Thermodynamics (XI) Assignment(Solution)

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1 SYLLABUS CUM COM./XI/03 4 hermodynamics (XI) Assignment(Solution) Comprehension ype Questions aragraph for Question -5 For an ideal gas, an illustration of three different paths A, (B + C) and (D + E) from an initial state,, to a final state,, is shown in the given figure. ath A represents a reversible isothermal expansion from, to,. ath (B + C) represents a reversible adiabatic expansion (B) from, to 3,, followed by reversible heating the gas at constant volume (C) from 3,, to,,. ath (D + E) represents a reversible expansion at constant pressure (D) from,, to,, 3 followed by a reversible cooling at constant volume (E) from,, 3 to,,. Q. What is q rev, for path (A)? (a) ( ) (b) nr ln (c) nr ln (d) nr ln (d) According to the first law of thermodynamics U = q w In isothermal process, U = 0 q = w or w = nr ln Q. What is q rev, for path (B + C)? (a) nr ln (b) nr ln (c) zero (d) nr ln ath (B + C) occurs in adiabatic conditions q = 0 We know that q S = rev q rev = S q rev = nr ln Q3. What is S for path A? (a) nr ln (b) ( ) (c) ( ) (d) nr( ) (a) For path A qrev nr S = ln nr ln Q4. What is q rev, for path (D + E)? (a) ( ) (b) Cv () d (c) nr ln (d) 3 C () d (a) w = q = ( ) 3 S.C.O. 08 IInd floor, Sec-36 D Chandigarh, h.30085, Broaden Your Horizon []

2 SYLLABUS CUM COM./XI/03 4 Q5. What is S for path (D + E)? (a) ( ) (b) nr ln (d) For path (D + E) S = nrln (c) + ( ) (d) nr ln aragraph for Question 6-0 A sample of ideal gas undergoes isothermal expansion in a reversible manner from volume to volume. he initial pressure is and the final pressure is. he same sample is then allowed to undergo reversible expansion under adiabatic conditions from volume to. he initial pressure being same but final pressure is. Q6. he work of expansion in adiabatic process (w adi ) is related to work of expansion in isothermal process (w iso ) as (a) w adi = w iso (b) w adi < w iso (c) w adi = w iso (d) w adi > w iso (b) Work in reversible isothermal expansion is greater than work done in adiabatic expansion. Q7. Which of the following is correct? (a) = (b) / = / (c) / = / 3 (d) = (a) = (Boyle s law) Q8. Which graphic representation is correct (b) Q9. If 3 and are equal, then (a) (adi) = (iso) (b) (adia) < (iso) (c) Both (adi) = (iso) < (d) (adi) > (iso) (d) Cp Q0. Which relation is correct? C (a) = 3 (b) = 3 (c) = (d) (c) aragraph for Question -5 he state of a mole of an ideal gas changed from state A at pressure and volume follows four different processes and finally returns to initial state A reversibly as shown below in the graph. By interpreting the graph, answer the following questions. Q. Which is the kind of process followed from state A to state B? (a) Isochoric expansion (b) Isobaric expansion (c) Isothermal reversible expansion (d) Isothermal irreversible compression (b) Expansion from state A to state B occurs at constant pressure (isobaric expansion) = S.C.O. 08 IInd floor, Sec-36 D Chandigarh, h.30085, Broaden Your Horizon []

3 SYLLABUS CUM COM./XI/03 4 Q. In state D to state A, what kind of process is followed? (a) Isobaric expansion (b) Isobaric compression (c) Isochoric process (d) Isothermal compression (c) Expansion from state D to state A occurs at constant volume (isochoric process)= Q3. What would the be total work done by the gas? (a) (b) (c) 0 (d) Zero (a) Work done = - Or -( ) = - Q4. What would be the work done in state B to C? (a) (b) (c) (d) Zero (d) In conversion from B to C, volume does not change or = 0, therefore w = 0. Q5. What would be the heat absorbed by the system in this cyclic process? (a) - (b) Zero (c) (d) (d) q = - w q = aragraph for Question A sample consisting of mol of a mono-atomic perfect gas C R is taken through the cycle as shown. Q6. emperature at points (), (), and (3), respectively is (a) 73 K, 546K, 73 K (b) 546 K, 73 K, 73 K (c) 73 K, 73 K, 73 K (d) 546 K, 546 K, 73 K (a) Q7. H for the overall cycle is (a) J (b) J (c) J (d) Zero (d) In cyclic process, H = 0. Q8. U for the process ( ) is (a) 0.00 J (b) kj (c) J (d) J (b) In expansion from state to state. = atm =.44 L = L C = 3 R = = = 73 K U = nc = = J Q9. U for the process ( 3) is (a) 0.00 J (b) kj (c) kj (d) None of these (c) In conversion from state to state 3 = = = - 73 K C = 3 R U = nc = (-73) = J S.C.O. 08 IInd floor, Sec-36 D Chandigarh, h.30085, Broaden Your Horizon [3]

