UNI # 6 HERMODYNMIS EXERISE #. m Zn.S Zn.( f i + m H O.S H O.( f i (6.8 gm (.4 J/g ( f + 8 gm (4. J/g ( f [(6.8 (.4 + 8(4.] f (6.8 (.4 ( + (8 (4. ( (6.8(.4( (8(4.( f 97. (6.8(.4 (8(4.. U q + w heat absorb (q 4 joule w 7 joule since Work done by the system. ( U q + w 4 7 joule. Decrease in internal energy U ( work done by the system w U w U w q he process is adiabatic. ( 4. t constant volume w U q first law. he energy due to external field is not included in i nternal energy like gravitat ional field, ear th's magnetic field etc. ( 6. Heat and work are path dependent, or indefinite quantity. ( 7. For monoatomic ideal gas total degree of freedom ( hree translational mode of motion ( R R r + R R 8. r monoatomic ideal gas R t constant pressure q H n q (. mole.(8.4 J / K mole (6 9. Reversible process-involve infinitesimally small driving force. Hence system and surrounding remains in equilibrium. (. temperature at 'a' R at (a at (c R...(i ( (4 8 R U n ( f i R(8 R U. R. Work done in adiabataic process ( U w n w ((.(. he case of irreversible adiabatic process. w ( f i n ( ( f i n R (f i (atm(l L n (R (L atm.8 (L atm k mole. B > and B > H n ( B > w ext ( B < 6. Heat of reaction at const. pressure r H Heat of reaction at const. pressure r U r H r U + n g R n g r H r U ( R B >
8. Fusion and vaporisation are example of isothermal processes - ( Q f < b Q f R b 9. H O(s q H O(l H O(l 6 q H fusion (kj/mole mole 8 q 6.. kj q ms (6 gm (4.8 J/K gm (6((4.8 q NE q + q. kj. When ice liquid : the process is reversible fusion. he fusion is isothermal process. ( : dq dq d. U q + w constant since H q p.6 U + atm ( litres R U.6 (L atm (. J/L atm U. For reversible adiabatic process ( q rev S system. S n ln ln 7 S ln R ln 98 H 4. Using S transition H S surr trans S trans 4.7 68 S surr 4.7 7 S total 4.7 68 transition transition X q, temperature of ice bath 7K + 4.7 7. Initial pressure R f R (.8 atm final pressure atm final volume ; f f i i i i ( R 4.6 L f S ln S R ln 4.6 6. for spontaneous reactions S total > ( S total S system + S surr X S total S system 98 S total > S sys + X 98 X S syst > 98 Hence S syst can be negative but numerically smaller than X 98. S syst X 98 7. For dissociation reactions ( H > and n g > r S > 8. G H S G [( 8] G + 8 spontaneous but less than certain temperature. 9. Formation of Fe O 4 Fe(s + O (g Fe O 4 (s ; G? G ( 9 kcal + ( 77 kcal 4. kcal/mole. H S H ( 66. S 97.6 + 7.8.74 4.49 bove this temperature the process becomes spontaneous. (
HERMODYNMIS EXERISE #. Intensive property ( extensive property ( extensive property ph concentration mole/volume though ph is a dimensionless number and intensive property (ph EMF energy Intensive property ch arg e Boiling point( b temperature intensive property entropy (s q extensive intensive extensive property. t constant, the molecule with maximum atoms have greatest internal energy.. q U W ( + 4 K 4. Irreverssible adiabatic process ( W ext atm ext atm K (R. R( W ( atm atm atm and W ( R( R( (7 R R. B B isochoric heating ( B isothermal expansion ( isobaric cooling B ( B isochoric heating B 6. B emperature at ( D D 4 ( /. B now D : U H U n remember for ideal gas H n 7. (,, Isochoric heating H B(,, 7 4 7 W (4 4 K H irr. adiabatic process (q R ( R W W B + W B + W D ( B B ln ( D B W ( ln 4 ( 4 W ln and q W ( U q ln (,, H overall H + H H U + ( H ( + ( H U + ( H ( + ( H H overall ( + ( + ( +
