Energy and root mean square velocity question

In summary, the discussion explains the discrepancy in using the equation ##V_{rms} = \sqrt{\frac{3RT}{M}}## for an ideal diatomic gas, as the average energy of a molecule is not equal to 1/2 mVrms2, but has an additional contribution from rotations. The correct equation to use is ##V_{rms} = \sqrt{\frac{5RT}{M}}##. There may be discrepancies in the given values, but it is important to understand the underlying concepts in order to solve problems effectively.
  • #1
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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  • #2
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.
 
  • #3
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?
 
  • #4
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 ##.
 
  • #5
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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Likes Navin and Charles Link

FAQ: Energy and root mean square velocity question

What is energy?

Energy is the capacity of a physical system to do work. It can exist in many forms, such as thermal, mechanical, electrical, chemical, and nuclear.

What is root mean square velocity?

Root mean square velocity is the square root of the average of the squared velocities of particles in a gas. It is a measure of the average speed of the particles in a gas.

How is energy related to root mean square velocity?

The kinetic energy of a gas is directly proportional to the root mean square velocity of its particles. This means that as root mean square velocity increases, so does the kinetic energy of the gas.

What factors affect root mean square velocity?

The root mean square velocity of gas particles is affected by temperature, molar mass, and the type of gas. As temperature increases, root mean square velocity also increases. Heavier molecules will have lower root mean square velocities compared to lighter molecules at the same temperature.

How is root mean square velocity calculated?

The root mean square velocity is calculated using the formula v(rms) = √(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas in kilograms per mole.

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