4 SYLLABUS CUM COM./XI/03 4 aragraph for Question 0-4 he pressure volume of varius thermodynamics processes is shown in graphs: Work is the mode of transference of energy. It has been observed that reversible work done by the system is the maximum obtainable work. w rev > w irr he works of isothermal and adiabatic processes are different from each other. w isothermal reversible =.303nRlog 0 =.303nR log 0 w adiabatic reversible = C v ( - ) Q0. If w, w, w 3, and w 4 are work done in isothermal, adiabatic, isobaric, and isochoric reversible processes, respectively then the correct sequence (for expansion) would be (a) w < w < w 3 < w 4 (b) w 3 = w = w = w 4 (c) w 3 < w < w 4 < w (d) w 3 > w > w > w 4 (d) w = Work done = Area under curve Q. A thermodynamic system goes in a cyclic process as represented in the following - diagram: he network done during the complete cycle is given by the area (a) Cycle ACBDA (b) AA B BDA (c) AA B B (d) Area not bonded by curve (a) Q. - plots for two gases during adiabatic processes are given in the given figure.: lot D and lot C should correspond to (a) He and O (b) He and Ar (c) O and He (d) O and N (c) Slopw of the adiabatic curve Slope of B > slope of A He ( =.66); O ( =.44) S.C.O. 08 IInd floor, Sec-36 D Chandigarh, h.30085, Broaden Your Horizon [4]

5 SYLLABUS CUM COM./XI/03 4 Q3. he q value and work done in isothermal reversible expansion of one mole of an ideal gas from initial pressure of bar to final pressure of 0. bar at constant temperature 73 K are: (a) 5. kj, -5. kj (b) +6. kj, 6. kj (c) 5. J, 5. J (d) -6. J, -6. J (a) w = nrlog = log = -5.7 kj 0. q = - w = kj for isothermal process Q4. Calculate work done when mole of an ideal gas is expanded reversibly from 30 L to 60 L at a constant temperature of 300 K (a) 8.78 kj (b) -.73 kj (c) 0.73 kj (d) kj (b) Matching ype Questions Q5. Match the following (a) Joule-homson coefficient (p) U 0 (b) Kirchhoff s equation (q) a b 0 R (c) Ideal gas (r) H C (d) Inversion temperature (s) ( H/ ) H C (a s) (b r) Kirchoff s equation relate the variation of enthalpy with temperature. U (c p) For ideal gas, 0 a (d q) Inversion temperature, i = Rb Q6. Match the following hermodynamic properties Relation (a) G (p) -R log K (b) H Θ (q) ln K R (c) S Θ (r) -nfe (d) G Θ (s) G - (a p, r) (b q) (c s) (d p, r) Multiple Correct Answer ype Questions Q7. Which is not correct relationship? dh du du (a) d d = (+ve) (b) d = 0 (for ideal gas) d nr (c) d = (for ideal gas) (d) All of these H U (a, c) R Q8. Which of the following are not correct at 98 K? (a) f G Θ element = 0 (b) f H Θ element = 0 (c) f S Θ element = 0 (d) f G Θ compound = 0 (b, c, d) S.C.O. 08 IInd floor, Sec-36 D Chandigarh, h.30085, Broaden Your Horizon [5]

