This was Exam 1 last year. It is presented here with, first, just the problems and then with the problems and their solutions. YOU WILL BENEFIT MOST if you attempt first just the problems as if you were taking the exam rather than turning to the solutions right away! 1. (10 + 10 points) Two quick calculations: (a) In 1938, a physical chemist named Ken Pitzer (who was the president of Stanford in the 1960 s, by the way) measured the entropy of diamond at 298.15 K and found S o = 2.448 J mol 1 K 1. The Debye temperature for diamond, Θ D, is 1890 K. How well does the Debye theory do at predicting Pitzer s 298.15 K value? (b) An experimental study of the synthesis of Ir 2 S 3 (s) from Ir(s) and S 2 (g) in 1972 yielded a reaction molar enthalpy change per mole of Ir 2 S 3 formed of 400.68 kj mol 1 at 298.15 K. Since S 2 (g) is not the standard state of sulfur at 298 K, this is not the standard molar enthalpy of formation of Ir 2 S 3 (s). What is ΔH f o for Ir2 S 3 (s) at 298.15 K? You ll need some or maybe all of the data below to answer this question. S(g) S(l) S 2 (g) S(s) ΔH o f /kj mol 1 277.17 1.85 128.60 0 2. (6 points each) Here are some questions about water, along with some data that will prove useful. Assume P = 1.00 atm and n = 1.00 mol for all parts of this problem. T fus = 273.15 K and T vap = 373.15 K At 273.15 K, ΔH fus = 6.007 kj mol 1. At 373.15 K, ΔH vap = 40.66 kj mol 1. C P (H 2 O(l)) = 75.29 J mol 1 K 1 S o (H 2 O(g), 373.15 K) = 196.68 J mol 1 K 1 S o (H 2 O(s), 273.15 K) = 41.36 J mol 1 K 1 (a) What is the enthalpy change if H 2 O(s) at 273.15 K is converted to H 2 O(g) at 373.15 K? (b) Calculate S o (H 2 O(l), 373.15 K) S o (H 2 O(l), 273.15 K). (c) What is the value of the quantity ΔH TΔS for the process H 2 O(l, 273.15 K) H 2 O(s, 273.15 K)?
3. (2 points each) For the following, circle the appropriate choice from among the four options given for each. (a) In order for a process to occur spontaneously in an isolated system, we must have ΔS sys > 0 ΔS sys < 0 q = 0 ΔS Univ < 0 (b) The difference C P C V can never be measured positive negative zero (c) An ideal gas at 400 K is adiabatically compressed from 2.00 atm to 2.50 atm. The enthalpy change of the gas is zero positive negative irreversible (d) To increase the molar entropy of an ideal gas held in a rigid container at 100 K by an amount R, one should change the temperature to 200 K 195 K 51 K 448 K (e) To compress a gas isothermally to a chosen final volume using the minimum amount of work, I should perform the compression at constant P very quickly very slowly at constant T (f) On a graph of P versus V for an ideal gas, lines of zero slope are isochores isobars isotherms adiabats (g) For an ideal gas, the quantity ( U/ V) S equals 0 T P nrt/v (h) Mercury is an unusually volatile metal, even at room temperature. The standard molar enthalpy of formation of Hg(g), ΔH o f,m, at 298.15 K must be 0. equal ΔH o vap (Hg(l)). be < 0. be > 0.
4. (4 + 4 + 4 + 6 points) Here is a P-V diagram for a three-step process on an ideal gas that starts in state 1 at T = 300 K, P = 2.50 atm, and V = 2.00 L. Each path of the three steps is shown by a dashed line, and each numbered point is an equilibrium state. 3.0 2.5 2.0 P/atm 1.5 1.0 0.5 0 0 1 2 3 4 5 6 V/L (a) What are the temperatures of states 2 and 3? at state 2: at state 3: (b) Find ΔS for the steps from 1 to 2 and from 3 back to 1. ΔS(1 2) = ΔS(3 1) = (c) Now find ΔS for the step from 2 to 3. This looks tricky, because P, V, and T are all changing, but it s really easy to calculate if you stop and think a minute. (d) Calculate w, q, and ΔU for the path 2 3. 5. (8 + 8 points) The general molar heat capacity C for an arbitrary reversible path in anything is given by C = U T V + U + P V dv T dt where the derivative dv/dt depends on the specific path. (a) Suppose the system is an ideal gas and the path is VT 3/2 = a constant. Show that C = 0 and tell me what this path is called. You ve seen it before!
