= T T V V T = V. By using the relation given in the problem, we can write this as: ( P + T ( P/ T)V ) = T

Transcription

1 hermodynamics: Examples for chapter Show that C / = 0 for a an ideal gas, b a van der Waals gas and c a gas following P = nr. Assume that the following result nb holds: U = P P Hint: In b and c, differentiate with respect to both temperature and volume and recall that for exact differentials the order of differentiation can be exchanged. he lecture notes give: C = U respect to volume: C = U. Differentiate this equation with = U By using the relation given in the problem, we can write this as: C = U P + P/ = P = Next we consider the various equations of state: a Ideal gas. P = nr/ from which the second derivative of pressure see above is zero and therefore C = 0. b For a van der Waals gas we have: P = nr n a. Differentiation nb of P with respect to once just gives nr/ nb. his does not depend on and hence C = 0. c Differentiation of P twice with respect to again gives zero and hence C = 0. 1

2 . Show that q rev is not a state function i.e. dq rev is not exact for a gas obeying the equation of state P = R, but that d b qrev/ is. Assume a reversible process and consider only P -work. Hint: you may proceed as follows: a Use the previous problem to calculate U/. b Use du = U d + U d to calculate du. c Use the first law of thermodynamics to get an expression for dq. d Substitute the equation of state into the above expression. e Apply the exactness test for differentials dq = M,d + N,d. Use results from the previous problem to differentiate M with respect to. f Repeat the same calculation for dq/. By using the result given in the first problem, we can obtain U = 0. he total differential for du now gives du = U d and hence du = C d. he first law of thermodynamics, du = dq + dw, gives dq = C d dw. Considering P -work, we can write: dq = C d +P ext d. Because the process is reversible, P ext = P and dq = C d +Pd. For dq to be exact we should have: C = P From the first problem we know that the left hand side is zero. he right hand side, however, is not zero: P = nr 0 nb dq hus dq is inexact. For dq/ we have: = C d + P d. Now the exactness test gives: C / = 0 is constant

3 nr/ nb hus dq/ is exact. = 0 the expression does not depend on 3. An ideal gas initially at P 1,, 1 undergoes a reversible isothermal expansion to P,, 1 path 1. he same change in state of the gas can be achieved by allowing it to expand adiabatically from P 1,, 1 to P 3,, and then heating it at constant volume to P,, 1 path. Note that has not been specified and you should find an equation that determines it. Show that the entropy change for the reversible isothermal expansion path 1 is the same as the sum of the entropy changes in the reversible adiabatic expansion and the reversible heating path. Because the two paths give the same result, it is probable that the integral is independent of path. his is not a complete proof why? Path 1: First we recall that for an ideal gas U = 3 nr see lecture notes. he temperature is constant along this path and thus change in internal energy must be zero. he first law of thermodynamics now states that U = q rev + w rev and hence q rev = w rev. Recall the following equation from the lecture notes: w rev = nr 1 ln Using the definition of entropy gives: S = q rev 1 = nr ln q rev = nr 1 ln Path : First consider the first segment from P 1,, 1 to P 3,,. he expansion is adiabatic i.e. no heat exchange with the surroundings and hence q rev = 0. For this reason the change in entropy is also zero S = q rev / = 0. he final temperature in an adiabatic d expansion is determined by integrating C = nrd see lecture notes: 3

4 1 C d = nr d = nr ln Along the second segment P 3,, to P,, 1, we have a constant volume process. he lecture notes now give dq rev = C d. he definition of entropy is ds = dqrev and thus: ds = C d S = 1 C d = 1 C d Comparison of this with the expression determining along the first segment, gives the final result: S = nr ln his is the same result that was obtained along path 1. For a complete proof of exactness, one would have to use the exactness test or consider infinitely many paths or rather, the exactness test. 4. Water is vaporized reversibly at 100 C and bar. he heat of vaporization is kj mol 1. a What is the value of S for the water? b What is the value of S for the water plus the heat reservoir at 100 C? he heat reservoir is thermally isolated from its surroundings. a S H O = q kj mol 1 = = J K 1 mol 1. Note that K sign means that water receives heat. b he reservoir loses heat to water exactly the same amount as above. he change in entropy for the heat reservoir is q/ = J K 1 mol 1. 4

