ssm_ch20 - Chapter 20 Student Solutions Manual 5(a Since...

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Chapter 20 – Student Solutions Manual 5. (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 Δ = = T , where Q is the heat absorbed. According to the first law of thermodynamics, Δ 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, Δ E int = 0 and Q = W . Thus, 3 9.22 10 J = = = 23.1 J/K. 400 K × Δ W S T (c) Δ S = 0 for all reversible adiabatic processes. 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, ( ) ( ) + = = + a a ai w w wi a a ai f w w f wi f a a w w m c T m c T m c T T m c T T T m c m c . 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 57.0 C 330 K.
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