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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 3551. |
Calculate the free eneryg change at 298 K for the reaction , `Br_(2)(l)+Cl_(2)(g) to BrCl(g)`. For the reaction `DeltaH^(@)=29.3 kJ`& |
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Answer» Correct Answer - `-1721.8 J` `DeltaS= 2 xx 239.7 - [152.3 +223]` `= 104.1 J"mol"^(-1)K^(-1)` `DeltaG = 29300- [298 xx104.1]` `=- 1721. 8J` |
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| 3552. |
Write the relation between standard free energy change and equilibrium constant Kp for a reversible reaction. |
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Answer» Relationship between standard free energy change ΔG° and the equilibrium constant ΔG° = -2.303 RT log Kp, where R is the gas constant = 8.314 JK-1. T is the temperature in Kelvin. |
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| 3553. |
Mention different types of process. |
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Answer» 1. Isothermal process 2. Adiabatic process 3. Isoehoric process 4. Isobasic process 5. Reversible process 6. Irriversible process 7. Cyclic process |
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| 3554. |
Calculate the free energy change on dissolving one mole (58.5 g) of sodium chloride at 25°C. Lattice energy = 777.8 kJ mol-1, hydration energy = -774.1 kJ mol-1 of NaCl, ΔS at 25°C = 0.043 kJ mol-1 K-1. |
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Answer» ΔHsolution = ΔHlatticet ΔHhydration = 777.8 kJ mol-1 + (-774.1 kJ mol-1) = 777.8 kJ mol-1 - 774.1 kJ mol-1 = 3.7 kJ mol-1 ΔG = ΔH - TΔS = 3.7 kJ mol-1 = (273 + 25) K x 0.043 kJ mol-1 K-1 = (3.7 kJ mol-1) - (298 K) x (0.043 kJ K-1 mol-1) = 3.7 kJ mol-1 - 12.814 kJ mol-1 = -9.114 kJ mol-1 |
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| 3555. |
What is exothermic reaction? Give example. |
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Answer» The reaction in which heat is evolved is called exothermic reaction. Example : C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ |
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| 3556. |
What is endothermic reaction? Give example. |
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Answer» The reaction in which heat is absorbed is called endothermic reaction. Example : N2(g) + O2(g) → 2NO(g) ΔH = +180.8 kJ |
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| 3557. |
A process must be spontaneous (feasible) if:A. Entropy of system increasesB. Energy of system decreasesC. Gibbs free energy decreasesD. Entropy of univese increases |
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Answer» Correct Answer - D |
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| 3558. |
Assertion : An adiabatic process is an isoentropic process. Reason : Change in entropy is zero in case of adiabatic processA. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of t he assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 3559. |
If enthalpy of neutralisation of a weak acid with strong base is -10.87 kacl mole then calculate enthalpy of ionisation of weak acid. [Given : `H^(+)(aq)+OH^(-)(aq)to H_(2)O(l)" " Delta H= -13.7 kcal]` |
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Answer» Correct Answer - 3 |
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| 3560. |
Calculate `DeltaH^(@)` for the reaction : `TiCl_(4)(g)+2H_(2)O(l)to TiO_(2)(s)+4Hcl(g)` `{:(DeltaH_(f)^(@),(KJ.mol,^(-1))),(TiCl_(4)(g),-763),(H_(2)O(l),-286),(TiO_(2)(s),-945),(HCl(g),-92):}`A. `-264KJ`B. `12KJ`C. `22KJ`D. `298KJ` |
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Answer» Correct Answer - c |
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| 3561. |
Sodium carbonate `(Na_(2)CO_(3)0`can be obtained by heating sodium hydrogen carbonate,`NaHCO_(3)` `2NaHCO_(3)(s) rarr Na_(2)CO_(3)(s) +H_(2)O(g) + CO_(2)(g)` Calculate the temperature above which `NaHCO_(3)` decomposes to form products at 1 bar . Given `Delta_(f)H^(@) ( kJ mol^(-1)) : NaHCO_(3)(s) = =- 947.7 , Na_(2)CO_(3)(s) = - 1130 .9`, `H_(2)O(g) = - 241.8 , CO_(2)(g) = - 393.5` `S^(@) (JK^(-1) mol^(-1)) : NaHCO_(3)(s) = 102.1 , Na_(2)CO_(3) (s)=136.0`, `H_(2)O(g) = 188.8, CO_(2)(g) 213.7` |
