Department of Chemistry Indian Institute of Technology Madras. CY 102 Chemistry II Assignment I Thermodynamics (First and second laws thermodynamics)

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1 Department of Chemistry Indian Institute of Technology Madras CY 102 Chemistry II Assignment I Thermodynamics (First and second laws thermodynamics) Jan 27, Two moles of N 2 gas at 25 0 C and 1 atm pressure are expanded reversibly and isothermally to a pressure of 0.5 atm. a) Calculate the work done (including the sign). b) Calculate the heat change and the internal energy change involved in the process. c) Calculate the work done in expansion against a constant external pressure of 0.5 atm (final state is 0.5 atm as in a). What is your inference of the result in comparison with the result of a)? Ans: a) Isothermal (reversible) expansion: Class formula: w = -nrt ln V 2 /V 1. = -2 mol x JK -1 mol -1 x K. x log P 1 /P 2 as PV = Constant at constant temperature. = -2 x x x log 2 J = J b) Heat Change and the internal energy change. Internal energy change is zero for an ideal gas undergoing process at constant temperature E = 0. Heat change = - work done by the gas. = J. (This much heat is absorbed by the system to do the work) c) Work done in expansion against a constant pressure of 0.5 atm. = -P V.

2 To Calculate V, calculate V 1 and V 2. V 1 = nrt/p 1 = 2 mol x L atm K -1 mol -1 x K/1 atm. = L V 2 = 2V 1 as P 2 = ½ P 1. w = -P V = 0.5 atm x L = L atm. = J. Observation: Expansion against constant pressure is NOT a reversible process. Therefore, the work done by the gas is not the maximum possible work done, but less than the maximum. It is a path-dependent quantity. 2. An ideal gas expands reversibly and isothermally from 20 atm to 1 atm at 300 K. What are the values for a) work done by the gas per mole? b) heat absorbed by the gas per mole? c) E per mole and d) H per mole? Ans: a) Reversible isothermal expansion from 20 atm to 1 atm at 300K. a) w = -nrt ln P 1 /P 2. b) Q = -w c) E = 0 d) H = q as the work done is reversible, under isothermal conditions. Ans: w = j. 3. The heat capacity of a gas is represented by the formula γ C == α + βt + p T 2 For O 2, α = J K -1, β = J K -2 and γ = x 10-5 J K. Calculate the amount of heat required to heat the gas from 300 K to 1000K. Ans: H = T1 T2 C P (T) dt = T 1 T2 (α + βt + γ/t 2 ) dt = α (T 2 T 1 ) + β/2 (T 2 2 T 1 2 ) - γ (1/T 2 1/T 1 )

3 For O 2 α = JK -1. β = JK -2. γ = x 10 5 JK T 1 = 300 K T 2 = 1000 K. Substitute: Answer H = kj α(t 2 T 1 ) = kj β/2 (T 2 2 T 1 2 ) = 1.90 kj. -γ(1/t 2 1/T 1 ) = 0.39 kj. 4. An ideal monatomic gas is expanded by a reversible adiabatic process from an initial pressure of 1 atm to a final pressure of 0.3 atm. The initial temperature of the gas is 300 K. Calculate the work done, internal energy change and the final temperature of the gas. Ans: Reversible adiabatic expansion. q = 0, E = q + w = w For adiabatic processes, PV γ = constant γ = C P /C v = 5/3 for an ideal monatomic gas. Also PV = nrt for all ideal gases. (V 1 /V 2 ) = P 2 /P 1 and (V 1 /V 2 ) = (T 1 /T 2 ) (P 2 /P1) or (T 1 /T 2 ) = (P 2 /P 1 ) 1-γ/γ substitute the values T 1, P 1 and P 2. T 2 is calculated as K. Work done = C v T = kj (T 2 < T 1, work done is done by the gas, hence negative). Internal energy charge E = kj. 5. An ideal monatomic gas at 1 atm pressure with a heat capacity at constant volume C v =3R/2 is expanded adiabatically against a constant external pressure of 0.5 atm. Calculate q, w, Eand H (per mole) Ans: For T use the same formula as above T 1 = 300 k, T 2 = k. E = -P V = J H = C p T = 5/2R x 72.7 J.

