How to Solve for Joule-Coefficient in a Gas System?

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Homework Statement


I have the following task:
U0ctjpj.png


Homework Equations

The Attempt at a Solution



I already managed to do a), but how do I solve b) ? What is this Joule-Coefficient and how can I calculate this for the two different kinds of gases without any given values?
 
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krootox217 said:

Homework Statement


I have the following task:
U0ctjpj.png


Homework Equations

The Attempt at a Solution



I already managed to do a), but how do I solve b) ? What is this Joule-Coefficient and how can I calculate this for the two different kinds of gases without any given values?
The equation for the Joule-Coefficient is given in the problem statement. Are you asking what it means physically?

As far as the two different kinds of gases are concerned, you are supposed to do it algebraically. For an ideal gas, I assume you know PV=nRT. Do you know the P-V-T equation for the van der Waals approximation to real gas behavior?

Chet
 
Hi,

Does this mean, that I have to do it like this:

rCjDah4.png


And now just simplify?
 
Thanks a lot!

Is this already the result, or can I simplify it even more?

ZmQeBtN.png
 
krootox217 said:
Thanks a lot!

Is this already the result, or can I simplify it even more?

ZmQeBtN.png
If pV=nRT, what is pV-nRT equal to?

If ##p=\frac{RT}{V_m-b}-\frac{a}{V_m^2}##, what is ##p-\frac{RT}{V_m-b}## equal to?

Chet
 
Chestermiller said:
If pV=nRT, what is pV-nRT equal to

0

Chestermiller said:
If p=RTVmbaV2mp=\frac{RT}{V_m-b}-\frac{a}{V_m^2}, what is pRTVmbp-\frac{RT}{V_m-b} equal to?

-a/(Vm)^2

And this means the coeficient is in the upper case 0 and in the lower case -a/(Vm)^2 and therefore these are the final results?
 
krootox217 said:
0

-a/(Vm)^2

And this means the coeficient is in the upper case 0 and in the lower case -a/(Vm)^2 and therefore these are the final results?
Yes, but don't forget to divide by Cv.
 
Oh I nearly forgot, thank you!

This means it is in the first case still 0 and in the second case -a/((Vm)^2*Cv)
 
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Btw. I'm new here. Is there a possibility to give a plus point or something like that for your help?
 
krootox217 said:
Btw. I'm new here. Is there a possibility to give a plus point or something like that for your help?
Please excuse my manners. Welcome to Physics Forums!

If you want to thank someone for their help, click Like at the bottom of their post.

Chet
 
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