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Tutorial 5 Notes - CHM2132 Tutorial notes for Oct 9th 13th...

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CHM2132 Tutorial notes for Oct. 9 th , 13 th 1) When 3.00 mol of a gas at 230 K and 150 kPa is subjected to isothermal compression, its entropy decreases by 15.0 J K -1 . Calculate (a) the final pressure of the gas and (b) Δ G for the compression. We are given: n =3.00 mol, T 1 =230K, p 1 = 150 kPa, S = -15.0 J K -1 , T = 0 a) What is p 2 ? We are given entropy and we have been told that this is an isothermal compression. The equation for this entropy change is: ! S = nR ln V 2 V 1 " # $ % & Since the process is isothermal, p 1 V 1 = p 2 V 2 and; ! S = nR ln p 1 p 2 " # $ % & Since we have been given p 1 , n and Δ S we can rearrange to find p 2 : ln p 1 p 2 ! " # $ % & = S nR ln p 2 = ( S nR + ln p 1 = ( ( 15 J K -1 mol -1 3.00 mol ( ) 8.314 J K -1 mol -1 ( ) + ln 150 ( ) = 5.61 p 2 = 274 kPa b) What is Δ G ? For isothermal processes, the pressure dependence of Gibbs is given by: ! G ! p " # $ % & T = V We can rearrange this to isolate the variables to solve for Gibbs: dG = Vdp Then integrate both sides: dG G 1 G 2 ! = Vdp p 1 p 2 ! Since the volume of the gas changes as a function of pressure we must cast V in terms of pressure by substituting in the ideal gas equation:
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dG G 1 G 2 ! = nRT p dp p 1 p 2 ! = nRT dp p p 1 p 2 ! ! G = nRT ln p 2 p 1 " # $ % & This is the Gibbs energy equation for the isothermal expansion of a gas. We can now substitute in the given values: ! G = 3.00 mol ( ) 8.314 J K -1 mol -1 ( ) 230 K ( ) ln 274 kPa 150 kPa " # $ % & = 3.45 kJ Note that since ! G = ! H " T ! S and Δ
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