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The equivalence of the ideal gas and thermodynamic temperature scales

The equivalence of the ideal gas and thermodynamic temperature scales. Choose the ideal gas as the working fluid for a Carnot engine or refrigerator and place one of the reservoir in thermal equilibrium with water at triple point. . Here,. Conclusion: T(t ideal gas )=t ideal gas

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The equivalence of the ideal gas and thermodynamic temperature scales

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  1. The equivalence of the ideal gas and thermodynamic temperature scales Choose the ideal gas as the working fluid for a Carnot engine or refrigerator and place one of the reservoir in thermal equilibrium with water at triple point. Here, Conclusion: T(tideal gas)=tideal gas (the two scales are identical)

  2. The maximum efficiency of a cyclic heat engine From the Carnot principles:

  3. The Kelvin Planck statement for a cycle with contact withone reservoir T1 Q1 C W • Sign convention: • heat flow into system (cyclic device C) is positive . • work done by system is positive

  4. Second Law for cyclic devices in contact with two reservoirs Using Carnot corollaries, the Kelvin-Planck statement for one reservoir can be extended to two reservoirs T1 Q1 C Q2 • Sign convention: heat flow into the system (cyclic device C) is positive. • C can be either engine/refrigerator • No arrow for work is shown in figure (to save • space). T2

  5. The second law for a cyclic device exchanging heat with one/two reservoirs T1 Q1 C Note : T (thermodynamic temperature in Kelvin scale) is always non-negative Q2 Is this also true for an integer N larger than 2? T2

  6. Procedure for extending the second law N reservoirs To prove that the statement is true for all positive integers N • Proof by induction. Steps: • Assume when any closed system undergoes a cyclic process, the statement • is true for all N ≥ 2 . • Show that if the statement for N is true, then the statement is also true for N+1. • Since, we already know the statement is true for N=2, the step above means it • is true for N=3, which again means the statement is true for N=4 and so on for all • positive integers N.

  7. T1 Choose Q’N+1 such that (N+1)th reservoir has no net heat exchange Q1 T2 Q2 2nd Law for reversible cycle R Qj QN For R+C system (blue curve): Tj TN Show that if the statement for N is true, then the statement is also true for N+1. TN+1 C QN+1 Q’N+1 R Q’N

  8. T1 Choose Q’N+1 such that (N+1)th reservoir has no net heat exchange Q1 T2 Q2 2nd Law for reversible cycle R Qj QN For R+C system (blue curve): Tj TN Show that if the statement for N is true, then the statement is also true for N+1. TN+1 C QN+1 Q’N+1 R Q’N Q’N

  9. Using Using and and T1 T1 T2 T2 Q1 Q1 Q2 Q2 R+C Qj Tj QN QN Q’N TN TN

  10. Render each reservoir Tj Idle by choosing For each reversible cycle Rj R1 R2 Rj RN T0 Method 2: Idea is to use the KP statement for one reservoir; This requires introducing construction to make each existing reservoir redundant by introducing suitable reversible engines connected to a common reservoir. T1 Q1 T2 Q2 C Qj QN Tj Since system C+R1+R2+..Rj+..RN is exchanging heat with one reservoir, KP statementQ0≤0 TN

  11. The Clausius inequality System T is the temperature of boundary through which heat transfer occurs. The integral with a circle is performed over a cycle and all parts of system boundary that have different temperatures. The circle is used to denote a cyclic process. Valid for both irreversible and reversible cycles. Actually is a way of stating second law for cycles.

  12. Some conclusions from Clausius inequality for an irreversible cycles for an internally reversible cycles (also known as Clausius theorem) is impossible

  13. Definition of a new property: entropy Regardless of how the reversible cycle is executed following integral over a cycle is zero.

  14. Definition of a new property: entropy Regardless of how the reversible cycle is executed the integral over a cycle is zero. The integrand must be signifying the differential of a property

  15. Definition of a new property: entropy Note: We now have a definition for entropy change and not for the absolute value of entropy. So when you see only S anywhere, understand S to be an entropy change calculated from some reference state (S-Sref). Regardless of how the reversible cycle is executed the integral over a cycle is zero. The integrand must be signifying the differential of a property Entropy change in a process:

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