4 6. Determine whether the following are exact differentials. a) xy 2 dx x 2 dx x ydy b) 2 dy y y b). Verify that dq = de +Pdv = C V dt + RTd ln V is not an exact differential but that dq C T = v d ln T + R d ln V is an exact differential. Ans: a). Let xy 2 dx x 2 y dy = df Then 2 f/ x y = 2 f/ y x must be satisfied. It is NOT. Therefore xy 2 dx x 2 y is NOT an exact differential. b) dq = C v dt + Rt d ln V. 2 q/ V T = / V (C V ) = for an ideal gas. 2 q/ T V = [ / T (RT/V)] V = R/V 2 q/ V T. Hence dq is NOT an exact differential. dq/t = C V /T dt + R/v dv = df 2 f/ V T = ( / T[R/V]) V = 0. 2 f/ V T = ( / V[C V /T]) T = 0 = 2 f/ T V. Hence dq/t is an exact differential. 7. Derive the relation C V = - E V V T T E From the expression for the total differential of E (T, V). Ans: E, V, T follow the cyclic relation ( E/ T) v ( T/ V) E ( V/ E) T = -1. Or ( E/ T) V = -1/( T/ V) E ( V/ E) T = -( V/ T) E ( E/ V) T More elaborately: E = E (V,T) de = ( E/ T) v dt + ( E/ V) T dv. = C V dt + ( E/ V) T dv.

5 Express V as a function of E and T and substitute for dv in the above expression. dv = ( V/ E) t de + ( V/ T) E dt. de = ( E/ V) T {( V/ E) T de + ( V/ T) E dt} + C V dt. 0 = {1-( E/ V) T ( V/ E) T } de + {C v + ( E/ V) T ( V/ T) E }dt The equality implies the coefficients of de and de are zero, as two of the three variables E, v, T are independent. Hence: ( E/ V) T ( V/ E) T = 1, C V = - ( E/ V) T ( V/ T) E. 8. Show that P V C p C v =T T V T P Using appropriate relations between partial derivatives, show also that C p C v = T V T V P P T 2 For water at 4 0 C, given that the density of water is maximum, what does the above result indicate. Also evaluate (C p C v ) for an ideal gas. Ans: de = ( E/ T) v dt + ( E/ V) T dv = C v dt + π T dv. π T internal pressure. ( E/ T) P = π T ( V/ T) P + C V. ( E/ T) P C V = π T ( V/ T) P Also, π T = T( P/ T) V - P C P C V = ( H/ T) P ( E/ T) V ; H = E + PV. = ( E/ T) P + P ( V/ T) P ( E/ T) V. = P( V/ T) P + π T ( V/ T) P. = (π T + P) ( V/ T) P = T( P/ T) V ( V/ T) P. Also P 1 T 1 V Form cyclic partial derivatives. ( P/ T) v = -1/( T/ V) P ( V/ P) T = - ( V/ T) P /( V/ P) T C P C V = -T( V/ T) [( V/ T) P /( V/ P) T ].

6 For an ideal gas Pv = nrt ( V/ T) P = nr/p. ( V/ P) T = -nrt/p 2 C P C V = nr. 9. For n moles of a Van der Waals gas, whose equation of state is given by 2 na P + V 2 ( V - nb) = nrt calculate the expression for the entropy change for the system from an initial state T 1, V 1 to the final state T 2, V 2. Ans: Use: du = tds PdV [i.e. u u(s,v)] = C V dt + [T( P/ T) V P] dv [u u(t,v)] Also, from ds = ( S/ T) V dt + S/ V) T dv and the firstline, ( S/ T) V = C V /T; ( S/ V) T = ( P/ T) V P/T + P/T = ( P/ T) V. Thus the equation for entropy change you need to use is ds = C v dt/t + ( P/ T) V dv. For a van der waals gas. (P + n 2 a/v 2 )(V nb) = nrt, P = nrt/v nb n 2 a/v 2 ( P/ T) v = nr/v nb ds = C v dt/t + nr/ V nb dv. S = T1V1 T2V2 ds = T1 T2 C V /dt + V1 V2 nr/v-nb dv = C V ln T 2 /T 1 nr ln V 2 nb/v 1 nb if C v is independent of temperature. 10. Without doing explicit calculations, predict the signs of P, V, T, Hand S for one mole of an ideal gas taken through each of the following four steps of a Carnot cycle. Assume C p and C v Constants. Step (a): Isothermal reversible expansion P 1 V 1 T 1 P 2 V 2 T 1 Step (b): Adiabatic reversible expansion P 2 V 2 T 1 P 3 V 3 T 2 Step (c): Isothermal reversible compression P 3 V 3 T 2 P 4 V 4 T 2 Step (d): Adiabatic reversible compression P V T P 1 V 1 T 1 Text book material.

