Calculating Heat Energy: Oxygen and Neon Gas Mixture (HELP)

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Discussion Overview

The discussion revolves around calculating the heat energy required to increase the temperature of a mixture of oxygen and neon gases from 298K to 398K, both at constant volume and under constant pressure conditions. Participants explore the application of the equipartition theorem and the relevant equations for heat capacity.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Participants discuss the use of the equipartition theorem and its implications for calculating heat energy, questioning what temperature should be used in the calculations.
  • Some participants propose using Q = Cv ΔT for constant volume and Q = Cp ΔT for constant pressure, while others suggest different values for Cv and Cp based on the molecular structure of the gases involved.
  • There is confusion regarding the degrees of freedom for diatomic and monatomic gases, with some participants asserting that oxygen has 5 degrees of freedom and neon has 3, while others challenge these claims.
  • Participants debate the correct application of formulas for internal energy and enthalpy, particularly regarding the addition of ΔU for each gas in the mixture and the implications of constant volume versus constant pressure scenarios.
  • Some participants express uncertainty about the calculations and request verification of their mathematical approaches, leading to further discussion about the correct values to use for heat capacities and the conversion from grams to moles.

Areas of Agreement / Disagreement

There is no consensus on the correct values for heat capacities or the degrees of freedom for the gases. Multiple competing views remain regarding the application of the equipartition theorem and the calculations involved in determining heat energy.

Contextual Notes

Participants note the importance of distinguishing between heat capacities for different types of gases and the need to convert mass to moles for accurate calculations. There are unresolved questions about the assumptions made in the calculations and the definitions of terms used in the discussion.

chocolatepie
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Calculating Heat Energy (HELP)

Homework Statement


A mixture consisting of 1.00g oxygen gas and 1.00g neon gas is trapped within a 1.0L container at an initial temperature of 298K. Calculate the heat energy required to increase the temperature of this mixture to 398K
(a) at a constant volume of 1.0L
(b) under conditions where the container is allowed to expand against a constant pressure pext. The pressure pext is equal to the initial pressure of the mixture at 298K.
* note: base your calculations on the equipartition theorem. (do not tabulate heat capacity values)

Homework Equations



C=q/ΔT
Um(T)=3/2RT (*)

The Attempt at a Solution


(a) ΔU = qv when volume is constant
When using equipartition theorem (*), what should I put for temperature?

(b) ΔH = qp when pressure is constant.
Should I use formula H = U + pV to calculate ΔH?

Thanks in advance!
 
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Hi chocolatepie! :smile:

I'm not really sure what is intended by the equipartition theorem.
Can you enlighten me?

I think you're supposed to use Q = Cv ΔT for constant volume and Q = Cp ΔT for constant pressure.

I guess you can use Um(T)=3/2RT for Neon, or rather Cv=3/2R and Cp=5/2R.
But I suspect you should use a different formula for Oxygen since it is diatomic (Cv=5/2R and Cp=7/2R).
And anyway, to use this, you would need to convert your masses to moles.
 
Ok, so becuase O2 is diatomic, I am going to use Um(T) = 6/2RT = 3RT

I don't understand why Um(T) became Q? (is Q heat? meaning q?)
 
chocolatepie said:
Ok, so becuase O2 is diatomic, I am going to use Um(T) = 6/2RT = 3RT

Errr... no.
Could you check your notes?

Um(T)=Cv T.
Cv depends on the type of molecule.
It is Cv=f/2 R where f is the number of degrees of freedom (max 3 translational and max 3 rotational).
chocolatepie said:
I don't understand why Um(T) became Q? (is Q heat? meaning q?)

Oh sorry, I always try to use the symbols from the OP, but I am used to Q myself instead of q.
I think that in your case they mean the same thing.
(Actually, I'm used to using a capital Q when referring to a mol, and a lowercase q when referring to a kg, but it seems you do not make that distinction.)
 
I like Serena said:
Errr... no.
Could you check your notes?

Um(T)=Cv T.
Cv depends on the type of molecule.
It is Cv=f/2 R where f is the number of degrees of freedom (max 3 translational and max 3 rotational).




Oh sorry, I always try to use the symbols from the OP, but I am used to Q myself instead of q.
I think that in your case they mean the same thing.
(Actually, I'm used to using a capital Q when referring to a mol, and a lowercase q when referring to a kg, but it seems you do not make that distinction.)


