#پاسخ_مسئله  

🔐    dS = 1/T dU + P/T dV

🔹If we now apply this to our problem, the change in entropy is then straightforwardly:

       =>ΔS = (1/T¹ - 1/T²) ΔU + (P¹/T¹ - P²/T²) ΔV

🔸The third law of thermodynamics shows that the entropy always increases in any physical process. Thus, when equilibrium is achieved, the entropy will have achieved a maximum, so that ΔS = 0. This means that the joint system cannot increase its entropy by further exchanging volume or internal energy between system 1 and system 2. ΔS = 0 can only be achieved when T¹ = T² and P² = P².
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📚Concepts in Thermal Physics
👥Stephen J. Blundell  and Katherine M. Blundell

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