6 SYLLABUS CUM COM./XI/03 4 S.C.O. 08 IInd floor, Sec-36 D Chandigarh, h.30085, Single Correct Answer ype Questions Q9. For the reaction; X O 4 (l) XO (g) U =. kcal, S = 0 cal K - at 300 K Hence G is (a).7 kcal (b) -.7 kcal (c) 9.3 kcal (d) -9.3 kcal (b) H = U + n g R = = 3.3 kcal G = H - S = (0.0) = -.7 kcal Q30. If G = H S and G = H + temperature, is given by: H G (a) (b) nf nf (c) On comparison: d( G) S = - d d( G) d S (c) nf then variation of EMF of a cell E, with (d) S nf d( nfe) de HereEis the S = - nf d d Emf of cell de S d nf Q3. mol of NH 3 gas at 7C is expanded under adiabatic condition to make volume 8 times ( =.33). Final temperature and work done, respectively, are (a) 50 K, 900 cal (b) 50 K, 400 cal (c) 50 K, 000 cal (d) 00 K, 800 cal (a) = = 50 K w = -C = -C ( - ) = - 3 (50-300) = 900 kcal Q3. he standard heat of combustion of Al is kj mol - at 5C. If Al reacts with O at 5C, which of the following release 50 kj of heat? (a) he reaction of 0.64 mol of Al (b) he formation of 0.64 mol of Al O 3 (c) he reaction of 0.3 mol of Al (d) he formation of 0.50 mol of Al O 3 (d) Al + 3 O Al O 3 7g H = kj mol - (a) 0.64 mol of Al = on combustion gave = 53 kj. Hence false. (b) Formation of 0.64 mol of Al O 3 gave = = 045 kj Hence, false. (c) 0.3 mol of Al on combustion gave = 6 kj. Hence false (d) Formation of 0.50 mol of Al O 3 give = 5.3 kj Hence, true. Q33. Inversion temperature is Rb a Rb a (a) (b) (c) (d) a Rb a Rb (b) Q34. For an ideal gas Joule-homson coefficient is: (a) ositive (b) Negative (c) Zero (d) Dependent on molecular weight Broaden Your Horizon [6]

7 SYLLABUS CUM COM./XI/03 4 (c) In an ideal gas, there is no force of attraction between molecules. Hence, no heat is lost in the expansion. Q35. H (g) + O (g) H O(l) BE (H H) = x ; BE (O = O) = x BE (O H) = x 3 Latent heat of vaporisation of water liquid into water vapour = x 4, then f H (heat of formation of liquid water) is x (a) x + x - x 3 + x 4 (b) x 3 x - - x 4 x (c) x + x - x 3 x 4 (d) x + - x 3 + x 4 (c) H = (BF) reactant (BE) products [But all the species must be in gaseous state. In product, [H O(l) H O(g)] H must be added. Hence, H (g) + O (g) H O(l) H = ( BE) H (BE = [(H) vap + (BE) O-H ] x = x + -[x 4 + x 3 ] H ) OO x = x + -x 4 x 3 Q36. If a certain mass of gas is made to undergo separately adiabatic and isothermal expansions to the same pressure, starting from the same initial conditions of temperature and pressure, then, as compared to that of isothermal expansion, in the case of adiabatic expansion, the final (a) olume and temperature will be higher. (b) olume and temperature will be lower. (c) emperature will be lower but the final volume will be higher. (d) olume will be lower but the final temperature will be higher. (b) Q37. he enthalpy of formation of hypothetical MgCl is -5 kj mol - and for MgCl is -64 kj mol -. What is the enthalpy of the disproportionation of MgCl. (a) 39 kj mol - (b) -39 kj mol - (c) -767 kj mol - (d) -57 kj mol - (b) (i) Mg(s) + Cl (g) MgCl (s) H = -64 (ii) Mg(s) + Cl (g) MgCl(s) H = - 5 (iii) MgCl MgCl + Mg H =? H = H - H = -64 (-5) = - 39 kj mol - Q38. he expression sub H Θ = fus H Θ + vap H Θ is true at all (a) emperature (b) ressures (c) emperatures and pressures (d) emperature and atm pressure conditions (d) Q39. he word standard in standard molar enthalpy change implies (a) emperature 98 K (b) ressure atm (c) emperature 98 K and pressure atm (d) All temperatures and all pressures (c) Standard state implies to atm pressure and 98 K temperature. S.C.O. 08 IInd floor, Sec-36 D Chandigarh, h.30085, Broaden Your Horizon [7]

