Chapter 20 Question

Question 2: An ideal gas, in contact with a controllable thermal reservoir, can be taken from initial state i to final state f along the four reversible paths in Fig. 20-19. Rank the paths according to the magnitudes of the resulting entropy changes of (a) the gas, (b) the reservoir, and (c) the gas-reservoir system, greatest first. p The initial points i and the final points f are the same, so all four paths have the same change of entropy of the gas, because the change entropy is path independent: ∆S1 = ∆S2 = ∆S3 = ∆S4. 6 i t - 4 (a) Therefore, the change of entropy of the gas is the same for all paths, R 3 ? 2 ∆Sgas,1 = ∆Sgas,2 = ∆Sgas,3 = ∆Sgas,4 1 t For any reversible process the total change of entropy is zero, so the f V change of entropy of the system (gas) is equal and opposite to the change Fig. 20-19. of entropy of the reservoir, ∆Sgas = −∆Sres.

(c) The gas-reservoir system does not change entropy for a reversible process, so the change of total entropy is zero for all four reversible paths.

Question 3: Point i in Fig. 20-20 represents the initial state of an ideal gas at temperature T . (a) Taking algebraic signs into account , rank the entropy changes that the gas undergoes as it moves, successively and reversibly, from point i to points a, b, c, and d, greatest first. (b) Do the same for the magnitude of the entropy change, |∆S|, greatest first. p (a) The p − V plot shows common initial point i with the temperature T , a and two isotherms having temperature either higher or lower, by the t same amount ∆T , than at i. There are four possible processes, two 6 of them go to the higher temperature (one isochoric and one isobaric) d t t - t b i T and two go to the lower temperature (one isobaric and one isochoric). t? T + ∆T c First, we know that a process that absorbs heat and increases the T − ∆T temperature of the gas, will increase its enstropy, so the change of V entropy of the gas is positive, ∆S > 0. A process that releases energy Fig. 20-20. in the form of heat will decrease the entropy of the gas, so ∆S < 0. Comparing isobaric and isochoric process to a higher temperature, in the case of the same number of moles of the same ideal gas and the same initial and final temperature, the only difference in the value for the entropy change will come from the molar specific heats: CV or Cp, where Cp > CV . For Tf Tf isobaric process, ∆Sgas = nCp ln , and for isochoric process, ∆Sgas = nCV ln , so the change of Ti Ti entropy is larger for the isobaric process.

In the case of temperature decrease, ∆Sp < ∆SV , because ∆Sp is larger negative number. Therefore, b > a > c > d.

(b) Comparing magnitudes of the entropy change for the two isobaric processes of heating up or cooling down, from the same initial temperature and by the same ∆T , one can use the approximation: Q ∆S ≈ in which the average temperature for heating up will always be larger than the average Tavg temperature for cooling down. Therefore, the magnitude of the entropy change is larger for the cooling down process because its average temperature is smaller. It is also true for the comparisson between two isochoric processes. The magnitude of the entropy change is larger for the cooling down process. So, using approximation relation we have: |∆Si,d| > |∆Si,b| and |∆Si,c| > |∆Si,a|. While comparring the isobaric and isochoric proceess between the same two temperatures, the magnitude of the entropy change is always larger for the isobaric process, because Cp > CV . So, using

Cp > CV we have: |∆Si,d| > |∆Si,c| and |∆Si,b| > |∆Si,a|. In order to ﬁnish comparisson, we need to check relation between |∆Si,c| and |∆Si,b| Ti + ∆T Ti − ∆T Ti |∆Si,b| = nCp ln and |∆Si,c| = |nCV ln | = nCV ln , where we used that Ti Ti Ti − ∆T T T ln f = −ln i . Ti Tf

Assuming that ∆T << Ti the ratio Ti/(Ti − ∆T ) ≈ (Ti + ∆T )/Ti. So, the major diﬀerence between

|∆Si,b| and |∆Si,c| is due to Cp and CV and therefore |∆Si,b| > |∆Si,c| So, the answer for (b) is: d > b > c > a.