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Chemistry 431 Problem Set 3 Fall 2018 Solutions 1

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Chemistry 431 Problem Set 3 Fall 2018 Solutions 1 Chemistry 431 Problem Set 3 Fall 2018 Solutions 1. One mole of an ideal monatomic gas at 25.◦C and 20. bar pressure is expanded adia- batically against a constant external pressure of 1.0 bar until equilibrium is reached. Calculate q; w; ∆U and ∆H for the process. Answer: ! 3 nRTf nRTi nR(Tf − 298 K) = −1:0 bar − 2 Pf Pi 3 298 K (T − 298 K) = − T − 2 f f 20 Tf = 185 K q = 0 3 w = ∆U = C ∆T = (1 mol)(8:3144 J mol−1K−1)(185 K − 298 K) = −1409 J V 2 5 ∆H = C ∆T = ∆U = −2349 J p 3 2. Three moles of an ideal monatomic gas at an initial pressure of 10.0 bar and an initial temperature of 100.◦C are expanded adiabatically against a constant external pressure of 2.0 bar until equilibrium is reached. Calculate the final temperature of the gas. Answer: w = −Pext(Vf − Vi) = ∆U = CV (Tf − Ti) nRT nR(373 K)! 3 −(2:0 bar) f − = nR(T − 373 K) 2:0 bar 10:0 bar 2 f 3 −(T − 74:6 K) = T − 559:5 K f 2 f 5 T = 634:1 K 2 f Tf = 253:6 K 1 3. Calculate q; w; ∆U and ∆H for the adiabatic reversible compression of 3.0 moles of an ideal diatomic gas from 5.0 liters at 35.◦C to 1.0 liter. Answer: γ−1 γ−1 TiVi = Tf Vf 7 γ = 5 308 K(5)2=5 = T (1)2=5 T = 586 K q = 0 5 w = ∆U = C ∆T = (3 mol)(8:3144 J mol−1K−1)(586 K − 308 K) = 17335 J V 2 7 ∆H = C ∆T = ∆U = 24270 J p 5 4. A cylinder fitted with a frictionless piston contains 2.00 moles of an ideal monatomic gas at an initial pressure of 5.00 bar and an initial temperature of 298 K. The gas is first expanded adiabatically against a constant external pressure of 1.00 bar until equilibrium is reached, followed by an adiabatic, reversible compression until the final gas pressure equals the initial gas pressure of 5.00 bar. Calculate q; w; ∆U and ∆H for the overall, two-step process. Answer: q = 0 Step 1 w1 = ∆U1 = −Pext∆V = CV ∆T nRT nR(298 K)! 3 −1:00 bar f − = nR(T − 298 K) T = 203 K 1:00 bar 5:00 bar 2 f f −1 −1 nRTf (2:00 mol)(0:08314 L bar mol K )(203 K) Vf = = = 33:7 L Pf 1:00 bar Step 2 Pi = 1:00 bar Vi = 33:7 L γ γ CP 5=2 nR 5 PiVi = Pf Vf with γ = = = CV 3=2 nR 3 5=3 5=3 (1:00 bar)(33:7 L) = (5:00 bar)Vf P V (5:00 bar)(12:8 L) V = 12:8 L T = f f = = 386: K f f nR (2:00 mol)(0:08314 L bar mol−1K−1) Overall q = 0 2 3 ∆U = w = C ∆T = (2:00 mol)(8:3144 J mol−1K−1)(386 K − 298 K) = 2195 J V 2 5 ∆H = C ∆T = ∆U = 3658 J P 3 5. One mole of an ideal monatomic gas at 1.0 bar pressure and a temperature of 25.◦C is taken through the following series of steps: (a) A heating at constant volume to 100.◦C; (b) An adiabatic compression against a constant external pressure of 25. bar until the volume is halved; (c) A cooling at constant pressure until the final temperature is 35.◦C. Calculate q; w; ∆U and ∆H for the overall process \a" + \b" + \c." [Hint: Think about state functions before solving this problem. Do not \brute force" the solution.] Answer: 3 ∆U = C ∆T = (1 mol)(8:3144 J mol−1K−1)(10 K) = 125 J V 2 5 ∆H = C ∆T = ∆U = 208 J p 3 −1 −1 nRTi (1 mol)(0:08314 L bar mol K )(298 K) Vi = = = 24:78 L Pi 1 bar 3 q = C ∆T = (1 mol)(8:3144 J mol−1K−1)(75 K) = 935 J a V 2 qb = 0 373 P = (1 bar) = 1:25 bar i 298 nRT (1 mol)(0:08314 L bar mol−1K−1)(373 K) V = = = 24:81 L i P 1:25 bar Vf = 0:5Vi = 12:39 L 3 −25 bar(12:39 L − 24:81 L) = (1 mol)(0:08314 L bar mol−1K−1)(T − 373 K) 2 T = 2863 K 5 q = C ∆T = (1 mol)(8:3144 J mol−1K−1)(308 K − 2863 K) = −53108 J c p 2 q = qa + qb + qc = −52173 J w = ∆U − q = (125 J + 52173 J) = 52298 J 3 6. 