What is the minimum temperature for an ideal gas following a circular path?

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SUMMARY

The discussion focuses on calculating the minimum temperature of an ideal gas following a circular path defined by the equation (V-10)^2 + (P-10)^2 = 25. The initial conditions are 2 moles of gas at P1 = 10 atm and V1 = 5 L, leading to an initial temperature calculation of T1 = 304.66 K. The maximum temperature is determined to be Tmax = 1116 K, while the minimum temperature is calculated as Tmin = 225 K. The participants confirm that the method of minimizing PV through graphical representation is valid and effective for this scenario.

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  • Graphing techniques for visualizing mathematical functions
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Homework Statement



2 moles of an ideal gas, P1 = 10 atm V1 = 5L, are taken reversibly in a clockwise direction around a circular path given by (V-10)^2 + (P-10)^2 = 25. All I have left to solve for is the minimum temperature.

Homework Equations



PV = nRT

The Attempt at a Solution



I found T1 = PV/nR = (10atm * 5L) / (2 moles * .08206) = 304.66K

From the equation I found out that the center of the circle is (V, P) = (10,10) with a Radius of 5.

From there I broke the circle up into 4 different sections starting at (5,10) and going clockwise. I found the max by saying that P = V and between (10, 15) and (15,10),

(V-10)^2 = (P-10)^2 = 12.5
P-10 = 3.536 Vmax = Pmax = 13.536 and Vmin = Pmin = 6.464

PV = nRT

Tmax = (13.536^2)/(2 moles * .08206) = 1116K (correct answer)
Tmin = (6.464^2)/(2moles * .08206) = 255K (Answer:Tmin = 225K)

Did I miscalculate where PV is a minimum between the section of (10,5) and (5,10) of the circle?
 
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kalbuskj31 said:
Did I miscalculate where PV is a minimum between the section of (10,5) and (5,10) of the circle?

No you're fine. Mathematically you are minimizing PV=(5\sin{t} +10)(5\cos{t}+10). Graphing is easiest. For any real data that's what you'd have to do anyway.
 

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