7 11. On a single entropy versus temperature diagram, draw the curves for the following reversible processes for an ideal gas ( the initial state is the same foe all processes). (a) isothermal expansion, (b) adiabatic expansion and (c) isochoric process in which heat is added. Ans: S a Logarithmic. c b a) Isothermal expansion. T T- constant b) adiabatic expansion. q = 0. s = 0 c) Isochoric with heat added. S = C v ln T 2 /T 1. T 1 is an initial point. So, S(T 2 ) = S 0 (at T 1 ) + C v ln T 2 C v ln T Sketch diagrams for the reversible Carnot cycle of an ideal gas with constant C and C ; P Vs S; E Vs P; T Vs P; H Vs P; V Vs T; S Vs V; V Vs E; V Vs H; E Vs T; H Vs T, T Vs S, E Vs H and H Vs S. Ans: discuss three cases. The others are similar. a) P. vs. S II S III I IV P

8 I Isothermal expansion. II adiabatic exp. III Isothermal compression. IV adiabatic compression. b) c) I II S II IV S III I III IV P T Others can be constructed similarly. 13. One mole of water at 0 0 C is brought into contact with a large heat reservoir at C. When the water reaches C calculate the entropy change of water, the reservoir and the universe(which consists of water and the reservoir). If the water was brought into contact with a heat reservoir at 75 0 C first and then water at 75 0 C is brought into contact with the reservoir at C so that the final state of the water is C, calculate the net entropy change of water, the reservoir and the universe. Assume that the heat capacity of water to be constant at 1.0 cal g -1 K -1. Ans: For a body at temp T 1 in contact with a reservoir at T 2 with T 1 < T 2. S for the body (for a suitable reversible path by which T 1 T 2 ) = nc v ln T 2 /T 1. S for the reservoir: heat is supplied at T 2 to heat a body from T 1 to T 2 with a constant heat capacity C v.

9 s = -nc V (T 2 T 1 )/T 2. s universe = nc V ln T 2 /T 1 nc V (T 2 T 1 )/T 2. > 0 if T 2 > T 1. = 0 if T 2 = T 1. (obvious). Apply this for each of the cases mentioned in the problem. T 1 = 273 k, T 2 = 373 k. C V = 18.0 cal mol -1 k -1 S = 18[2.303 log373/ /373] = cal. Please understand the underlying assumption. Reservoir is so huge that the process of heat transfer is assumed to be reversible! Otherwise S heat/t. Process 2: S universe is calculated for two processes. T 1 = 273 k. T 2 = 348 k And T 1 = 348 k, T 2 = 373 k. S = 18 [2.303 log 348/273 75/348] + [2.303 log 373/348 25/373] = cal. What does this tell you? If you divide this further and further, (i.e. infinitely many such intermediate temperatures and attain equilibrium at each stage, what will be the value of S universe? [zero, in the end!]. 14. A mass m of liquid at temp T 1 is mixed with an equal mass of the same liquid at a temperature T 2. If the system is isolated show that the entropy change of the universe is ( T1+ T2)/2 2mC p ln TT 1 2 Where C p is the specific heat capacity of the liquid and is assumed to be constant over the temperature range. Ans: Heat capacity of the liquid is mc P. Let T 1 > T 2, T be an intermediate temp. (equilibrium). Then: mc P (T 1 T) = mc P (T T 2 ). Heat lost by the hot liquid. Heat gained by the hot liquid.

10 T 1 T = T T 2 or T = T 1 + T 2. It is also trivially true. Entropy change in the liquid. (hot) = m C P ln (T 1 + T 2 )/2/T 1 Entropy change in the cold liquid = mc P ln (T 1 + T 2 )/2/T 2. Total entropy change = 2m C P ln [(T 1 + T 2 )/2 / T 1 T 2 ] Always positive. So long as T 1 and T 2 are different. 15. Two moles of an ideal gas are expanded isothermally at 298 K from a volume V to a final volume of 2.5 V. Find the value of Sgas and Stotal for the following: a) Reversible expansion b) Irreversible expansion in which the heat absorbed is 400 J mol -1 less than the reversible expansion and c) Free expansion. Ans: Reversible expansion. s gas = nr ln V 2 /V 1 = 2 x ln 2.5 = JK -1. S surr = JK -1 as the heat transfer takes places reversibly from surroundings at the same temperature. S tot = 0. b) S gas = JK -1. Remember the initial and final states of the gas are exactly the same as in the previous case, therefore S gas = only still. S surr = -(nrt ln V 2 /V 1 400)/T! = JK -1. S tot = 1.34 JK -1 > 0. (irreversible path). c) free expansion: system does not draw any heat from the surroundings. S surr = 0. S tot = JK The molar entropy of an ideal gas at 298 K is 146 J K -1 Mol -1. Find the value of entropy of the gas at 500 K when its pressure is kept constant. Ans: Missing statement: (C P of the gas is a constant and is JK -1 mol -1.) use this to calculate. S 500 = S C P ln T 2 /T 1

11 = x x log 500/298. = JK -1 mol How much does the entropy of an ideal gas increase when one mole is heated from 298 K to 1000 K at a constant pressure and then expanded to five times the volume at the higher temperature? C = 20.9 J K -1 mol -1 Ans: S = C P ln T 2 /T 1 + R ln V 2 /V 1. = 2.9 x log 1000/ x x log 5 = = 38.7 JK -1 mol -1.