ch20 - Chapter 20 1. (a) Since the gas is ideal, its...

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837 Chapter 20 1. (a) 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 . The work done by the gas during the isothermal expansion is 2 2 1 1 2 1 ln . V V V V dV V W p dV n RT n RT V V We substitute V 2 = 2.00 V 1 to obtain     3 = ln2.00 = 4.00 mol 8.31 J/mol K 400 K ln2.00 = 9.22 10 J. W n RT (b) Since the expansion is isothermal, the change in entropy is given by   1 S T dQ Q T   , where Q is the heat absorbed. According to the first law of thermodynamics, E int = 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, E int = 0 and Q = W . Thus, (c) S = 0 for all reversible adiabatic processes. 2. An isothermal process is one in which T i = T f , which implies ln ( T f / T i ) = 0. Therefore, Eq. 20-4 leads to     22.0 = ln = = 2.41 mol. 8.31 ln 3.9/1.3 f i V S nR n V 3. An isothermal process is one in which T i = T f , which implies ln( T f / T i ) = 0. Therefore, with V f / V i = 2.00, Eq. 20-4 leads to      = ln = 2.50 mol 8.31 J/mol K ln 2.00 =14.4 J/K. f i V S nR V 4. From Eq. 20-2, we obtain 3 9.22 10 J = = = 23.1 J/K. 400 K W S T
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CHAPTER 20 838    4 = = 453 K 46.0 J/K = 2.08 10 J. Q T S 5. We use the following relation derived in Sample Problem — “Entropy change of two blocks coming to equilibrium:” = ln . f i T S mc T (a) The energy absorbed as heat is given by Eq. 19-14. Using Table 19-3, we find    4 J = = 386 2.00 kg 75 K = 5.79 10 J kg K Q cm T where we have used the fact that a change in Kelvin temperature is equivalent to a change in Celsius degrees. (b) With T f = 373.15 K and T i = 298.15 K, we obtain   J 373.15 = 2.00 kg 386 ln =173 J/K. kg K 298.15 S 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,    333 J/g 15.0 g = =18.3 J/K. 273 K S (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 T f satisfies m 1 c 1 ( T f T i ,1 ) + m 2 c 2 ( T f T i 2 ) = 0, which we solve for T f :               1 1 ,1 2 2 ,2 1 1 2 2 + 50.0 g 386 J/kg K 400 K + 100 g 128 J/kg K 200 K + 50.0 g 386 J/kg K + 100 g 128 J/kg K 320 K. i
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ch20 - Chapter 20 1. (a) Since the gas is ideal, its...

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