Problem from Fermi's Thermodynamics

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SUMMARY

The discussion focuses on solving a thermodynamics problem from Fermi's Thermodynamics, specifically from chapter 4, involving the equation of state pV1.2 = 109T1.1. The thermal capacity is constant at 0.1 calories/K for a constant volume of 100 L. The first law of thermodynamics is applied to derive expressions for internal energy U as a function of temperature T and volume V. The final expression derived is U = (0.1 cal/K)T plus a function of T and V that vanishes when V = 100 L.

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Chasing_Time
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Hi all, I am getting stuck on this problem from chapter 4 of Fermi's Thermodynamics:

Homework Statement



"A body obeys the equation of state:

pV^{1.2} = 10^{9}T^{1.1}

A measurement of its thermal capacity inside a container having the constant volume of 100 L shows that under these conditions, the thermal capacity (heat) capacity is constant and equal to 0.1 calories / K. Express the energy of the system as a function of T and V."

Homework Equations



The first law: dQ = dU + dW.

Since we want U=U(T,V), we can express the first law as:

<br /> \left(\frac{\partial Q}{\partial T}\right)_V dT + \left(\frac{\partial Q}{\partial V}\right)_T dV = \left(\frac{\partial U}{\partial T}\right)_V dT + [(\left(\frac{\partial U}{\partial V}\right)_T + (\frac{10^9*T^{1.1}}{V^{1.2}})] dV<br />

We also know:

<br /> \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p<br />

where again p can be substituted from the defined equation of state.

The Attempt at a Solution



I'm tempted to say that \left(\frac{\partial Q}{\partial T}\right)_V = 0.1 cal / K with the corresponding V = 100 L. If I do this and assume all dV = 0 for constant volume I get what appears to be a trivial result. I feel like I need to get a perfect differential to solve for U(T,V), but I don't see how. Other than that I haven't done much more than toy with the above equations. Thanks for your time.
 
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I have looked at the problem again recently but still don't have much more insight. I see that, by the first law, since no work is being done (constant volume):

<br /> <br /> \left(\frac{\partial Q}{\partial T}\right)_V = C_v = \left(\frac{\partial U}{\partial T}\right)_V<br /> <br />

Using <br /> <br /> \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p<br /> <br />, setting up a perfect differential and taking the derivatives, we can get:

dU = (0.1 cal / K) dT + \frac{10^{8}T^{1.1}}{V^{1.2}}dV, but I don't see how to go further. Any ideas?
 
Last edited:
Try expressing the constant-volume heat capacity as

C_V=T\left(\frac{\partial S}{\partial T}\right)_V

Then determine S by integration (note that a function of V appears that needs to be evaluated via a Maxwell relation), and use

U=TS-PV+\mu N

where \mu needs to be evaluated by using the heat capacity condition.

This was pretty difficult to work through. There may be an easier way. I eventually got U=(0.1\,\mathrm{cal}/\mathrm{K})T plus a function of T and V that vanished when V=100\,\mathrm{L}.
 

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