# Initial and Final Thermal Energies of a Gas

1. Jan 12, 2006

### G01

A 100cm^3 box (1X10^-4 m^3) contains helium at a pressure of 2.0 atm and at a temperature of 100C (373K) It is placed in thermal contact with a 200cm^3 (2x10^-4 m^3) box containing argon at a pressure of 4.0atm and a temp. of 400C (673K)
a) What is the initial thermal energy of each gas?
b) final thermal energy?
c)How much heat is transferred?
d)Final Temp?
e)Final pressure of each?

OK, I got part a).

First I used $$PV = nRT$$ to solve for the mols of each gas. Then I used

$$E_{th} = 3/2nRT$$

to find the thermal energy of each gas.
Helium- .007mol, 30.4 J
Argon- .0145 mol, 121.6J
These answers for the initial thermal energy are right. My problem starts in part B.

To find the thermal energy of gas A you'd use the formula:

$$E_A = \frac{n_A}{n_A+n_B}E_{total}$$

When I use this formula to find the final thermal energy of Helium I get:

$$E_{He} = \frac{n_{He}}{n_{He}+n_{Ar}}E_{total}$$

$$\frac{.007}{.0215}(152J) = 49.5 J$$

That would also give me 102.5 J for Argon. Both of these final thermal energies are wrong according to the book but I can't see my mistake. Because of this I get part c, d, and e wrong also. Can someone help me out here? Thanks.

2. Jan 12, 2006

### civil_dude

I wonder, since the gases aren't mixing, why use the formula for mixed gases.

3. Jan 12, 2006

### G01

My textbook tells me that that formula is not only for mixing gases but also for when two gasous systems thermally interact without mixing. I se no reason, according to my textbook for why I shouldn't use the formula:

$$E_{He} = \frac{n_He}{n{He}+n{Ar}}(E_{total}$$

The equilibrium temperature will be reached when the average kinetic energy of the helium atoms ($$\epsilon$$) is equal to the average kinetic energy of the Argon atoms:

$$\epsilon_{He} = \epsilon_{Ar}= \epsilon_{total}$$

If N is the number of atoms then this implies:

$$\frac{E_{He}}{N_{He}} = \frac{E_{Ar}}{N_{Ar}} = \frac{E_{total}}{N_{He}+N_{Ar}}$$

Now if we forget about Argon for now and solve for The final energy of Helium:

$$E_{He} = \frac{N_{He}}{N_{He}+N_{Ar}}(E_total)$$

Now if we divide numerator and denominator of that fraction by Avogadro's Number we get (if n is mols):

$$E_{He} = \frac{n_{He}}{n_{He}+n_{Ar}}(E_{total})$$

As you can see the only information needed to derive the equation I used was the fact that the gases were at the same temperature and thus had the same average kinetic energy per molecule. It doesn't matter whether the gases are mixed or not. I know I must have done something wrong. But I can't find it. (Unless of course, my understanding of this formula is wrong and everything I just typed was a collossal waste of time!)

Last edited: Jan 12, 2006
4. Jan 12, 2006

### G01

5. Jan 12, 2006

### G01

Bumping again

6. Jan 12, 2006

### lightgrav

well, you only kept one digit for the amount of Helium ...
I got 104.75 J and 47.25J ... if the book is VERY far off, it is wrong.
(we know that there's about twice as many Argons as Heliums,
and we know how much Energy there is to split between them ...
these are both monatomic; there's no expansion, nothing subtle here.

7. Jan 12, 2006

### G01

Your answers are exactly what the book got. I'm not doubting they are right . I just can't find what I did wrong. Did you use the formulas I used or different ones. I have no idea why my method didn't work. What did you do that was different from what I did?

Last edited: Jan 12, 2006
8. Jan 12, 2006

### lightgrav

keep more sig figs ... 3 is usually okay, 4 is better. ONE is NEVER enough.
You're within 3% of the book's answer, what do you want for roundoff?!

9. Jan 12, 2006

### G01

SERIOUSLY IS IT sig figs that have been screwing me up this entire time!!!!!!!!!!!!!!!!!!!!!!! OK I'm getting answers much closer now thank you. Man I hate sig figs...................... funny I feel stupid because I ended up knowing how to do the problem the entire time lol.