Thermodynamics: Compression of an Adiabatic Gas

AI Thread Summary
The discussion focuses on the compression of a monatomic ideal gas, specifically calculating initial and final pressures, and work input for both isothermal and adiabatic processes. The initial pressure was determined to be 11.6 atm, with a final pressure of 34.8 atm, and the work for the isothermal process calculated as 3.88 kJ. Participants clarify that adiabatic conditions imply no heat exchange (q = 0), but do not mean no change in internal energy (ΔU ≠ 0). The need for a different equation to calculate work for the adiabatic process is emphasized, as the standard equations for isothermal processes do not apply. The discussion highlights the complexities of thermodynamic calculations in varying conditions.
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Homework Statement


Assume 1.500 mol of a monatomic ideal gas is compressed from 3.00 L to 1.00 L.

a. If the initial and final temperature is 10.0 °C, what are the initial and final pressures (in atm)?

b. How much work input (in kJ) is required if a reversible isothermal path at 10.0 °C is followed?

c. How much work input (in kJ) is required if the compression is adiabatic rather than isothermal? Assume the initial temperature is 10.0 °C.

Homework Equations


PV=nRT
w=-nRTln(vf/vi)
U=q+w

The Attempt at a Solution


I got both a and b using the first two equations I listed. I got the inital pressure to be 11.6 atm and the final pressure to be 34.8 atm. For the second part I got the work to be 3.88 kJ.

I'm stuck on the third part. I know adiabatic means that there no change in U, so q=-w but I don't know how that applies to the question.
 
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You're missing an equation for the energy U of an ideal gas. And adiabatic does not mean ΔU = 0, it means q = 0.
 
So ΔU=-PΔV, where U=w

But the problem isn't isobaric, so that equation wouldn't work. I can't seem to find another equation though...
 
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