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Answer» `Delta_(r)H^(@) = ( -1130.9) + ( -241.8) + ( -393.5)- 2 xx ( -947.7 )= 129.2 kJ mol^(-1)` `Delta_(r)S^(@) = ( 136.0 +188.8 +213.7) - 2( 102.1) = 334.3 JK^(-1) mol^(-1)` `Delta_(r)G^(@)= Delta_(r)H^(@)-TDelta_(r)S^(@)`, At equilibrium , `DeltaG^(@)= 0, T - ( Delta_(r)H^(@))/( Delta_(r)S^(@))= ( 129200)/( 334.3) = 386.5K` Thus, above 386.5 K , `Delta_(r)G^(@)` will be negative and the reaction will be spontaneous. |
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| 3562. |
Water can be lifted into the water tank at the top of the house with the help of a pump . Then why is not considered to be spontaneous ? |
| Answer» A spontaneous process should continue taking place by itself after initiation . Bt this is not so in the given case because water will go up so long as the pump is working. | |
| 3563. |
Calculate the resonance energy of isoprene `(C_(5)H_(8))` from the data given. The standard heat of sublimation of graphite is `718 K kJ"mole"^(-1)` and heat of formation `C_(5)H_(8)(g)` is `79 kJ` mole. (Give your answer in kcal `"mole"^(-1)` , approximate integer.) |
| Answer» Correct Answer - 5 | |
| 3564. |
The enthyalpy change for the process `C_("(graphite)") rarr C_((g)), DeltaH= + x kJ` represents enthalpy ofA. FusionB. SublimationC. CombustionD. Vapourisation |
| Answer» Correct Answer - B | |
| 3565. |
Given that `C_("(graphite)") rarr C_((g)), DeltaH=+716.7 kJ`. `C_("(diamond)") rarr C(g), DeltaH=714.8 kJ`. The `DeltaH` for the following reaction is `C_("(graphite)") rarr C_("(diamond)")`A. 1.9 kJB. `-1.9 kJ`C. ZeroD. 714.8 kJ |
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Answer» Correct Answer - A `DeltaH=H_("(Diamond)")-H_("(Graphite)")` |
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| 3566. |
Statement-1: At low temperatures, DH is the dominant factor for spontaneity of a reaction. Statement-2: At low temperatures, the opposing factor TDS remains very small.A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 3567. |
A reaction has an equillibrium constant of 0.98 at 300K and 1.2 at 400K. Select the incorrect option:A. The reaction in endothermic.B. At standard conditions, reaction will be non spontaneous at 300K and spontaneous at 400K.C. Enthaply change of reaction, `DeltaH_(R)^(@)=1200RxxIn ((1.2)/(0.98))`D. `DeltaS_(R)^(@)=3R In (1.2)/(0.98)` |
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Answer» Correct Answer - D |
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| 3568. |
For the transition `C_("(diamond)") rarr C_("(graphite)"), DeltaH=-1.5 kJ`. It follows thatA. Graphite is stable than diamondB. Diamond is stable than graphiteC. Graphite is endothermic substanceD. Diamond is exothermic substance |
| Answer» Correct Answer - A | |
| 3569. |
A reaction has an equillibrium constant of 0.98 at 300K and 1.2 at 400K. Select the incorrect option:A. The reaction in endothermic.B. At standard conditions, reaction will be non spontaneous at 300K and spontaneous at 400K.C. Enthaply change of reaction, `delta H_(R)^(@)=1200RxxIn ((1.2)/(0.98))`D. `deltaS_(R)^(@)=3RIn (1.2)/(0.98)` [Assuming `deltaH^(@) and deltaS^(@)` to be temperature independent] |
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Answer» Correct Answer - B |
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| 3570. |
The magnitude of ethalphy changes for reversible adiabatic expansion of a gas from volume `V_(1)` to `V_(2)`(in `L)`is `DeltaH_(1)` and for irreversible adiabatic expansion for the same expansion is `DeltaH_(2)`. Then when `DeltaU_(1)` and `DeltaU_(2)` are the changes in mahnitudes for the internal energy of gas in the two expansions.A. `DeltaH_(1) gt DeltaH_(2)`B. `DeltaH_(1) lt DeltaH_(2)`C. `DeltaH_(1) = DeltaH_(2)`D. `DeltaH_(1) = DeltaU_(1)` and `DeltaH_(2) = DeltaU_(2)` |
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Answer» We know that work done in a reversible expansion process is always greater than in a irreversible expansion process Using the first law of thermodynamics, `q = 0 = DeltaU +(-w)` Thus, decrease in `U` will be large in reversible adiabatic than irreversible adiabatic expansion and since `DeltaU = nC_(V)DeltaT`, we can conclude that `DeltaT` in reversible adiabatic expansion will be greater than in irreversible adiabatic expansion. Now, `DeltaH_(1) = nC_(P)DeltaT_(1)`, and `DeltaH_(2) = nC_(P)DetaT_(2) rArr DeltaH_(1) gt DeltaH_(2)` |
