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Unformatted text preview: 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 pdV n RT n RT V V We substitute V 2 = 2.00 V 1 to obtain ( 29 ( 29 ( 29 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 ( 29 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, 3 9.22 10 J = = = 23.1 J/K. 400 K × ∆ W S T (c) ∆ S = 0 for all reversible adiabatic processes. 2. From Eq. 202, we obtain ( 29 ( 29 4 = = 405 K 46.0 J/K =1.86 10 J. ∆ × Q T S 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. 204 leads to ( 29 ( 29 ( 29 = ln = 2.50 mol 8.31 J/mol K ln 2.00 =14.4 J/K. f i V S nR V ∆ 4. (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 194. Consequently, ∆ S = 333 12.0 273 =14.6 J / g g K J / K. a fa f 859 CHAPTER 20 (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. 5. We use the following relation derived in Sample Problem 202: = ln . ∆ f i T S mc T (a) The energy absorbed as heat is given by Eq. 1914. Using Table 193, we find ( 29 ( 29 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 ( 29 J 373.15 = 2.00 kg 386 ln =173 J/K. kg K 298.15 ∆ ⋅ S 6. An isothermal process is one in which T i = T f which implies ln ( T f / T i ) = 0. Therefore, Eq. 204 leads to ( 29 ( 29 22.0 = ln = = 2.75 mol. 8.31 ln 3.4/1.3 ∆ ⇒ f i V S nR n V 7. (a) The energy that leaves the aluminum as heat has magnitude Q = m a c a ( T ai T f ), where m a is the mass of the aluminum, c a is the specific heat of aluminum, T ai is the initial temperature of the aluminum, and T f is the final temperature of the aluminumwater system. The energy that enters the water as heat has magnitude Q = m w c w ( T f T wi ), where m w is the mass of the water, c w is the specific heat of water, and T wi is the initial temperature of the water. The two energies are the same in magnitude since no energy is lost. Thus, ( 29 ( 29 + = = ....
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 Fall '08
 Many
 Physics, Thermodynamics, Work, Heat, Eq.

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