(b) Now suppose the system is still an ideal gas, but the path is V/T = a constant. What is C for this path? 6. (3 points each) For any pure substance, we can construct an equation of state of the form P = P(V, T), i.e., we can consider P to be some function of two independent state variables V and T. (a) Write the total differential dp for such a function. (Note: this is just math, not thermodynamics!) (b) Now, using your expression in (a), set dp = 0 and write an expression for ( P/ T) V. (Again, this is just math!) (c) Next, introduce β and κ, defined on the front page, into your expression from part (b). (d) Verify your general result in (c) for the specific case of an ideal gas.
Now for the solutions: 1. (10 + 10 points) Two quick calculations: (a) In 1938, a physical chemist named Ken Pitzer (who was the president of Stanford in the 1960 s, by the way) measured the entropy of diamond at 298.15 K and found S m o = 2.448 J mol 1 K 1. The Debye temperature for diamond, Θ D, is 1890 K. How well does the Debye theory do at predicting Pitzer s 298.15 K value? The Debye theory for a non-metal says C(T) = at 3 where a is a constant. The entropy at 298.15 K is ths given by the expression S o 298.15 K 298.15 K C(T)dT = = at 3 dt = 0 T 0 T at 3 3 = 12π4 R 298.15 K 3 = 2.54 J mol 1 K 15 1890 K 1 which is in pretty good agreement with the experimental value! (b) An experimental study of the synthesis of Ir 2 S 3 (s) from Ir(s) and S 2 (g) in 1972 yielded a reaction molar enthalpy change per mole of Ir 2 S 3 formed of 400.68 kj mol 1 at 298.15 K. Since S 2 (g) is not the standard state of sulfur at 298 K, this is not the standard molar enthalpy of formation of Ir 2 S 3 (s). What is ΔH f o for Ir2 S 3 (s) at 298.15 K? You ll need some or maybe all of the data below to answer this question. S(g) S(l) S 2 (g) S(s) ΔH o f /kj mol 1 277.17 1.85 128.60 0 The experimental value, ΔH o = 400.68 kj mol 1, is for the reaction 2Ir(s) + 3 2 S 2(g) Ir 2 S 3 (s) but since the standard state of sulfur is not the diatomic gas but rather the solid, we need to correct this enthalpy by that of the reaction that produces S 2 (g) from S(s): 3 S(s) 3 2 S 2(g) ΔH o = 3 2 ΔH f,m o (S 2 (g)) ΔH o f,m(s(s)) = 3 (128.60 kj mol 1 ) = 192.90 kj mol 1 2 which gives us the Ir 2 S 3 formation enthalpy: 2Ir(s) + 3 S(s) Ir 2 S 3 (s) ΔH o = ΔH o f,m(ir2 S 2 (s)) = 192.90 kj mol 1 + ( 400.68 kj mol 1 ) = 207.78 kj mol 1 2. (6 points each) Here are some questions about water, along with some data that will prove useful. Assume P = 1.00 atm and n = 1.00 mol for all parts of this problem. T fus = 273.15 K and T vap = 373.15 K At 273.15 K, ΔH fus = 6.007 kj mol 1. At 373.15 K, ΔH vap = 40.66 kj mol 1. C P (H 2 O(l)) = 75.29 J mol 1 K 1 S o (H 2 O(g), 373.15 K) = 196.68 J mol 1 K 1
S o (H 2 O(s), 273.15 K) = 41.36 J mol 1 K 1 (a) What is the enthalpy change if H 2 O(s) at 273.15 K is converted to H 2 O(g) at 373.15 K? We follow the path H 2 O(s), 273.15 K H 2 O(l), 273.15 K H 2 O(l), 373.15 K H 2 O(g), 373.15 K, and add up the enthalpy changes at each step: ΔH = 6.007 kj mol 1 + (75.29 J mol 1 K 1 ) (100 K) + 40.66 kj mol 1 = 54.196 kj mol 1 (b) Calculate S o (H 2 O(l), 373.15 K) S o (H 2 O(l), 273.15 K). There are two ways (at least!) to do this calculation. The first simply calculates the entropy change directly on taking water from melting point to boiling point: ΔS = 373.15 K 273.15 C P,m dt T = (75.29 J mol 1 K 1 )ln 373.15 K 273.15 K = 23.49 J mol 1 K 1 The second approach takes advantage of the absolute molar entropies we are given for ice and steam, converts those into entropies of the liquid phase at the melting and boiling points, and subtracts