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| 3571. |
For the transition `C` (diamond) `rarr C` (graphite), `Delta H = - 1.5 kJ` it follows thatA. diamond is more stable than graphiteB. graphite is more stable than diamondC. both diamond and graphite are equally stabelD. nothing can be predicted from this reaction |
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Answer» Correct Answer - B Thermodynamical equation idicates that the reaction (transition) is exothermic, i.e., energy is released out when graphite is formed from diamond. Thus, graphite with lower energy content than diamond is relatively more stable. |
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| 3572. |
The magnitude of enthalpy changes for irreversible adiabiatic expansion of a gas from 1 L to 2L is `DeltaH_(1)` and for reversible adiabatic expansion for the same expansion is `DeltaH_(2)`. Then :A. `DeltaH_(1)gtDeltaH_(2)`B. `DeltaH_(1)ltDeltaH_(2)`C. `DeltaH_(1)ltDeltaH_(2)`, enthalapy being a state functionD. `DeltaH_(1)=DeltaE_(1)`, and `DeltaH_(2)=DeltaE_(2)` Where `DeltaE_(1) "and" DeltaE_(2)` are magnitudes changes in internal energy of gas in three expansions respectively . |
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Answer» Correct Answer - B |
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| 3573. |
For the charge : `R to P: " "DeltaH=-ve` If in above reaction , the randomnes of system increases then:A. `DeltaH_(f)(P)gtDeltaH_(f)(R)`B. `DeltaH_(C)(P)gtDeltaH_(C)(R)`C. `DeltaG_(f)(P)gtDeltaG_(f)(R)`D. `DeltaS_(f)(P)gtDeltaS_(f)(R)` |
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Answer» Correct Answer - b,d |
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| 3574. |
The value of `Delta H_(transition)` of C(diamond) is 1.9 kJ/mol at `25^(@)`C. Entropy graphite id higher than entropy of diamond . The implies that.A. C(diamond) is thermodynamically more satble than (graphite) at `25^(@)`C.B. C(graphite) is thermodynamically more satble than (diamond) at `25^(@)`C.C. diamond will provide more heat om complete sombustion at `25^(@)` C.D. `DeltaG_(transition) ` of ` C("diamond") "to" C("graphite")` is -ve |
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Answer» Correct Answer - b,c,d |
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| 3575. |
The value of `DeltaH_("transition")` of C (graphite) `rarr` C (diamond) is 1.9 kJ/mol at `25^(@)C`. Entropy of graphite is higher than entropy of diamond. This implies that :A. C (diamond) is more thermodynamically stable than C (graphite) at `25^(@)C`B. C (graphite) is more thermodynamically stable than C (diamond) at `25^(@)C`C. diamond will provide more heat on complete combustion at `25^(@)C`D. `DeltaG_("transition")` of C (diamond) `rarr` C (graphite) is - ve |
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Answer» Correct Answer - B::C::D `C("graphite")rarrC("diamond")` `DeltaG = DeltaH - TDeltaS` `=1.9 -298 (-ve)=+ve` Graphite is more stable than diamond thermodynamically `Delta_(r)H=(Delta_(C)H)_(G)-(Delta_(C)H)_(D),(Delta_(C)H)_(G)=-x(Delta_(C)H)_(D)=-y` `1.9 =-x+yimplies y=1.9 +x` Diamond provides more heat on complete combustion in comparison of graphite. |
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| 3576. |
Select the correct statement for an ideal gas undergoinf reversible or irreversible adiabatic process from same initial processA. For same final pressure, final ktemperature is more in irreversible processB. For same final volume, final temperature is more in irreversible processC. For same final temperature final pressure is more in irreversible processD. For same final temperature final volume is more in irreversible process |
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Answer» Correct Answer - a,b,d |
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| 3577. |
