Thermodynamics Answers to Tutorial # 1

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1 Thermodynamics Answers to Tutorial # 1 1. (I) Work done in free expansion is Zero as P ex = 0 (II) Irreversible expansion against constant external pressure w = P ex (V 2 V 1 ) V 2 = nrt P 2 V 1 = nrt P 1 w = P ex nrt 1 P 2 1 P 1 w = 4atm (2 mol)(8.314 J mol 1 K 1 )(400 K) atm w = 4x2x 8.314x w = J = ( )3.332 kj Note: By the surrounding on the gas according to the definitions is (+) ve (III) See derivation w = 2.303nRT log 10 P 2 P 1 w = 2.303x 4 mol x(8.314j mol 1 K 1 ) log w = 7.0 kj 2. This conversion occurs under a constant external pressure of k Pa (i.e x 10 5 Nm -2 ) w = P ex (V 2 V 1 ) where, P ex = x 10 5 Nm -2 V 1 = V liquid water = 18 x 10-6 m 3 for 1 mole V 2 = V water vapor

2 Since at standard temperature and pressure (STP i.e. at x 10 5 Nm -2 and 273 K) I mole of any ideal gas occupies 22.4 x 10-3 m 3. V 2 = V water vapor = 22.4 x x 10-3 m 3 for one mole of water vapor at 373 K = 30.6 x 10-3 m 3 for one mole Hence, w = x 10 5 Nm x x 10 6 m 3 w = kj 3. For the decomposition of 1 mole of TNT, n gas = = 7 mol U = (-) 3.3 kj H = U + RT ( n gas ) = (-) 3.3 kj + (8.314x 10-3 kj mol -1 ) x (300 K) x (7 mol) = 14.0 kj 4. The equation relevant to the Haber process for the formation of gaseous NH 3 is 3/2 H 2 (g) + 1/2 N 2 (g) NH 3 (g) n gas = = - 1 mol Applying; H = U + RT ( n) U = 46.1 kj x 10 3 kj K 1 x 300 K x ( 1 mol) = (-) 43.6 KJ 5. Applying the first law of thermodynamics U = q + w And using the sign convention Q = (+) kj W = (-) kj U = (+) (-) = kj 6.

3 I. Process should be isochoric II. All other type of work is also not possible III. System is homogenous (i.e. No phase change is possible during the change in temperature.) IV. C, is independent of temperature. 7. Using, U = n C, m T =(10 mol) (30 J K -1 mol -1 ) ( ) K = + 30 kj 8. I. Process is Isobaric II. Other type of work other than P V work (i.e.-dw ) is not possible. III. System is homogeneous (i. e. No phase change is possible) IV. The isobaric thermal capacity does not vary with temperature. 9. When water is heated, from 353 K to 393 K, a phase change (liquid to vapor) occurs. Therefore we cannot use a single thermal capacity to calculate H. we have to break the process to 3 steps and calculate H associated with each step. I. Heating water from 353 K to 373 K ( i.e. H ) II. Vaporization of water at 373 K ( H vaporation ) III. Heating of water vapor from 373 K to 393 K. ( H ) H = H ( ) + H evaporate + H ( ) = C p,m dt + H evap + C p,m dt = dt dt 373

4 = [75.0 x ( )] ( ) = 49.5 kj 10. Since P ex = Constant = 3 atm = W = (-) (V B V A ) = n C v,m (T B ) = ΔU This is an adiabatic process. Therefore calculations of final temperature is required. (-) (V B V A ) = n C v,m (T B ) Also, V B = nr T B, V A = nr P A Substituting for V B andv A we have, - nr T B nr P A = n C v,m (T B ) -nr T B P A. = n C v,m (T B ) Substituting for = 300 K, = 3 atm, P A = 6 atm C v,m =3R/2 ( monoatomic gas) -nr T B K = 6 5 T B = 1200 K T B = 240 K T = T B = -60 K 3 nr 2 (T B 300 K) To calculate w and U we make use of the equation, W = U = nc v,m (T B ) = (2 mol) (3/2 x J K -1 mol -1 )( )K = (-) 1.5 k J For an ideal gas under all conditions H = n C p,m T Since C p,m = 5R/2 ( monoatomic gas) T = - 60 K

