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

• chocolatepie
In summary: I forget how many, but it's less than Neon?Yes, oxygen has 6 degrees of freedom and Ne has...well I forget how many, but it's less than Neon.
chocolatepie
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)

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?

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Hi chocolatepie!

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.

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.

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?

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

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.
What was it?

Could you explain why it is 5?

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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?

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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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. )

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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.

Now I have:
(a) qv = 3325.6J
(b) qp = 3525.79J

?!

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?

## 1. How do you calculate the heat energy of an oxygen and neon gas mixture?

The heat energy of a gas mixture can be calculated by multiplying the mass of the mixture by the specific heat capacity of the mixture and the change in temperature. This formula can be written as Q = m * c * ΔT, where Q is the heat energy, m is the mass of the mixture, c is the specific heat capacity, and ΔT is the change in temperature.

## 2. What is the specific heat capacity of an oxygen and neon gas mixture?

The specific heat capacity of a gas mixture is the amount of heat energy required to raise the temperature of 1 gram of the mixture by 1 degree Celsius. The specific heat capacity of an oxygen and neon gas mixture can be calculated by taking the weighted average of the specific heat capacities of oxygen and neon, based on their respective masses in the mixture.

## 3. How do you determine the mass of oxygen and neon in a gas mixture?

The mass of oxygen and neon in a gas mixture can be determined by using the ideal gas law, which states that the number of moles of a gas is equal to its mass divided by its molar mass. By knowing the number of moles of oxygen and neon in the mixture, their masses can be calculated using their respective molar masses.

## 4. Can the heat energy of an oxygen and neon gas mixture be negative?

Yes, the heat energy of a gas mixture can be negative if the temperature of the mixture decreases. This would indicate a loss of heat energy from the mixture.

## 5. How does the temperature affect the heat energy of an oxygen and neon gas mixture?

The temperature has a direct effect on the heat energy of a gas mixture. As the temperature increases, the heat energy of the mixture also increases. Similarly, as the temperature decreases, the heat energy of the mixture decreases. This is because the change in temperature is a key factor in the calculation of heat energy.

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