those two absolute entropies: S o (l, 273.15 K) = S o (s, 273.15 K) + ΔS fus = 41.36 J mol 1 K 1 + 6007 J mol 1 273.15 K = 63.85 J mol 1 K 1 S o (l, 373.15 K) = S o (g, 273.15 K) ΔS vap = 198.68 J mol 1 K 1 40,660 J mol 1 373.15 K = 87.72 J mol 1 K 1 ΔS = 87.72 J mol 1 K 1 63.35 J mol 1 K 1 = 24.36 J mol 1 K 1 which is in pretty good agreement with the first approach. (c) What is the value of the quantity ΔH TΔS for the process H 2 O(l, 273.15 K) H 2 O(s, 273.15 K)? We ve seen this expression before in the form ΔH fus = TΔS fus. Thus ΔH TΔS must equal zero. 3. (2 points each) For the following, circle the appropriate choice from among the four options given for each.
(a) In order for a process to occur spontaneously in an isolated system, we must have ΔS sys > 0 ΔS sys < 0 q = 0 ΔS Univ < 0 In general, the entropy of the Universe must increase. But for an isolated system, the system is the entire Universe! (b) The difference C P C V can never be measured positive negative zero This difference is VTβ 2 /κ, which can be zero (if β is zero), but can never be negative (because κ is always positive). (c) An ideal gas at 400 K is adiabatically compressed from 2.00 atm to 2.50 atm. The enthalpy change of the gas is zero positive negative irreversible Adiabatic compression raises the gas temperature (w = ΔU > 0), and that insures ΔH > 0. (d) To increase the molar entropy of an ideal gas held in a rigid container at 100 K by an amount R, one should change the temperature to 200 K 195 K 51 K 448 K We re at constant volume going from 100 K to some T f. We write ΔS = R = C V ln T f 100 K = 3 2 R ln T f 100 K or T f = (100 K) e 2/3 = 195 K (e) To compress a gas isothermally to a chosen final volume using the minimum amount of work, I should perform the compression at constant P very quickly very slowly at constant T Very slowly (approaching reversibly) is the best answer, but I accepted at constant T as well, because I wasn t super clear in stating the situation by throwing in the word isothermally. (f) On a graph of P versus V for an ideal gas, lines of zero slope are isochores isobars isotherms adiabats
An isobar is the line of constant pressure, and on a P, V graph, these are horizontal lines of zero slope. (g) For an ideal gas, the quantity ( U/ V) S equals 0 T P nrt/v From the master equation: du = TdS PdV. The quantity ( U/ V) S is the coefficient of dv, which is P. (h) Mercury is an unusually volatile metal, even at room temperature. The standard molar enthalpy of formation of Hg(g), ΔH f o, at 298.15 K must be 0. equal ΔH o vap (Hg(l)). be < 0. be > 0. ΔH o f is the enthalpy change for Hg(l) Hg(g) at 298.15 K, which must be >0 (it s endothermic), but it s not equal to ΔH o vap (Hg(l)) because that quantity refers to the Hg(g) at standard (1 atm) pressure, which is way greater than the vapor pressure of Hg(l) at 298.15 K. 4. (4 + 4 + 4 + 6 points) Here is a P-V diagram for a three-step process on an ideal gas that starts in state 1 at T = 300 K, P = 2.50 atm, and V = 2.00 L. Each path of the three steps is shown by a dashed line, and each numbered point is an equilibrium state. 3.0 2.5 2.0 P/atm 1.5 1.0 0.5 0 0 1 2 3 4 5 6 V/L (a) What are the temperatures of states 2 and 3? at state 2: 120 K at state 3: 750 K