5 H = (2 mol) (5/2 x J K -1 mol -1 )(-60 K ) = (-) 2.5 kj Required solutions are q = 0, U = w = (-) 1.5 kj ; T = - 60 K, H =(-) 2.5 kj 11. In both cases the final pressure would be half the initial pressure (i.e. 3 atm). In the reversible case the final temperature was found to be 240 K. The question we have to answer is whether the final temperature in the case of the reversible expansion will be different from 240 K Note that maximum work is always obtainable in the case of a reversible process. It follows therefore that the work done in the reversible adiabatic expansion should be greater. Since w = nc v,m dt, there should be a proportionate increase in the numerical value of T in the case of the reversible expansion. The final temperature in the case of the reversible adiabatic expansion therefore has to below than in the case of the irreversible expansion. It follows that the final state in the two expansions will therefore be different. 12. They can be applied only for reversible and adiabatic expansions of IDEAL gases together with the ideal gas equation. There are 4 equations, that enables us to calculate the final temperature, final pressure and final volume in any such gas expansions. 13. P 1 γ T γ = Constant P A 1 γ = T B γ P A = 6 atm = 300 K = 6 atm T B =? γ = C p,m C v,m = 5/3 1-γ = (-) 2/3

6 = T B T B = 227 K T = [ ] K = -73 K For an ideal gas under all conditions, W = U = nc v,m T = (2 mol) (3/2)(8.314 J K -1 mol -1 )(-73 K) = -1.8 kj H = nc p,m T = (2 mol)(5/2) )(8.314 J K -1 mol -1 )(-73 K) = -3.0 kj 14. Calculate the heat change that occurs for the process carried out reversibly from the given initial state to given final state. Then calculate entropy change S using the following relationship S = B A dq rev T Initial state = state A Final state = state B 15. The definition for entropy change ds is given below ds = dq rev /T Heat change dq rev is an extensive thermodynamic property since it depends on the amount of matter present. The thermodynamic property T is an intensive property. The ratio of the two, therefore is an extensive property. Entropy therefore is an extensive property. SI unit is J K The transformation of liquid water to water vapor at its boiling point is a typical example for a univariant phase change (It occurs reversibly under constant pressure.)

7 Under reversible condition the process of evaporation of liquid water at its boiling point is in equilibrium. G = Applying ; G = H-TS At constant temperature G = H T S G = (34.7 kj mol -1 ) (300 K x J K -1 mol -1 ) = (-) 5.1 kj mol -1 Since G < 0 it follows that ammonium chloride will dissolve in water spontaneously at constant temperature and pressure at 300 K 18. (I) T 2 =( ) K = 1000 K T 1 = ( ) K = 300 K S P =n C p,m ln T 2 T 1 = (5 mol) x (29.1 J K -1 mol -1 ) (2.303 log = J K -1 (II) S T = nrt T ln V 2 V 1 = (5 mol) (8.314 J K -1 mol -1 ) (2.303log 3V 1 V 1 = J K S = nc p,m ln T B + nr ln P A C p,m = 5R/2 ( ideal monoatomic gas) = 273 K P A = k Nm -2 T B = 1000 K P A = 1013 k Nm -2 S = nc p,m ln T B + nr ln P A = (1 mol) (5/2)( J K -1 mol -1 )ln (1 mol)( J 273 K-1 mol -1 )ln = (-19.15) J K -1 = 7.83 J K See the answer for question 11. q,w, U, H and T have already been calculated. Final temperature was found to be 240 K for that process.

8 In order to calculate S for the above process it is necessary to find a reversible path for the change. Reason for this is that ENTROPY is a stable function. Its values do not depend on the path. It will be same for all the other processes as long as the same initial and same final conditions are present. We already have deduced a general equation for entropy changes of ideal gases.one form of that is given below. S = nc p,m ln T B + nr ln V B V A It is also possible to prove that S = nc p,m ln T B + nr ln P A The problem gives P A,, and. We need to calculate T B (Use the value 240 K calculated in the previous problem) n = 2 molc p,m = 5R/2 = 300 K T B = 240 K P A = 6 atm P A = 3 atm S = (2 mol) (5/2)( J K -1 mol -1 )ln 240 = [ (-9.276)] J K -1 = 2.25 J K (2 mol)( J K-1 mol -1 )ln For 1 mole of liquid water the enthalpy of vaporization = kj mol -1 = kj mol K S trans = H trans T = 0.11 kj K -1 mol -1 = 110 J K -1 mol -1

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