Translational KE, specific heats at constant volume and pressure

In summary, the conversation discusses the calculation of average translational kinetic energy, molar specific heat at constant volume and constant pressure, and the value of γ for two gases, helium and nitrogen. The equation Kave=\frac{3}{2}kBT is used to calculate the average translational kinetic energy for both gases, and the values for molar specific heat and γ are obtained from a table. However, there seems to be some confusion regarding the correct values for helium, with the given answers marked as incorrect. The professor later informs that there is an error in the homework assignment.
  • #1
castrodisastro
82
0

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 = [itex]\frac{3}{2}[/itex]R = 12.5 J/(mol*K) monatomic gas
Cp = [itex]\frac{5}{2}[/itex]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 = [itex]\frac{3}{2}[/itex]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 [itex]\frac{1}{2}[/itex]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=[itex]\frac{3}{2}[/itex]kBT

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

Kave=[itex]6.278\ \times\ 10^{-21}\ J[/itex] ←INCORRECT

Nitrogen

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

Kave=[itex]6.278\ \times\ 10^{-21}\ J[/itex] ←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 [itex]\frac{3}{2}[/itex]. 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.
 
Physics news on Phys.org
  • #2
How do you know the answers you marked as incorrect are incorrect?
 
  • #3
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.
 
  • #4
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##. :tongue2:
 
  • #5
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=[itex]\frac{3}{2}[/itex]kBT is written this way specifically for a monatomic gas. N would equal 1.
 
  • #6
Try converting T using K = 273.15 + C°

AM
 
  • #7
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
 
  • #8
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
 
  • #9
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.
 

1. What is translational kinetic energy?

Translational kinetic energy is the energy an object possesses due to its motion in a straight line. It is directly proportional to the mass and square of the velocity of the object.

2. How is translational kinetic energy related to temperature?

According to the kinetic theory of gases, the average kinetic energy of gas particles is proportional to the temperature of the gas. This means that an increase in translational kinetic energy of the particles results in an increase in temperature.

3. What is specific heat at constant volume?

Specific heat at constant volume is the amount of heat required to raise the temperature of a substance by one degree Celsius at constant volume. It is a measure of the amount of internal energy a substance possesses.

4. How is specific heat at constant volume different from specific heat at constant pressure?

Specific heat at constant volume is measured at a constant volume, meaning the substance cannot expand or do work. On the other hand, specific heat at constant pressure is measured at a constant pressure, allowing the substance to expand and do work. This results in a difference in the amount of heat required to raise the temperature of the substance.

5. Why is the specific heat at constant pressure greater than the specific heat at constant volume?

The specific heat at constant pressure is greater because at constant pressure, heat energy is not only used to increase the internal energy of the substance, but also to do work by expanding the substance. This means that more heat energy is required to raise the temperature of the substance compared to when it is held at a constant volume.

Similar threads

Replies
1
Views
1K
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
7K
  • Mechanical Engineering
Replies
3
Views
960
  • Introductory Physics Homework Help
Replies
6
Views
11K
  • Introductory Physics Homework Help
Replies
12
Views
1K
Back
Top