ch20 - 1. (a) Since the gas is ideal, its pressure p is...

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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 22 11 2 1 ln . == = ³³ VV dV V Wp d V n R T n R T We substitute V 2 = 2.00 V 1 to obtain () ( ) ( ) 3 =l n2.00 = 4.00 mol 8.31 J/mol K 400 K ln2.00 = 9.22 10 J. Wn R T ⋅× (b) Since the expansion is isothermal, the change in entropy is given by 1 ST d Q Q T ∆= = ³ , 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.
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2. From Eq. 20-2, we obtain () ( ) 4 == 405 K 46.0 J/K =1.86 10 J. ∆× QTS
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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 () ( ) ( ) =l n= 2.50 mol 8.31 J/mol K ln 2.00 =14.4 J/K. f i V Sn R V §· ∆⋅ ¨¸ ©¹
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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 19-4. Consequently, S = 333 12.0 273 =14.6 J/g g K J/K. af a f (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.
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5. We use the following relation derived in Sample Problem 20-2: =l n . §· ¨¸ ©¹ f i T Sm c 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 ∆× Qc mT 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
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6. 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 =l n = = 2 . 7 5 m o l . 8.31 ln 3.4/1.3 §· ¡ ¨¸ ©¹ f i V Sn R n V
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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 aluminum- water 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, () ( ) + == . + −− ¡ aaa i www i aa a i f ww f w i f mcT mc T T T T mc The specific heat of aluminum is 900 J/kg K and the specific heat of water is 4190 J/kg K. Thus, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.200 kg 900 J/kg K 100 C + 0.0500 kg 4190 J/kg K 20 C = 0.200 kg 900 J/kg K + 0.0500 kg 4190 J/kg K = 57.0 C or 330 K. f T ⋅° ° ⋅⋅ ° (b) Now temperatures must be given in Kelvins: T ai = 393 K, T wi = 293 K, and T f = 330 K.
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This homework help was uploaded on 09/30/2007 for the course PHYS 2213 taught by Professor Perelstein,m during the Fall '07 term at Cornell University (Engineering School).

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ch20 - 1. (a) Since the gas is ideal, its pressure p is...

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