Question

1. (a) Write the total differentials of s-S(V, T),S- S(P, T). Get used to it. This is the first step in many derivations in thermodynamics. We'll actually come back to these particular ones in the next lecture or so. (b) Now, from the 1st law we know that du dq+ dw. () Take the differential along a reversible path: dU - dgrev+ dwre. [Question: why don't we write dUrev? Yes, you must answer this and the other questions in blue.] (ü) Next, divide both sides by T, then substitute the 2nd law definition of entropy, ds -dare/T and for dwrev-PdV (li) Multiply by T. dU should now be expressed entirely in terms of state functions (S, T, P, and Finally, identify the partial derivatives as (e) Tand() -P, a thermodynamic such as(5) V) or their differentials. Why is this important? (c) Now find the total differential of U-u(s,v). Equate dU from part (b) to dU from part (c). avis definition of temperature and pressure. Can you rationalize why temperature and pressure can be defined as partial derivatives of energy with respect to entropy and volume, respectively? Do you think you could find a route another definition of temperature that involves enthalp ,-T? such as T)

Answer 1

(a) Total differentiates of s = s (V, T) is ds = (partial s/partial v)_T dv + (partial s/partial T)_v dT Total differential of s = s (P, T) is ds = (partial s/partial p)_T dP + (partial s/partial T)_p dT (b) 1st law: du = dq + dw-(1) (i) In differential along a reversible path du = dq_rev + dw_rev we don't write du_rev because unlike q (heat) and W (work) , u is state quantity, q & w depends on the path taken by the system to moue from some initial state I to final state f . But U_f -U_i (difference between internal energies of final & initial states) is path independent. (ii)Dividing both sides by T, we get du/T = da_rev/T + dw_rev/T Using, ds = dq_rev/T, dw_rev = -pdv du/T = ds -p/T dv (iii) Multiplying both sides by T du = Tds-(2)