8 SYLLABUS CUM COM./XI/03 4 Q40. For hypothetical reversible reaction /A (g) + 3/B (g) AB 3 (g); H = -0 kj if standard entropies of A, B, and AB 3 are 60, 40, and 50 J K - mol -, respectively. he above reaction will be equilibrium at (a) 400 K (b) 500 K (c) 50 K (d) 00 K (b) At equilibrium, S total = 0 S Θ - S R Θ = 0 Q4. Standard molar enthalpy of formation of CO is equal to (a) Zero (b) he standard molar enthalpy of combustion of gaseous carbon. (c) he sum of standard molar enthalpies of formation of CO and O (d) he standard molar enthalpy of combustion of carbon (graphite). (d) Standard molar enthalpy of formation CO and the standard molar enthalpy of combustion of carbon (graphite) refer to the same chemical equation: C(graphite) + O (g) CO (g) Q4. In thermodynamics, a process is called reversible when (a) he surroundings and system change into each other. (b) here is no boundary between the system and surroundings (c) he surroundings are always in equilibrium with the system. (d) he system changes into the surroundings spontaneously. (c) Q43. he products of combustion of an aliphatic thiol (RSH) at 98 K are (a) CO (g), H O(g), and SO (g) (c) CO (l), H O(l), and SO (g) (b) CO (g), H O(l), and SO (g) (d) CO (g), H O(l), and SO (l) (b) C H 5 SH + 9 O (g) CO (g) + 3H O(l) + SO (g) S.C.O. 08 IInd floor, Sec-36 D Chandigarh, h.30085, Single Integer ype Question Q44. Bond dissociation energy of XY, X, and Y (all diatomic molecules) are in the ratio of ::0.5 and H f of XY is -00 kj mol -. he bond dissociation energy of X is 00x. Find the value of x. (4) X Y XY; H = - 00 kj Let the bond dissociation energy of X, Y, and XY be a : a : a (the given ratio) kj mol -, respectively. a a a = a = x = 400 x = 4 Q45. he polymerisation of propene to linear polypropene is represented by the reaction n CH 3 CH CH CH CH 3 CH n where n has large integral value, the average enthalpies of bond dissociation for (C = C) and (C C) at 98 K are +590 and +33 kj mol -, respectively. he enthalpy of polymerisation is -360 kj mol -. Find the value of n. Broaden Your Horizon [8]

9 SYLLABUS CUM COM./XI/03 4 sol. (5) Energy released = Energy due to formation of two single bonds = 33 = 66 kj mol - of propene H polymerisation/mol = = - 7 kj mol - H polymerisation = - 7 n = N = 5 Q46. he lattice energy of solid KCl is 8 kcal mol - and the enthalpy of solution of KCl in H O is.0 kcal mol -. If the hydration enthalpies of K + and Cl - ions are in the ratio of : then the enthalpy of hydration of K + is -0x K cal mol -. Find the value of x. (6) KCl(s) K + (g) + Cl - (g), H = 8 kcal mol - KCl(s) + aq K + (aq) + Cl - (aq), H =.0 kcal mol - Let the enthalpy of hydration of K + is Cl kcal mol - K + (g) + aq K + (aq), H 3 = a Cl - (g) + aq Cl - (aq), H 4 = a H 3 = - H + H - H 4 ª = a 3ª = - 80, a = -60 hyd H Θ of K + = ª = - 60 = - 0-0x = - 0 x = 6 Q47. A heated iron block at 7C loses 300 J of heat to the surroundings which are at a temperature of 7. his process is 0.05x J K -. Find the value of x. qsys 300 (5) sys S = 73 7 sys = JK surr S = q sys surr = 300 JK 300 total S or universe S = sys S + surr S = JK x = 0.5 x = 5 S.C.O. 08 IInd floor, Sec-36 D Chandigarh, h.30085, Broaden Your Horizon [9]

I. Multiple Choice Questions (Type-I)

I. Multiple Choice Questions (Type-I) 1. Thermodynamics is not concerned about. (i) energy changes involved in a chemical reaction. the extent to which a chemical reaction proceeds. the rate at which a