First, we calculate for future reference the amount of gas, n: n = PV/RT = 0.203 mol. Next we find the two other temperatures from PV/T = constant given that P 1 = 2.5 atm, V 1 = 2 L, and T 1 = 300 K. We find the values shown above. (b) Find ΔS for the steps from 1 to 2 and from 3 back to 1. ΔS(1 2) = 2.32 J K 1 ΔS(3 1) = 3.87 J K 1 Step 1 2 is at constant V; so, we find ΔS from the expression ΔS = C V ln T f = T 3 nr ln 120 K i 2 300 K = 2.32 J K 1 but step 3 1 is at constant P; so, we have here ΔS = C P ln T f = T 5 nr ln 300 K i 2 750 K = 3.87 J K 1 If you used L atm energy units instead of J units, that was fine, by the way. In both of these expressions, we used n = 0.203 mol, which we found earlier. (c) Now find ΔS for the step from 2 to 3. This looks tricky, because P, V, and T are all changing, but it s really easy to calculate if you stop and think a minute. We note that the trip 1 2 3 1 is a three-step cycle. Thus, the total ΔS for all three steps must add to zero! We have therefore ΔS(2 3) = ΔS(1 2) + ΔS(3 1) = 6.19 J K 1 (d) Calculate w, q, and ΔU for the path 2 3. First, note that this path is an expansion, which means the work must be negative. We can calculate the work from the graph by evaluating the area under the diagonal path line. That area is w, and we find w = 5.25 L atm = 532 J. Next, we should find ΔU because it s easy to find and because q = ΔU w. We find ΔU = (3/2)nRΔT = 1.596 kj (for ΔT = 750 K 120 K = 630 K and n = 0.203 mol). This gives us q = ΔU w = 2.128 kj. 5. (8 + 8 points) The general molar heat capacity C for an arbitrary reversible path in anything is given by C = U T V + U + P V dv T dt where the derivative dv/dt depends on the specific path.
(a) Suppose the system is an ideal gas and the path is VT 3/2 = a constant. Show that C = 0 and tell me what this path is called. You ve seen it before! Let s be sure we understand exactly what we ve been asked to do here. We want to take a given path, use it in the general expression for C, and then interpret the value for C to tell us something about the path. In reality, we may very well be able to deduce C from the physical nature of the path, which we might recognize, but that s not what we were asked to do here. So let s do the first path: Note that our target expression for C has the simple derivative dv/dt, which is how we specify what the path is. Here, we have VT 3/2 = constant, or V = (constant)/t 3/2, so that we can write dv/dt = ( 3/2) (constant) T 5/2 = ( 3/2)(V/T) (because the constant = VT 3/2 ). We also remember that ( U/ V) T = 0 and ( U/ T) V = C V = 3nR/2. This gives us C = U T V + U + P V dv T dt = 3 2 nr + 0 + P 3 2 V T = 3 2 nr 3 2 PV T = 0 The heat capacity is zero along an adiabatic, reversible path. (b) Now suppose the system is still an ideal gas, but the path is V/T = a constant. What is C for this path? Here, we have V/T = constant, or V = (constant)t, so that we can write dv/dt = constant = V/T = nr/p. We find C = U U + + P T V V dv T dt = 3 2 nr + 0 + P constant = 3 nr + PnR 2 P = 5 2 nr = C P Note again that we first found dv/dt based on the path, then deduced C from the general expression, and finally said, Ah ha! I know that value! That s C P! 6. (3 points each) For any pure substance, we can construct an equation of state of the form P = P(V, T), i.e., we can consider P to be some function of two independent state variables V and T. (a) Write the total differential dp for such a function. (Note: this is just math, not thermodynamics!) If P = P(V,T), then dp = P V T dv + P T V dt
(b) Now, using your expression in (a), set dp = 0 and write an expression for ( P/ T) V. (Again, this is just math!) If dp = 0, then P V T dv = P T V dt or P = P V T V V T T P (c) Next, introduce β and κ, defined on the front page, into your expression from part (b). P = P V = 1 T V V T T P Vκ βv = β κ (d) Verify your general result in (c) for the specific case of an ideal gas. For the ideal gas, β = (1/V)(nR/P) = 1/T, and κ = (1/V)( nrt/p 2 ) = 1/P. We have P = d(nrt/v) T V dt = nr V = P T = β κ = (1/T) (1/P) = P T