In a temperature range near some absolute temperature T, the tension force F of a stretched plastic rod is related its length l by the expression

F = aT 2(L-Lo) where a and Lo are positive constants, Lo being the unstretched length of the rod. When L Lo, the heat capacity to its length L by the expression CL of the rod (measured at constant length) is given by the relation CL bT, where b is a constant. A. Write down the fundamental thermodynamic relation for this system, expressing dS in terms of dL and dE. B. The entropy S(T,L) of the rod is a function of T and L. Compute c. Knowing S(To, Lo), find S(T, L) at any other temperature T and length L. (It is most convenient to calculate first the change of entropy with temperature at the length Lo where the heat capacity is known. ) d. If you start at T-T, and L = Li and stretch the thermally insulated rod quasi-statically until it attains the length L, what is the final temperature Ty? e. Calculate the heat capacity CL(L, T) of the rod when its length is L instead of Lo f. Calculate S(T,L) by writing S(T, L)-S(TyLo) = [S(T, L)-S(Tp, L)] + [S(Tp, L)-S(TO, Lo)] and using the result of part (e) to compute the first term in square brackets. Show that the final answer agrees with the one found in part (c)

a) The fundamental thermodynamic relation for this system is given by dS = (1/T)dE + (F/T)dL, where S is the entropy, E is the internal energy, T is the temperature, F is the tension force, and L is the length of the rod.

b) To compute the entropy S(T, L), we need to integrate dS. Since the heat capacity CL is given by CL = bT, we have dE = CL(T,L)dT. Substituting this in the fundamental relation, we get dS = (b/T)L(T,L)dT + (aT/T)L(T,L)dL. Integrating both sides gives S(T, L) = S(To, Lo) + bL[ln(T/To)] + a/2L[ln(L/Lo)].

c) To find S(T, L), we first find the change in entropy with temperature at the length Lo where the heat capacity is known: dS = (b/T)Lo dT. Integrating this expression from To to T gives S(T,Lo) - S(To,Lo) = bLo[ln(T/To)], which we can use to find S(T,L) using the expression in part (b).

d) Since the rod is thermally insulated, we have dE = 0, so the fundamental relation reduces to dS = (F/T)dL. Integrating this expression from Li to L gives S(Ty, L) - S(T-T, Li) = [tex][a/2(Ty^2 - (T-T)^2)[/tex]- F(Li-L)]/T, where Ty is the final temperature.

e) The heat capacity CL(L, T) is given by CL = bT, where b is a constant.

f) Using the result from part (e), we have CL(T, L) = [tex]CL(Ty, Lo)(Lo/L)^2[/tex]. Substituting this expression in the equation in part (b) gives S(T, L) - S(Ty, Lo) = bLo[ln(T/To) - 2ln(L/Lo)] + a/2[ln(L/Lo)]. Using the result from part (c) to simplify the first term, we get S(T, L) - S(Ty, Lo) = bL[ln(T/Ty)] + a/2[ln(L/Lo)], which agrees with the result in part (b).

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