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Calculating Heat Energy

  • #1
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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Answers and Replies

  • #2
I like Serena
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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.
 
  • #3
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?)
 
  • #4
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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).


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.)
 
  • #5
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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  • #6
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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.
 
  • #7
Could you check my math in my previous post?
 
  • #9
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
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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.


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
 
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  • #14
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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
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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.


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
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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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