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One mole of monoatomic gas undergoes a linear process A to B shown in P-V diagram. Volume of gas from where process turn from an endothermic to an exothermic is?

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I tried to calculate the point where the rate of change of temperature wrt volume would become 0 but my answer didn't match. The solution given was: Process will turn from endothermic to exothermic at point where slope of P-V graph satisfy the slope of adiabatic curve for same gas. I couldn't understand why this is true. Could anyone please explain. Thank you.

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  • $\begingroup$ dT/dV <> 0 for not reacting ideal gas during adiabatic processes. $\endgroup$
    – Poutnik
    Commented May 20 at 10:29

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From the first law of thermodynamics, $$dU=dQ-dW$$ The process goes form endothermic to exothermic when dQ=0. This is also tangent to some adiabatic reversible curve for the gas.

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Basically the slope of the adiabatic curve would satisfy the slope of the PV diagram, at a point where the curve and the PV line intersect (The PV line becomes tangent to the adiabatic curve).

pV diagram

Based on this graph,

$ -P0/V0 = - γP/V $

$ -m = - (γ(-mv + P0))m/V = + P0/V0 $

$ mv(1+γ) = γP0 $

$ V = γP0/m((1+γ) = 5V0/8 $

Hence the volume at which it will switch would be $ 5V0/8 $.

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