My note says that Um(T)=(x/2)RT
Don't we write C for heat capacity? so does it mean x/2 is a heat capacity?


So far, I did the following:
because V= const, ΔU = qv (v meant to remind that it was at constant V)
ΔUm(T) = 6/2RΔT = 3RΔT = 2494.2 J for oxygen
ΔUm(T) = 3RΔT = 1247.1J for Ne

To calculate heat energy required, should I add these ΔUm(T) to use ΔH=ΔU +pΔV (where pΔV is gone because ΔV is 0)?
 
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chocolatepie said:
My note says that Um(T)=(x/2)RT
Don't we write C for heat capacity? so does it mean x/2 is a heat capacity?


So far, I did the following:
because V= const, ΔU = qv (v meant to remind that it was at constant V)
ΔUm(T) = 6/2RΔT = 3RΔT = 2494.2 J for oxygen
ΔUm(T) = 3RΔT = 1247.1J for Ne

To calculate heat energy required, should I add these ΔUm(T) to use ΔH=ΔU +pΔV (where pΔV is gone because ΔV is 0)?

Yes, we write C for heat capacity (for 1 mol).
More specifically, we write Cv for the heat capacity at constant volume, and Cp for the heat capacity at constant pressure.
Furthermore the heat capacity is Cv = (x/2)R.
(You need to include the R.)

For an ideal gas Um(T)=Cv ΔT.
The value of Cv depends on the type of gas.
We distinguish 3 varieties: monatomic (like Neon), diatomic (like Oxygen), and more complex molecules that extend in 3 dimensions (like NH3).
Each has a different Cv.
 
Could you check my math in my previous post?
 
Doesn't oxygen have 6 degrees of freedom and Ne has 3?
 
  • #10
To calculate heat energy required, should I add these ΔUm(T) to use ΔH=ΔU +pΔV (where pΔV is gone because ΔV is 0)?
 
  • #11
chocolatepie said:
Doesn't oxygen have 6 degrees of freedom and Ne has 3?

No, oxygen has 5 degrees of freedom.

And yes, Ne has 3 degrees, but you did not substitute this value correctly for x.


chocolatepie said:
To calculate heat energy required, should I add these ΔUm(T) to use ΔH=ΔU +pΔV (where pΔV is gone because ΔV is 0)?

Not quite.

You have the energy of a mol, but you need the energy of a gram of each.

Furthermore, the question asks for the heat required at constant volume.
That is qv.
You already had a formula for that.
What was it?
 
  • #12
Could you explain why it is 5?
 
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  • #13
At constant V, I had a formula ΔU = qv

Does it mean I need to add ΔU for each gas since they are in a mixture?

Also, I am using ΔU +PΔV =q for the part (b). There is no work value given.. and so is pressure
 
Last edited:
  • #14
chocolatepie said:
Could you explain why it is 5?

Each molecule has 3 translational degrees of freedom.
The rotational degrees of freedom is what this is about.

If you have a complex molecule like NH3, there are 3 rotational axes to which the molecule will resist movement so to speak.
This gives a total of 6 degrees of freedom.

With a diatomic molecule, the atoms are lined up on an axis.
If you try to rotate it, there is 1 axis to which it will not "resist" movement (the axis along the atoms).
And there are 2 axes to which it will "resist" movement.
So a diatomic molecule has 3+2=5 degrees of freedom.

A monatomic molecule does not "resist" any rotation, so it only has the 3 translational degrees of freedom.You can find it for instance here:
http://en.wikipedia.org/wiki/Ideal_gas#Heat_capacity
(But not so detailed I'm afraid. o:))
 
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  • #15
chocolatepie said:
At constant V, I had a formula ΔU = qv

Does it mean I need to add ΔU for each gas since they are in a mixture?

Yes.

The total internal energy is the sum of the internal energy of each of the components.


chocolatepie said:
Also, I am using ΔU +PΔV =q for the part (b). There is no work value given.. and so is pressure

Yes, you can use that.
 
  • #16
Now I have:
(a) qv = 3325.6J
(b) qp = 3525.79J

?!
 
  • #17
chocolatepie said:
Now I have:
(a) qv = 3325.6J
(b) qp = 3525.79J

?!

I get something different.
How did you calculate it?

Did you take into account that you have 1 gram of each component and not 1 mole?
 

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