CHEM 305 Solutions for assignment #2

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1 CHEM 05 Solutions for assignent #. (a) Starting fro C C show that C C Substitute the result into the original expression for C C : C C (b) Using the result fro (a), evaluate C C for an ideal gas. a. Both expressions have the derivative in coon, so the best way to proceed is to try to express in ters of other partial derivatives using the cyclic relationship: solve for the desired derivative, write the derivative in red in ters of its reciprocal to get the desired expression b. o evaluate this for an ideal gas, just evaluate each of the partial differentials using = NR/: NR NR note that this is not the second derivative it is the NR first derivative squared the second derivative is zero: NR and NR 0 Substitute these into the expression for C C : NR C C NR NR. Argon is allowed to expand isotherally and reversibly at 98 K fro an initial olar volue of 5 L ol - to a final olar volue of 0 L ol -. (a) Calculate the work done by treating it as an ideal gas. (b) Calculate the work for the sae process by treating it as a van der Waals gas. he van der Waals coefficients for argon can be found in able.. Give your final answers in kj ol -. a. he work of isotheral expansion of an ideal gas is given by eq.0:, w R ln, Lol kjk ol 98Kln - 5Lol -.77kJol Copyright 08 by Eric B. Brauns

2 CHEM 05 Solutions for assignent # b. Work is w,, d substitute the van der Waals equation for, R a d b,,, R a d d b,, R,, and a are all constants,, R d a d b,, b, R ln a b,,, evaluate using the van der Waals constants fro able. and the given conditions - - 0Lol.0 0 L ol kjk ol 98Kln - - 5Lol.00 Lol. a 0.05kJ 55 tl ol 0Lol 5Lol Lat Latol the second ter does not have the sae units as the first and ust be converted to kj ol before it can be subtracted fro the first ter.7kjol he result is alost the sae as in (a) because argon behaves nearly ideally under these conditions.. Express each of the following units as a cobination of eters, kilogras, and seconds: (a) Newton, (b) Joule, (c) ascal, (d) liter, and (e) Watt. a. Newton N kgs b. Joule J N and N kgs J kg s c. ascal a N and N kgs a kg s d. Liter L 0 c 0 L 0 0 c e. Watt W Js kg s.4 Specific heat capacities can be easured in a drop calorieter where a heated saple is dropped into the calorieter and the final teperature is easured. When 45.0 g of soe etal at an initial teperature of 4 K is dropped into 4.0 g of water (with C, water = 4.9 J K - g - ) initially at 8 K in an insulated container, the final teperature at equilibriu is 9 K. (a) Assuing that the pressure reained constant, find the specific heat capacity of the etal, C,etal. (b) How uch heat flowed fro the etal to the water? Note that in (a) we are finding the average specific heat capacity over the teperature range of the experient. a. he container is insulated so we know fro the first law that the energy absorbed by the water is equal to the energy released by the etal as it cooled: Copyright 08 by Eric B. Brauns

3 CHEM 05 Solutions for assignent # du du W M U U We also know fro the first law that du = q + w = q d. For solids and liquids over a sall teperature range, we can assue that the volue change for each will be sall enough that it can be neglected allowing us to equate du with q directly: du q U q However, when pressure is constant, the heat exchanged is equal to the enthalpy: q H which allows us to write the first law as H H W M (It s iportant to reeber that this is only true when pressure is constant in general, enthalpy is not conserved.) Enthalpy and constant pressure specific heat capacity are related according to H C,s where is the ass of the aterial. If we separate variables and integrate (treating the heat capacity as independent of teperature over this sall teperature range), we obtain H C,s Using this for ΔH W and ΔH M we have C C C W W,W f i,w M,M f i,m Now solve for C,M (the only unknown quantity):,m M for convenience, use the subscripts W and M for water and etal C W,W f i,w M f i,m 4.0g4.9JK g 9K 8K 45.0g9K 4K JK g b. Since pressure is constant, the heat transferred fro the etal to the water is equal to the enthalpy change for the etal: q H M M C M,M 45.0 g0.447 JK g 9K 4K.0kJ It s negative because the etal lost this aount of energy in the for of heat..5 A saple consisting of ol of an ideal gas (for which C, = R) is taken through the cycle shown to the below. (a) Deterine the teperatures at the points,, and. (b) Calculate ΔU, ΔH, q, and w for each step and for the overall cycle. (he figure is for illustration purposes only and is not necessarily to scale.) Copyright 08 by Eric B. Brauns

