Chapter 20 1 THINK If the expansion of the gas is reversible and isothermal then there s no change in internal energy However if the process is reversible and adiabatic then there would be no change in entropy EXPRESS Since the gas is ideal its pressure p is given in terms of the number of moles n the volume V and the temperature T by p nRT V If the expansion is isothermal the work done by the gas is and the corresponding change in entropy is where Q is the heat absorbed see Eq 20 2 ANALYZE a With V2 2 00V1 and T 400 K we obtain b According to the first law of thermodynamics Eint Q W Now the internal energy of an ideal gas depends only on the temperature and not on the pressure and volume Since the expansion is isothermal Eint 0 and Q W Thus c The change in entropy S is zero for all reversible adiabatic processes LEARN The general expression for S for reversible processes is given by Eq 20 4 Note that S does not depend on how the gas changes from its initial state i to the final state f 2 An isothermal process is one in which Ti Tf which implies ln Tf Ti 0 Therefore Eq 20 4 leads to 925 221121lnVVVVVdVWpdVnRTnRTVV 1STdQQT 3 ln2 00 4 00 mol8 31 J molK400 Kln2 00 9 2210 J WnRT 39 2210J 23 1 J K 400 K WSTlnlnfffiViiVTSSSnRnCVT 22 0 ln 2 75 mol 8 31ln3 4 1 3 fiVSnRnV 926 CHAPTER 20 3 An isothermal process is one in which Ti Tf which implies ln Tf Ti 0 Therefore with Vf Vi 2 00 Eq 20 4 leads to 4 From Eq 20 2 we obtain 5 We use the following relation derived in Sample Problem 20 01 Entropy change of two blocks coming to equilibrium a The energy absorbed as heat is given by Eq 19 14 Using Table 19 3 we find where we have used the fact that a change in Kelvin temperature is equivalent to a change in Celsius degrees b With Tf 373 15 K and Ti 298 15 K we obtain 6 a This may be considered a reversible process as well as isothermal so we use S Q T where Q Lm with L 333 J g from Table 19 4 Consequently b The situation is similar to that described in part a except with L 2256 J g m 5 00 g and T 373 K We therefore find S 30 2 J K 7 a We refer to the copper block as block 1 and the lead block as block 2 The equilibrium temperature Tf satisfies m1c1 Tf Ti 1 m2c2 Tf Ti2 0 which we solve for Tf ln 2 50 mol8 31 J molKln2 00 14 4 J K fiVSnRV 4 405 K46 0 J K 1 8610 J QTS ln fiSmcTT 4J 3862 00 kg75 K 5 7910 JkgK QcmT J373 15 2 00 kg386ln 173 J K kgK298 15 S S 33312 0273 14 6 J g g K J K afaf 11 122 21122 50 0 g386 J kgK400 K 100 g128 J kgK200 K 50 0 g386 J kgK 100 g128 J kgK320K iifmcTmcTTmcmc 927 b Since the two block system in thermally insulated from the environment the change in internal energy of the system is zero c The change in entropy is 8 We use Eq 20 1 9 The ice warms to 0 C then melts and the resulting water warms to the temperature of the lake water which is 15 C As the ice warms the energy it receives as heat when the temperature changes by dT is dQ mcI dT where m is the mass of the ice and cI is the specific heat of ice If Ti 263 K is the initial temperature and Tf 273 K is the final temperature then the change in its entropy is Melting is an isothermal process The energy leaving the ice as heat is mLF where LF is the heat of fusion for ice Thus S Q T mLF T 0 010 kg 333 103 J kg 273 K 12 20 J K For the warming of the water from the melted ice the change in entropy is where cw is the specific heat of water 4190 J kg K Thus The total change in entropy for the ice and the water it becomes is 121122 1 2lnln320 K320 K50 0 g386 J kgKln100 g128 J kgKln400 K200 K1 72JK ffiiTTSSSmcmcTT 10 02335 00 10 0 5 00 0 0368 J K 3VnCdTnASnATdTT 273 Kln0 010 kg2220 J kgKln 0 828 J K 263 KfiTfIITiTdQdTSmcmcTTT ln fwiTSmcT 288 K 0 010 kg4190 J kgKln 2 24 J K 273 KS 0 828 J K 12 20 J K 2 24 J K 15 27 J K S 928 CHAPTER 20 Since the temperature of the lake does not change significantly when the ice melts the change in its entropy is S Q T where Q is the energy it receives as heat the negative of the energy it supplies the ice and T is its temperature When the ice warms to 0 C When the ice melts When the water from the ice warms The total energy leaving the lake water is The change in entropy is Q 222 J 3 33 103 J 6 29 102 J 4 18 103 J The change in the entropy of the ice lake system is S 15 27 14 51 J K 0 76 J K 10 We follow the method shown in Sample Problem 20 01 Entropy change of two blocks coming to equilibrium Since S mc ln Tf Ti then with S 50 J K Tf 380 K Ti 280 K and m 0 364 kg we obtain c 4 5 102 J kg K 11 THINK The aluminum sample gives off energy as heat to water Thermal equilibrium is reached when both the aluminum and the water come to a common final temperature Tf EXPRESS The energy that leaves the aluminum as heat has magnitude Q maca Tai Tf where ma is the mass of the aluminum ca is the specific heat of aluminum Tai is the initial temperature of the aluminum and Tf is the final temperature of the aluminum water system The energy that enters the water as heat has magnitude Q mwcw Tf Twi where mw is the mass of the water cw is the specific heat of water and Twi is the initial temperature of the water The two energies are the same in magnitude since no energy is lost Thus 0 010 kg2220 J kg K10 K 222 J IfiQmcTT QmLF 0 01033310 3 3310 33 kg J kg JafchQmcTTwfi 0 010419015 629 chafafaf kg J kg K K J34 1810 J 14 51 J K 288 KS fiTTdTmcT 929 The change in entropy is ANALYZE a The specific heat of aluminum is 900 J kg K and the specific heat of water is 4190 J kg K Thus b Now temperatures must be given …
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