8. 7 98 7 7 7 9. W q by source. H q p W by (n (q source ( kcal.8 kcal since H is state function H will remain same from both path-isobaric and non-isobaric q p isobaric path non-isobaric path q But q q p only when path was isobaric. B H q p only for isobaric path.. Greater the n g greater the value of S.. S n ln for isochoric change. S 7 R ln 47 7 S R ln 47. he net heat absorbed by hot and cold body is equal to zero. q H + q Let is the total heat capacity of hot and cold body. ( f + ( f H f H H Entropy change S otal S hot body + S cold body f S hot body.ln H f S cold body.ln f f S otal ln ln H f ln H. (f S otal ln 4 H. 4. G H S y + mx H m S from intercept > H > m < S < S >. Melting of H O(s at and atm is a reversibly procers S otal aporisation of H O(l at 7 K is a reversible process S otal Below H O(s H O(l is non spontaneous not feasible. H O(l H O(s (freezing is feasible above H O(s H O(l feasibly G ive for melting process S otal increases 9. G H + E G + 6 cal. H O(l H O(g 7 K 7 K atm S H vap atm G H f H i
HERMODYNMIS EXERISE # OMREHENSION #. U Here U Kinetic energy of ideal gas ( U n. U q + w J q 8J q 8 J OMREHENSION # n...(i sub. ( in (...(ii. r S S (H OH, g [S (O, g + S (H, g]. R.R w d ( (. ( / w ( ( R d ( 4 98 9 6 J/K-mole. r H f H (H OH, g f H (O, g (4 87 kj/mol. rs rs ln,m (H OH,m (O,m (H 44 (9.4 + 8.8 4 J/K-mol work done by the gas w ( (. For diatomic gas with no vibrational degree of freedom R + R rs ( 6 ( 4 ln rs. J/K-mol 4. rh rh r ( / R U w ( ( 7 J ( 7 w ( (J 8J ( ( rh ( 87 4( rh 87.86 kj/mol. r G rh S 8786 (. 8.9 kj/mol
HERMODYNMIS.(i H O (g H O( (g > ( W ext ( ( (g W +ve (ii H O (s H O(g (g > (s W ve (iii H O ( H O(s (s > ( W ext ( (s ( ve (iv H (g + N (g NH (g W ext ( ext (n R n R n g R W +e (v ao (s ao (s + O (g W ext > ve. E 6 J w J q w 6 4 J. H glucose 88 kj mol (a Energy need to climb m Mgh 6. q 87 Joule Now useful energy from mole of glucose 88 4 7 kj 87 No. of mole of glucose required 7.67 mole grams of glucose 8.67.487 gm (b Energy need tp climb m will be time. wt. should be time.487 kg 4. q p q v + n g R 4.66 q v + 8.4 7 q v (8. 6. kj q v 7.8 kj. H kcal H E + EXERISE # 4[] 4.8 E +. E (48.9 Joule E.99 kcal 449.6 cal 4.8 6. NH 4 N(s + O (g N (g + O (g + H O( H 98 E + n g R 74.7 + 8.4 98 74.7 +.9 H 98 74.46 7. H 44 cal H E + ( 44E+. (.8.96 E 44..6 E 44.68 8. W extd (. (. W. J 9. Zn(s + H + (aq Zn + (aq + H (g H 6. kj Q E W 6. kj E +. ( 6 E 9. kj W ext (. 6 W.. W irr 8.4 4 W irr 9. kj W rev. og W rev. 8.4 log4 W rev 7.9 kj and q E W at E H W rev q 7.9 kj
. n 7 7 K diabatic condition Q EWn ( W W. kj/mole. rocess reversibly adiabatic 98. K 48.44 K v K. K 98.. 48.44. log. log. ( log. log.6 Now n ( R ( m ( m 8.4 m.6.6. W. log 4.. 8.4 98 W.988 kj 44.8L.4L 7K B 46K able- log State.4L State - State - atm atm.4 44.8 atm atm 46 atm 7 7 46K 46K State -.4 44.8 atm Step Name of process q w E H Isochoric R(7 R(7 R(7 B Isotherm 46R ln 46R ln Isotherm R(7 R(7 R(7 R(7 overall State - (Isochoric W q E R(7 H n p d H R(7 State - B (Isothermal E H H E + is Q w +. 8.4 log( 46 Q 46 R ln W 46 R ln E n W ( State - (Isobaric q E w E + ( or H E + R( 7. H (g + O (g H O( p p H O( p H (g reaction p O (g 7. 8.8 9.6 p reaction.9 kj H 7 H 98 + np ( 8.76 +.9 7 kj H 7 84. kj 6..4 + 48. JK mol H n n p d (.4 48. d.4 98 + H.64 kj/mole 48. 98 98 W ( ( W 477 E.64.477 E.87 kj
