Calculation of variation of entropy knowing state's equation

AI Thread Summary
The discussion revolves around calculating the entropy change of 1.00 mol of an ideal diatomic gas as it expands from pressure P and volume V to pressure 2P and volume 2V. The initial calculations using Maxwell relations and the ideal gas law led to an unexpected result of zero entropy change, prompting a reevaluation of the process. It is noted that to achieve this change, the temperature must increase by four times, indicating that heat must be added to the system. The participants express a need for a clearer relationship between entropy, volume, and pressure changes, particularly for ideal gases. The conversation highlights the complexities of thermodynamic calculations and the variations in notation from different courses.
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Homework Statement



A sample of 1.00 mol of an ideal diatomic gas, initially at pressure P and volume V, expands until it has a pressure of 2P and a volume of 2V. What's the entropy change in the gas on this process?

Homework Equations



2nd Maxwell relation:
7edca29d9b641071f66f0bf13db5aeb8.png


4th Maxwell relation:
4679c4ddd8322c720f4d1d30c44b724b.png


PV = nRT

The Attempt at a Solution



\frac{\partial S}{\partial V}= \frac{nR}{V}
\frac{\partial S}{\partial P} = -\frac{nR}{P}

I found out S by doing the standard procedure: integrating the first one and deriving in relation to P to find the function in relation to P. The expression for S i found was:

S=nR(lnV-lnP)

But the variation of S is always 0 this way, and that's not the solution...
 
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Did you get it yet? Think about what has to happen to double your volume yet double your pressure. Ordinarily when you double your volume what would you expect to happen to the pressure? And if you double the pressure, what would you expect to happen to the volume? So for these to happen together something else has to happen also.

What do both of these changes do to the entropy of the system?
 
Last edited:
netgypsy said:
Did you get it yet? Think about what has to happen to double your volume yet double your pressure. Ordinarily when you double your volume what would you expect to happen to the pressure? And if you double the pressure, what would you expect to happen to the volume? So for these to happen together something else has to happen also.

What do both of these changes do to the entropy of the system?

No, I didn't find it yet. When the pressure and volume both double up, the temperature has to rise by 4x?
 
Exactly. So intuitively you have to add heat to make system do this. Entropy rises with both an increase in volume and an increase in pressure. All the relations I've found though use specific heat. There are so many things you can calculate but I'm looking for something cut and dried relating entropy to both increase in volume and pressure although you can calculate the relative increase in temperature as you did and try to go from there. So I'm continuing to look and hoping someone who's more current on thermo for ideal gases will see this. another problem I'm running into is that I took physical chemistry (a year course) as a substitute for thermo (a semester course not nearly as rigorous) so there is a bit of a notation difference also. I'll keep trying and get back to you.
 
Do you know the correct answer to this question?
 
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