314$ J/mol K. The $C_p=9.3$ and $C_V=6.34$ look incorrect.