Please help me answer question 19.27 thank you very much of the system decreases. What can you con-processclude ab the sign and magnitude of ASu? (c) Duringa certainversible process, the surroundings undergo anentropy ange, ASurr-781/K. What is the entropychange of the system for this process?19.3819.27 a) What sign for AS do you expect when the volume of0.200 mol of an ideal gas at 27 °C is increased isothermallyfrom an initial volume of 10.0 L? (b) If the final volumeis 18.5 L, calculate the entropy change for the process.(c) Do you need to specify the temperature to calculate theentropy change?19.3919.28 (a) What sign for AS do you expect when the pressure on0.600 mol of an ideal gas at 350 K is increased isothermallyfrom an initial pressure of 76.0 kPa? (b) If the final pres-sure on the gas is 121.6 kPa, calculate the entropy changefor the process. (c) Do you need to specify the temperatureto calculate the entropy change?19.46The Molecular Interpretation of Entropy and theThird Law of Thermodynamics (Section 19.3)19.29 For the isothermal expansion of a gas into a vacuum,AE 0,q 0, and w 0. (a) Is this a spontaneous pro-"driving force" for the expan-SESS? (b) Explain why no work is done by the system during

Entropy change : It may be defined as the change in the disorder of a thermodynamic system.The greater the disorderliness the more is the entropy.(a) We know for a reversible isothermal process , change in entropy is given as : ∆S=nR lnV2V1Where,∆S=change in entropyn = number of moles R= gas constantV2= final volumeV1= initial volume As here the volume is increasing , i.e V2>V1 {as volume is increasing from initial to 10.0 L } So , lnV2V1>0 So, the equation , ∆S=nR ln(V2V1) is greater than zero Hence the sign for , ∆S is (+) ve (b) Given, Initial volume , V1=10.0 LFinal volume , V2=18.5 Lmoles, n = 0.20temperature , T =(27+273)K=300K Putting the value(s) in equation : ∆S=nR ln(V2V1) ⇒∆S = 0.20 mol ×8.314Jmol K×ln (18.510.0)⇒∆S = 1.023 JK Entropy change = 1.023 JK(c)For an ideal gas , change in entropy is given as : ∆S = nCVlnT2T1+nR ln V2V1Where,CV =Specific heat capacity at constant volume T2 and T1 are final and initial temperature respectively. From the above equation we can see that temperature is needed to calculate the change in entropy( it is not needed when the process is isothermal )