5 Note: Since water + heat reservoir is isolated from the rest of the world, the total change its entropy is 0. he total entropy in the system is conserved. 5. Assuming that CO is an ideal gas, calculate H and S for the following process: CO g, K, 1 bar CO g, 1000 K, 1 bar Consider 1 mol of gas and a reversible process. Given: C P = units: J K 1 mol 1. Here refers to the standard state. For gases this is 1 bar pressure but note that this does not specify temperature. he overbar denotes that these are molar quantities i.e. per mole. o calculate change in enthalpy, we integrate the heat capacity over temperature see lecture notes: H = 1 C P d = 1000 / K K [ = kj mol Furthermore, at constant pressure we can apply equations: dq = C P d and ds = dq rev / = C P /d: ] S = 1 C P d = 1000 / K K [ ln = 55.4 J K 1 mol 1 ]

6 6. Ammonia considered to be an ideal gas initially at 5 C and 1 bar pressure is heated at constant pressure until the volume has trebled. Assume reversible process. Calculate: a q per mole, b w per mole, c H, d Ū, e S given C P = in J K 1 mol 1. ripling of the ideal gas volume P = nr 1 leads to = P3 /nr, which means that the temperature will be three times higher; = 3 1. Note that pressure is constant. a q = CP d = kj mol K 98 K d = b w = P = R = J K 1 mol K = 4.96 kj mol 1. c Because pressure is constant, we have H = q P = 6.4 kj mol 1. d Ū = q + w = kj mol 1 = 1.4 kj mol 1. e S = 1 C P d = J K 1 mol K 98 K d = 7. hree moles of an ideal gas expand isothermally and reversibly from 90 L to 300 L at 300 K. a calculate U, S, w and q for this system, b calculate Ū, S, w per mole and q per mole, c If the expansion is carried out irreversibly by allowing the gas to expand into and evacuated container, what are the values of Ū, S, w per mole and q per mole? a Change in internal energy is zero because the gas is ideal. Recall that the internal energy for an ideal gas depends only on temperature. Here we have an isothermal process and hence no change 6

7 in internal energy occurs, U = 0. Note also that the 1st law now states that q = w. For an isothermal process see the lecture notes we have: w rev = nr ln q rev = nr ln = 3 mol J K 1 mol K ln 300 L 90 L = 9.01 kj and q = 9.01 kj. By using the definition of entropy, we can calculate the change in entropy: S = q rev = 9.01 kj 300 K = J K 1 b Divide everything by 3 mol to get per mole quantities: Ū = 0 kj mol 1, S = J K 1 mol 1, w = 3.00 kj mol 1,q = 3.00 kj mol 1 c Since the temperature is constant, we have Ū = 0. Since the gas is expanding into vacuum, P ext = 0 and thus w = 0. By the first law, q = 0. he entropy is the same as in b because its value depends only on endpoints of the path. Note that q along the present irreversible path cannot be used in calculating entropy one must always use a reversible path for example that in b above. For this reason S = J K 1 mol 1, which is the same value as in b. 8. An ideal gas at 98 K expands isothermally from a pressure of 10 bar to 1 bar. What are the values of w per mole, q per mole, Ū, S in the following cases? a he expansion is reversible, b he expansion is free irreversible, c he gas and its surroundings form an isolated system, and the expansion is reversible and d he gas and its surroundings form an isolated system, and the expansion is free irreversible. 7

8 a his is a reversible process and constant temperature implies that Ū = 0. Also the enthalpy change for an ideal gas depends only on temperature H = 0. By using the 1st law and the expression for reversible expansion see lecture notes, we get: w rev = R ln = R ln P P 1 = q rev By plugging in the values, we get w rev = 5.71 kj mol 1 and q rev = 5.71 kj mol 1. Now the definition of entropy gives the entropy change: S = q rev = 19.1 J K 1 mol 1 b Irreversible process free expansion. No external pressure no work done w = 0. hus by the first law q = 0 i.e. no heat exchanged with the surroundings since Ū = 0 and H = 0. Entropy depends only on the endpoints of the path and hence S = qrev = 19.1 J K 1 mol 1, where q rev is from part a. Note that only reversible paths can be used in calculating entropy! c Isolated system, which here means that system + surroundings is isolated from the rest of the world. For a reversible process we have ds tot = ds syst + ds surr = 0. For the system we have: ds syst = dq rev and for the surroundings ds surr = dq rev Since temperature is constant, U = 0 and H = 0. herefore: q rev = w = R ln P P 1 from previous calculations hus S syst = R lnp /P 1 and S surr = R lnp /P 1. hus the total change of entropy system + surroundings is zero. Also q tot = q sys + q surr = R lnp /P 1 R lnp /P 1 = 0. For the same reason, the total work w tot = w sys + w surr = 0. Note that the system + surroundings is isolated from the rest of the world and therefore the total change in heat q tot and work w tot must clearly be zero. 8

