Student solutions manual for Physical chemistry by Peter William Atkins

By Peter William Atkins

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A detailed analysis of the entropy changes associated with this device follows. 1 CV and Cp are the temperature dependent heat capacities of water INSTRUCTOR’S MANUAL 58 Three things must be considered in an analysis of the geothermal heat pump: Is it forbidden by the first law? Is it forbidden by the second law? Is it efficient? Etot = Ewater + Eground + Eenvironment Ewater = 0 Eground = −CV (Th ){Th − Tc } Eenvironment = −CV (Th ){Th − Tc } adding terms, we find that and Tc . Stot = Swater + Etot = 0 which means that the first law is satisfied for any value of Th Sground + Senvironment Swater = 0 qground −Cp (Th ){Th − Tc } = Th Th Cp (Tc ){Th − Tc } qenvironment Senvironment = = Tc Tc Sground = adding terms and estimating that Cp (Th ) Cp (Tc ) = Cp , we find that 1 1 − Tc Th Stot = Cp {Th − Tc } This expression satisfies the second law ( Stot > 0) only when Th > Tc .

6. It does not matter whether the piston between chambers 2 and 3 is diathermic or adiabatic as long as the piston between chambers 1 and 2 is adiabatic. 6. However, if both pistons are diathermic, the result is different. The solution for both pistons being diathermic follows. See Fig. 1. 1 Initial equilibrium state. 3 L = constant Final equilibrium state. The diathermic walls allow the passage of heat. Consequently, at equlibrium all chambers will have the same temperature T1 = T2 = T3 = 348 K.

P R ; = ∂T V V −b then, dp = ∂p RT 2a =− + 3 ∂V T (V − b)2 V R V −b dT + RT 2a − V3 (V − b)2 dV ∂V is more readily evaluated with the use ∂T p Because the van der Waals equation is a cubic in V , of the permuter. ∂p R ∂T V ∂V V −b =− =− ∂p ∂T p + V2a3 − (VRT ∂V T −b)2 RV 3 (V − b) RT V 3 − 2a(V − b)2 = For path a T2 ,V2 T1 ,V1 dp = = T2 T1 V2 R RT2 2a − + 3 dT + 2 V1 − b (V − b) V V1 dV R RT2 RT2 (T2 − T1 ) + − −a V1 − b (V2 − b) (V1 − b) =− RT1 RT2 + −a V1 − b V 2 − b 1 V22 − 1 V22 − 1 V12 1 V12 For path b T2 ,V2 T1 ,V1 dp = = V2 V1 − T2 RT1 R 2a dV + + dT 2 3 (V − b) V T1 V2 − b RT1 RT1 − −a V 2 − b V1 − b =− RT1 RT2 + −a V1 − b V 2 − b 1 V22 1 − 1 V22 V12 − + R (T2 − T1 ) V2 − b 1 V12 Thus, they are the same and dp satisfies the condition of an exact differential, namely, that its integral between limits is independent of path.

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