Physics degrees of freedom problem

AI Thread Summary
The discussion revolves around calculating work and heat transfer for a 1.12 mol sample of an ideal diatomic gas undergoing a process with changing pressure and temperature. Participants express confusion over how to find work or heat (q) when neither pressure, volume, nor temperature are constant. The connection between degrees of freedom and heat capacity is highlighted as crucial for understanding the system's energy changes. Additionally, the linear relationship between pressure and temperature is noted as a potential avenue for analysis, suggesting that graphing pressure versus volume could help determine work done. Overall, the challenge lies in integrating these concepts to derive the necessary thermodynamic quantities.
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A 1.12 mol sample of an ideal diatomic gas at a pressure of 1.00 atm and temperature of 491 K undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 735 K and 1.39 atm. Assume 5 active degrees of freedom.


Neither pressure nor volume nor temp are constant, so I'm confused to how I'm supposed to find work or q.
 
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Have you seen this formula before?

\Delta Q = \Delta U \pm W, the \pm is there depending on how you define when is work positive. If I recall correctly U = \frac {q}{2} n R T where q is the numbers of degree of freedom.
 
squib said:
A 1.12 mol sample of an ideal diatomic gas at a pressure of 1.00 atm and temperature of 491 K undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 735 K and 1.39 atm. Assume 5 active degrees of freedom.


Neither pressure nor volume nor temp are constant, so I'm confused to how I'm supposed to find work or q.

A couple of things you need to key on. What is the connection between degrees of freedom and heat capacity? And what can you do with the fact that the P vs T curve is linear?
 
I assume I could graph P vs V and find work, but that seems like more work then should be neccesary. I can find the change in U, or overall energy of the system, the only problem I'm having is with work, which should give me heat since I know U.
 
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