Translational KE, specific heats at constant volume and pressure

AI Thread Summary
The discussion revolves around calculating the average translational kinetic energy and specific heats of helium and nitrogen gases contained in copper cylinders. Participants attempted to compute the average kinetic energy using the formula Kave = (3/2)kBT but encountered discrepancies in their results, particularly for helium. The specific heats for both gases were confirmed as correct, but the calculated ratio γ (Cp/CV) for helium was questioned due to conflicting results. Ultimately, it was revealed that the homework assignment contained an error, leading the professor to drop the problem from grading. This highlights the importance of verifying calculations and understanding the underlying principles of thermodynamics.
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Homework Statement


Two copper cylinders, immersed in a water tank at 30.3 °C , contain helium and nitrogen, respectively. The helium-filled cylinder has a volume twice as large as the nitrogen-filled cylinder.

a) Calculate the average translational kinetic energy of a helium molecule and the average translational kinetic energy of a nitrogen molecule.

b) Determine the molar specific heat at constant volume (CV) and at constant pressure (Cp) for the two gases.

c) Find γ for the two gases.

Given/Known information
Cp = Cv+R
CV = \frac{3}{2}R = 12.5 J/(mol*K) monatomic gas
Cp = \frac{5}{2}R = 20.8 J/(mol*K) monatomic gas
CV = 20.7 J/(mol*K) Nitrogen gas
Cp = 29.1 J/(mol*K) Nitrogen gas
T = 30.3 °C = 303.3 K

Homework Equations



Kave = \frac{3}{2}kBT
γ = Cp/CV

The Attempt at a Solution



Part a)
Helium is a monatomic molecule.
The average energy per degree of freedom is given by \frac{1}{2}kB for each gas molecule. I was asked for the average translational kinetic energy, since a monatomic molecule has 3 translational degrees of freedom, and Nitrogen has 3 translational degrees of freedom and 2 rotational degrees, then the average kinetic energy is equal to,

Helium

Kave=\frac{3}{2}kBT

Kave=\frac{3}{2}(1.38\ \times\ 10^{-23}\ J\ K^{-1})(303.3 K)

Kave=6.278\ \times\ 10^{-21}\ J ←INCORRECT

Nitrogen

Kave=\frac{3}{2}(1.38\ \times\ 10^{-23}\ J\ K^{-1})(303.3 K)

Kave=6.278\ \times\ 10^{-21}\ J ←INCORRECT

I don't know if the two gases being inside of copper cylinders affects the calculations somehow since there is no temperature change, no reaction going on, or anything that will change the equation. I know that for a polyatomic molecule you need to consider the additional degrees of freedom available, but for both He and N2, there are 3 translational degrees of freedom, giving us the \frac{3}{2}. I don't know why that one is wrong.

Part b)
Values were obtained from a table in the book on Molar Specific Heats at constant volume and constant pressure. Also on this table was a column for the values of γ=Cp/CV for some gases.

CV, He = 12.5 J/(mol*K) ←CORRECT
Cp, He = 20.8 J/(mol*K) ←CORRECT
CV, N2 = 20.7 J/(mol*K) ←CORRECT
Cp, N2 = 29.1 J/(mol*K) ←CORRECT


Part c)
γ is stated to be the ratio Cp/CV

γHe = Cp, He/CV, He
γHe = (20.8 J/(mol*K))/(12.5 J/(mol*K))
γHe = 1.664 ≈ 1.67
INCORRECT

γN2 = Cp, N2/CV, N2
γN2 = (29.1 J/(mol*K))/(20.7 J/(mol*K))
γN2 = 1.4058 ≈ 1.4 ←CORRECT


I really don't know what other value γHe could be if we are told γHe=Cp/CV. Using values that I verified are correct for part b, taking Cp/CV for He should give me the correct answer, which the table in the book also verifies that it is a value of 1.67. Even more confusing is how doing that calculation for N2 gives me the correct answer for γN2.

All help is greatly appreciated.

Thanks in advance

Please let me know if something is confusing or if something looks like a typo, this took me a little while to proofread and edit and I went back numerous times to make it easier to read so I may have missed something.
 
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How do you know the answers you marked as incorrect are incorrect?
 
Isn't the equation kave=3/2*N*kb*T where N is the number of molecules in the vessel? What is N? I'm guessing it isn't 1.
 
RTW69 said:
Isn't the equation kave=3/2*N*kb*T where N is the number of molecules in the vessel? What is N? I'm guessing it isn't 1.

castrodisastro said:
a) Calculate the average translational kinetic energy of a helium molecule and the average translational kinetic energy of a nitrogen molecule.


I read "a molecule" as ##N=1##. :-p
 
DrClaude said:
How do you know the answers you marked as incorrect are incorrect?

This online homework assignment is designed in a way that for every question, in one field I type the answer in numerical form, and in a separate field I type in the units. After every time I fill it in I can hit a button that says check my answer and it just tells me if it's wrong or right.

RTW69 said:
Isn't the equation kave=3/2*N*kb*T where N is the number of molecules in the vessel? What is N? I'm guessing it isn't 1.

The question states that I have "two copper cylinders that contain helium and nitrogen respectively" In one container, the one with the helium, would be one molecule. That would be N=1.

The equation Kave=\frac{3}{2}kBT is written this way specifically for a monatomic gas. N would equal 1.
 
Try converting T using K = 273.15 + C°

AM
 
Andrew Mason said:
Try converting T using K = 273.15 + C°

AM

All my calculations were done using 303.5 K as my temp. Which is (30.3 °C) + 273 = 303.5 K
 
castrodisastro said:
All my calculations were done using 303.5 K as my temp. Which is (30.3 °C) + 273 = 303.5 K
You are using 303.3 K as the temperature. I suggested that you use 303.45K as the temperature and see if your answers match the given "correct" answers. You could also try rounding your final answer to the correct significant figures.

AM
 
Actually, our professor informed us that the homework assignment has an error on this problem, preventing it from giving students full credit. So he dropped this one. Thanks for the help though.
 
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