4 CHEM 05 Solutions for assignent # a. he gas is ideal, so = /NR: atl ol8.060 LatK ol at44l ol8.060 LatK ol 0.5at44L ol8.060 LatK ol 68.K NR K - - NR 68.K - - NR which is the sae as since is isotheral b. For : U NC, N R - - ol 8.4JK ol 56.K68.K.4 0 J.4kJ H NC, if C, = R/, then C, = 5R/ since C, C, = R 5 N R ol 8.4JK ol 56.K68.K J 5.57 kj Since this is a constant pressure process, q is equal to ΔH: q H 5.57kJ w - at44ll0.jl at.0 J. kj Note that q + w = ΔU in accord with the first law. For : U NC, N R.4 0 J.4kJ H NC ol 8.4JK ol 68.K 56.K, 5 N R 5 ol 8.4JK ol 68.K56.K J don t forget to convert L at to J Copyright 08 by Eric B. Brauns 4

5 CHEM 05 Solutions for assignent # Since this is a constant volue process, q is equal to ΔU: q U.4kJ he volue is constant, so no work is perfored: w 0 again, note that q + w = ΔU For : his step involves the isotheral copression of an ideal gas so the internal energy and enthalpy do not change: U H 0 With ΔU = 0, the first law says that q w NR ln - - L ol8.4jk ol 7.5Kln 44L.55 0 J.55kJ w.55kj he following table suarizes the results (including the values for the overall cycle): ath ΔU (kj) ΔH (kj) q (kj) w (kj) Cycle work for reversible copression of an ideal gas is - - ol8. 0 L atk ol 00K 49.L w NR ln f i.6 One ole of heliu gas expands reversibly fro 4.6 L and 00 K to 49. L. Calculate the final pressure (in at) and teperature if the expansion is (a) isotheral or (b) adiabatic. (c) Sketch these two processes on a diagra. You ay the gas as ideal and that C, = R/ and is independent of teperature. a. It s isotheral so,iso = = 00 K. he gas is ideal and we also know, so NR,iso,iso 0.5 at b. o find,ad, we need the final teperature,,ad. For an adiabatic expansion,,ad is related to,, and according to,ad where = C, /C,. Since C, is given and C, C, = R for an ideal gas, C, = 5R/. his akes = 5/ and = -/: 49.L,ad 00K o avoid rounding errors, it s best to use -/ rather than L 89.0K Now that we have,ad, we can use the ideal gas equation to deterine,ad : - - NR ol8. 0 L atk ol 89.0K,ad 0.5 at,ad 49.L Copyright 08 by Eric B. Brauns 5

6 CHEM 05 Solutions for assignent # c. he corresponding diagra for each:.00 (at) 0.50 isother 0.5 adiabat 4.6 (L) At low to oderate pressures, and between 00 K and 400 K, N (g) can be regarded as an ideal gas with C, = 9. J K - ol -. (a) Calculate ΔU, ΔH, q, and w for the reversible adiabatic copression of 0.04 ol of N fro 400 torr and L to a final volue of 0.5 L. (b) If a saple of N at 98 K and 760 torr is cooled to 00 K via a reversible adiabatic expansion, what is the final pressure? a. It s adiabatic so q = 0. he forulas we need for the reaining quantities are: U NC, H NC, and w U,, o use these, we need,, and C,. We are given C, and treating the gas as ideal, therefore we can easily calculate C, : Note that we are treating N C C R as ideal, but C, is ~7R/, not 5R/. he,, difference is that N is diatoic and diatoic gases have higher heat capacities than onatoic gases. We ll discuss this in detail next seester. C C R 0.8JK ol,, Since,, and N are given, and the gas is ideal, we can calculate : 400 torrl 60.K - - NR 0.04ol6.6L torrk ol o find we use: where C, 9.JK ol C, 0.8JK ol his gives L 60.K 79.K L Now we can calculate the desired quantities: U NC, 0.04 ol0.8jk ol 79.K 60.K 98.8J H NC, 0.04ol9.JK ol 78.9K60.K 8.J w U 98.8J b. We want for an adiabatic expansion given = 760 torr, = 98 K, and = 00 K. he pressures and volues associated with an adiabatic process are given by solve for : o use this forula, we need and. We can find using the ideal gas equation: Copyright 08 by Eric B. Brauns 6

7 CHEM 05 Solutions for assignent # NR substitute the s cancel 0.04ol 6.6L torrk ol 98K 760torr 0.978L We can find fro Solve for : 00K 0.978L 98K 4.99L Now we can find : 0.978L 760 torr 4.99L 6.6torr Note that we could have found without first finding by substituting the expression for (the red expression above) into the expression for (blue expression above): note: dividing by is the sae as ultiplying by Copyright 08 by Eric B. Brauns 7

Heat Capacity: measures the ability of the substance to pick up heat

Heat Capacity: measures the ability of the substance to pick up heat Calorietry: easures flow of heat across boundaries Heat Capacity: easures the ability of the substance to pick up heat Heat capacity under constant pressure, C p euals the aount of heat reuired to raise