Energy and root mean square velocity question

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Homework Help Overview

The discussion revolves around the application of kinetic theory to an ideal diatomic gas, specifically addressing the relationship between internal energy, degrees of freedom, and root mean square velocity. Participants explore the implications of using different degrees of freedom in the context of the equations for internal energy and velocity.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants examine the distinction between translational and rotational degrees of freedom for diatomic molecules. There is a focus on the implications of using the expression for root mean square velocity and how it relates to the degrees of freedom. Questions arise regarding discrepancies in numerical examples from textbooks and the correctness of given values for specific heat capacities.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of degrees of freedom and their relevance to the problem. Some participants have raised concerns about the accuracy of specific heat values presented in the problem, while others reflect on the challenges posed by exam formats and textbook examples.

Contextual Notes

There are noted discrepancies in the values of specific heat capacities provided in the problem, which may affect the calculations. Participants are also considering the implications of the assumptions made in the context of kinetic theory and the behavior of ideal gases.

Krushnaraj Pandya
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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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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.
Krushnaraj Pandya said:
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.
 
Chandra Prayaga said:
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 ##.
 
Charles Link said:
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.
Yes, I felt the same thing but there are solved examples which show you're supposed to do it this way in the book- so basically they wanted to test if we knew meyer's relation or had just memorized the value of R I think.
The JEE exams are notorious in the way that they've to think of new things since a million students appear for the exam and 12,000 have to be selected so sometimes they twist stuff way too much. We have to play along though if we want to get into a really good grad school
 
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