2.00 moles of an ideal diatomic gas at a temperature of 298.K and a pressure of 5.00 bar are expanded adiabatically against a constant external pressure of 2.00 bar until equilibrium is reached. The gas is then compressed adiabatically and reversibly until the change in the enthalpy ∆H for the two step process (first expansion, then com- pression) is 0. Calculate the final volume after the completion of the two-step process. Answer: Step 1 nRTf nRTi ∆U = CV (Tf − Ti) = w = −Pext − Pext Pi 5 Pext nR(Tf − 298: K) = nR −Tf + Ti 2 Pi 5 T − 745: K = −T + 119:2 K T = 247: K 2 f f f nRT (2:00 mol)(0:08314 L bar mol−1K−1)(247: K) After Step 1 V = = = 20:5 L P 2:00 bar Overall Process, final temperature: ∆H = 0 for the overall process, so that Tf = 298: K Reversible, adiabatic step γ−1 γ−1 TiVi = Tf Vf 7 2 T = 247: K;T = 298: K;V = 20:5 L; γ − 1 = − 1 = i f i 5 5 2=5 2=5 247:(20:5 L) = 298:Vf Vf = 12:8 L 7. Two containers are joined together as in the Joule experiment. One contains 1.0 moles ◦ ◦ of He at 75. C, and the other contains 2.0 moles of N2 at 25. C. The gases are allowed to diffuse into each other adiabatically. Assuming the gases to be ideal, what is the final temperature of the system? Answer: ∆U = 0 = CV;He∆THe + CV;N2 ∆TN2 3 5 (1 mol)R(T − 348 K) + (2 mol)R(T − 298 K) = 0 2 f 2 f Tf = 310 K 8. The pressure of a Van der Waals gas as a function of T and V is given by RT 1 2 P = − a V − b V where a and b are constants and the number of moles has been set to 1 for this problem. The internal energy U as a function of T and V for a Van der Waals gas is given by 1 U = C T − a V V 4 (a) Explictly evaluate (@T=@V )P for the Van der Waals gas and derive an expression for (@V=@T )P and from the explicitly evaluated derivative. Answer: a RT P + = V 2 V − b V − b a T = P + R V 2 @T ! 1 a 2a V − b! = P + 2 − 3 @V P R V V R @V ! " 1 a 2a V − b!#−1 = P + 2 − 3 @T P R V V R (b) Explicitly calculate (@U=@T )V ,(@T=@V )U and (@V=@U)T for the Van der Waals gas. Demonstrate that @U ! @T ! @V ! = −1 @T V @V U @U T Answer: a U = C T − V V @U ! = CV @T V U + a=V T = CV @T ! a = − 2 @V U CV V a = C T − U V V 1 V = a CV T − U @V ! a = 2 @U T (CV T − U) @U ! @T ! @V ! a a ! = CV − 2 2 @T V @V U @U T CV V (CV T − U) a2 = − 2 2 V (CV T − U) But a C T − U = V V 5 so that @U ! @T ! @V ! a2 V 2 = − 2 2 = −1 @T V @V U @U T V a (c) Evaluate (@U=@T )P for the Van der Waals gas. Answer: a dU = C dT + dV V V 2 @U ! a @V ! = CV + 2 @T P V @T P a " 1 a 2a V − b!#−1 = C + P + − V V 2 R V 2 V 3 R 9. (a) Give the total differential of the internal energy U of a system where U is taken to be a function of T and V . Answer: @U ! @U ! dU = dT + dV @T V @V T (b) Consider a system where only PV -work is done and Pext = P . Use the result from part \a" to show the expression for the heat is given by @U ! " @U ! # dq¯ = dT + + P dV @T V @V T Answer: dU =dq ¯ +dw ¯ =dq ¯ − P dV or @U ! " @U ! # dq¯ = dT + + P dV @T V @V T (c) Use part \b" to prove that the expression fordq ¯ is not an exact differential. Answer: @ @U ! ! @ " @U ! #! 6= + P @V @T V T @T @V T V 10. For any substance, show that @U ! κC = V @P β V Answer: 1 @V ! κ = − V @P T 6 1 @V ! β = V @T P κ "(@V=@P ) # @T ! = − T = β (@V=@T )P @P V κ @U ! @T ! @U ! CV = = β @T V @P V @P V 11. Show that @U ! @P ! @U ! = CV + κV @T P @T V @V T where CV is the constant volume heat capacity and κ is the isothermal compressibility.
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