9 d If we want to calculate the entropy change for the system, we need a reversible path for the calculation. Above this was done: S syst = R lnp /P 1. In free expansion, no work is done, w sys = 0 and w surr = 0. Since U = w sys +q sys and U = 0, q sys = 0 as well also then q surr = 0. hus the system is not interacting with the surroundings at all in this process. his means that the entropy of the surroundings does not change either, S surr = 0. he total entropy is then S tot = S syst + S surr = R lnp /P 1 = 19.1 J K 1 mol One mole of gas A at 1 bar and one mole of gas B at bar are separated by a partition and surrounded by a heat reservoir i.e. the temperature is constant. When the partition is withdrawn, how much does the entropy change? Both gases behave according to the ideal gas law. Hint: consider the calculation in three steps: I the initial entropy difference from standard state, II change in entropy due to expansion/compression of gases at constant temperature and finally III entropy change due to mixing of the gases. Note that the pressures of the two gases are different and thus the results in the lecture notes cannot be directly applied. I he initial i.e. before mixing entropies for the gases are: S A = S A nr ln S B = S B nr ln P ini A = S PA A nr ln 1 bar = SA 1 bar P ini B = S PB B nr ln II Let both gases expand from their initial pressures to the final pressure at constant temperature. his changes entropy of both gases according to: S A = S A nr ln Ptotal P A and S B = S B nr ln Ptotal P B 9

10 where P total is the final pressure after mixing. he final volume after mixing is: total = A + B = nr P A + nr P B = nr P A + nr P A = 3nR P A he total pressure after mixing is then: P total = nr total = 4 3 P A = 4 3 bar he entropy change due to expansion for both gases is: S A = S A nr ln 4 3 and S B = S B nr ln 4 3 Combining 1 and, we have: S A = n A R ln 4 3 and SB = n B R ln 4 3 ln. III Finally we must include the entropy change due to mixing n = 1: mix S = R ln 1 mol R ln mol 1 mol = R ln mol he total entropy change is then : S total = S A + S B + S mix = R ln = 1.51 J K 1 mol R ln R ln Calculate the change in molar entropy of aluminum that is heated from 600 to 700 C. he melting point of aluminum is 660 C, the heat of fusion is 393 J g 1 the molar mass for aluminum is 7 g mol 1, and the heat capacities at constant pressure of the solid and the liquid may be taken as 31.8 J K 1 mol 1 and 34.4 J K 1 mol 1 independent of temperature, respectively. 10

11 Note that 600 C is 873 K, 660 C is 933 K, and 700 C is 973 K. Use the following equation see lecture notes: S = fusion initial = C P s ln C P s d + H fusion fusion initial = 31.8 J K 1 mol 1 ln J K 1 mol 1 ln fusion + + H fusion fusion final fusion C P l d final + C P l ln fusion 933 K + 7 g mol J g K 933 K 973 K = 19.9 J K 1 mol K 11. Steam is condensed at 100 C, and the water is cooled to 0 C and frozen to ice. What is the molar entropy change of the water? Consider that the average specific heat of liquid water is 4. J K 1 g 1 the molar mass of water is g mol 1. he enthalpy of vaporization at the boiling point and the enthalpy of fusion at the freezing point are 58.1 J g 1 and J g 1, respectively. Use the same cycle as in the previous problem note that the cycle goes from high temperature to low temperature and thus the signs are reversed!: S = H vap vap K K C P l d H fus fus 11 = 58.1 J g g mol K

12 4. J K 1 mol g mol K ln K J g g mol K = J K 1 mol 1 1. Calculate the increase in the molar entropy of nitrogen when it is heated from 5 C to 1000 C at constant pressure with: CP = in J K 1 mol 1. Nitrogen is gaseous in the temperature range. he entropy change is then given by: S = K K C P g d = / K K ln = J K 1 mol