7. S 8.8 J/K H. kj 8. S H S n pd. (. 8.8 HERMODYNMIS 9 K +.6 4. 7 d.. log +.6 4. (6 7 S.68 kj 9. Br ( + l (g Brl(g, H 9. kj S Br. S l (g. S Brl(g 9.7 J mol K S R 9.7. 4.4 r G H S 9 98 4.4 7.8 J. l 4 ( l 4 (g atm 98 K S 94.98 JK r H H p H r ( 6.7 + 9. kj.6 kj/mol r G H R S r G.6 98 94.98 r G 4.96 kj/mol. Step- (- Ice ( K Ice (7 K ( K (7 K S m p ln.9 ln 7 S 6. J/ Step- (- Ice (7 K Water (7 K H S f.4.44 J/ 7 7 Step- (- Water (7 K Water (7 K S 4.8 ln 7 7 Step-4 (-4 4.6 J/ Water (7 K Steam (7 K (7 K (7 K S 4 H v 7.6 7 68.98 J/. H O (g + O(s H (g + O (g (i r H 98 r H (g + r H O (g r H O (s r H O(g 94. + 6.4 + 7.8 9.8 k cal/mol (ii r G 94.4 + +.79 + 4.64 (iii (iv (v r G 6.8 r G r H r S 6.8 9.8 98 r S 9.8 6.8 rs 98 r S. cal/ mole but at constant H E + n g r H r E 98 S 98 [H O(g] 9.8 k cal/mol r S s H O(g + S O r S H r S O. r S H O(g + 47. +. +. r S H (g 4. cal/ K mole O Step- (- EXERISE # 4[B] Steam (7 K Steam (4 K (7 K (4 K S.9 ln 4 46.6 J/ 7 S 98.4 J/., atm, K, atm (a constant ( constant ( / constant ( / ( / / 9.8 w U n v d 8.4 9.8 w 9.7 J 6.L
(b U w. 8.4 ( R R. (. 4 K 7.4L w U. 8.4 (4 w 496. J. L, 7/, 67 K,. Ma atm,.7 Ma 7 atm n R. (i U H (ii q w ln. 8.4 67 ln 7 q 7. kj w 7. kj 7( ( (. /7 48.9 L 47.46 K q, w U.. 8.4(47.46 67 w U.4 kj H.. 8.4 (47.46 67 4.7kJ (iii q w U H (iv q, U w.. 8.4 ( 67...8 7 67.79 96.4 6.9 K w U.. 8.4 ( 6 7. kj H.. 8.4 ( 6 9.9 kj (v U H 4.(i..8 67 69 L w ( 49 98. L-atm w 98.. 994.9 J w 9.94 kj q w 9.94 kj he entropy change of the system S sys. will be same in all the three process as it is state function. (S S sys. ln 8.4 ln 9.4 J/K For reversible process ( S S surr. S sys. 9.4 J/K (ii (iii S sys. 9.4 J/K q S surr. irrev. q rev. 86.6 98 S sys. +.87 S S sys. + ( S sys. +.87.87 J/K For free expansion system doesn't absorb any heat so q ( q S sys. S S sys..87 J/K.(i S sys., S surr., S (ii U w, n v ( ((.. 8.4 ( 47 47...8 47. 9.49 K 94.6 S sys. n p ln R ln 9.49.. 8.4 ln 8.4 ln 47 S sys..97 J/K since no heat is transfered ( q S surr. S S sys..97 J/K (iii In free expansion ( q w U is constant. ( S sys. ln. 8.4 ln.8 JK S surr. S S sys.8 J/K 6. atm, L atm,.99l Let a + b On finding a, b so ( w d ( d w ( + ( w (. + (.99. L-atm w J U w J H U + ( + (.99 99J H 99. kj
7.(i S sys. n v ln R ln R ln S S surr. S sys. (Reversible process ( R ln 8.68 qh.( q H 4.4 kj v p R 4.8 + 8.68 (ii S sys. R ln w q U R (9 q R (9 S surr. R (.9 U n v d 9.9 kj (iii w q U 9.9 kj H n p d 4.4 kj. t 98 K, S R ln R (.9 R (.4 8. G H S U + S dg du + d + d ds Sd w, d, d dq ds so dg ds + d ds Sd dg d Sd G d ( 4 4 S d d 4.6 (4/ 8. L-atm 8 J Sd (. d( +.( 9. n Sd ( +. (4 G 8 J.8 49.6 L 49.6 9. L & so t K G 6 kj/mole H 77 kj/mole G H S S kj/mole G 77 67 kj/mole dditional non- work (- G G 4 kj/mole. r p. 7. 4.7 J/K mole r S d d( r S r p H 469 8.9 J/K mole r S 7 r S r p ln 7 r S 7 8.9 4.7 ln J/K mole (i w (9. 49.6 4. L-atm w 4.. 48.6 J 4. k q H n d p 8.68.( d ( r H r p d r H 7 r H 4.7 ( r H 7 479. J/mole r G 479. ( 94. J.9 kj/mole