The value of `DeltaH_("transition")` of C (graphite) `rarr` C (diamond) is 1.9 kJ/mol at `25^(@)C`. Entropy of graphite is higher than entropy of diamond. This implies that :A. C(diamond) is more themodynamically stable then C(graphite) at `25^(@)`CB. C(graphite)is more thermodynamically stable than C (diamond) at `25^(@)C`C. diamond will provide more heat on complete conbustion at `25^(@)C`D. `DeltaG_("transition")` of C(diamond) `rarr` C(graphite) is -ve |
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Answer» Correct Answer - B::C::D More heat is evolved due to combustion of less stable form. |
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| 3578. |
Select the correct statement (s).A. System having non-permeable and adiabatic boundry must be isolated .B. Molarity , normality and molality all are intensive properties .C. `DeltaU=q+ w` can be used for any system.D. Closed system can transfer energy only . |
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Answer» Correct Answer - b,d |
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| 3579. |
From the following date , mark the opation(s) where `Delta H` is correctly written for the given reaction . Given: `H+(aq) +OH-(aq) to H_(2)O(l),` `DeltaH=-57.3 kJ` `DeltaH_(solution) HA(g)=-70.7 kJ"mole"` `DeltaH_(solution) BOH(g)=- 20 kJ"mole"` `DeltaH_("ionzatoin") ` of `HA=15 kJ//"mole"` and BOH is a strong base. A. `{:("Reaction",DeltaH_(r)(KJ//"mol")),(HA(aq) + BOH(aq) rarr BA(aq) + H_(2)O,-42.3):}`B. `{:("Reaction",DeltaH_(r)(KJ//"mol")),(HA(q) + BOH(q) rarr BA(aq) + H_(2)O,-93):}`C. `{:("Reaction",DeltaH_(r)(KJ//"mol")),(HA(q) rarr H^(+)(aq)=A^(-)(aq),-55.7):}`D. `{:("Reaction",DeltaH_(r)(KJ//"mol")),(B^(+)(aq)+OH^(-)(aq) rarr BOH(aq),-20):}` |
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Answer» Correct Answer - A::B::C (D) BOH(g) + aqueous `rarr B^(+) (aq) + OH^(-)(aq) " " DeltaH=-20 KJ//"mole"` |
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| 3580. |
From the following date , mark the opation(s) where `Delta H` is correctly written for the given reaction . Given: `H+(aq) +OH-(aq) to H_(2)O(l),` `DeltaH=-57.3 kJ` `DeltaH_(solution) HA(g)=-70.7 kJ"mole"` `DeltaH_(solution) BOH(g)=- 20 kJ"mole"` `DeltaH_("ionzatoin") ` of `HA=15 kJ//"mole"` and BOH is a strong base. |
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Answer» Correct Answer - ab,c |
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| 3581. |
From the following data, mark the option (s) where `Delta H` is correctlt written for the given reaction. Given `H^(+)(aq) + OH^(-)(aq) rarr H_(2)O(l), Delta H = -57.3 kJ` `Delta H_("solution")` of HA(g) = -70.7kJ/mol `Delta H_("solution")` of BOH(g) = 20kJ/mol `Delta H_("ionization")` of HA = 15 kJ/mol and BOH is a strong base Reaction `Delta H_(r)(kJ//mol)`A. `HA(aq)+BOH(aq) rarr BA(aq)+H_(2)O - 42.3`B. `HA(g) + BOH(g) rarr BA(aq)+H_(2)O -93.0`C. `HA(g) rarr H^(+)(aq)+A^(-)(aq)-55.7`D. `B^(+)(aq)+OH^(-)(aq) rarr BOH (aq) - 20.0` |
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Answer» Correct Answer - A::B::C (a) `-57.3 - (-15)` , (b) `-70.7 + 20 + 15 - 57.3` (c) `-70.7 + 15` |
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| 3582. |
Which of the following statement(s) is/are true?A. `DeltaU = 0` for combustion of `C_(2)H_(6)`(g) in a sealed rigid adiabatic containerB. `Delta_(f)H^(@)` (S, monoclinic) `ne 0`C. If dissociation energy of `CH_(4)(g)` is `1656 kJ//"mol"` and `C_(2)H_(6)` (g) is 2812 kJ/mol, then value of C-C bond energy will be 328 kJ/molD. If `DeltaH_(f)(H_(2)O,g)=-242 " kJ"//"mol", DeltaH_("vap")(H_(2)O,l)=44 " kJ"//"mol"` then, `Delta_(f)H^(@)(OH^(-),aq.)` will be -142 kJ/mol |
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Answer» Correct Answer - A:C `(a) DeltaE=q+W=0+0=0` `(b) (Delta_(F)H^(@))` (S, rhombic) = 0 `(c )(C-C)+(C-H)=2812` `(C-C)=2812-6xx(1656)/(4)=328` `(d) H_(2)O(l)rarrH_(2)O(g)` `44=-(Delta_(F)H)_(H_(2)O(l))-242` `(Delta_(F)H)_(H_(2)O(l))=-286` `H^(+)(aq)+OH^(-)(aq)rarrH_(2)O(l)` `-57.3 =- 286 -4 (Delta_(f)H)_(OH^(-))(aq)(Delta_(f)H)_(OH^(-))=-228.7` |
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| 3583. |
What is the change in internal energy when a gas constants from 377 ml to 177 ml under a constant pressure of 1520 torr , while at the same time being cooled by removing 124 J heat ? [Take :(1 L atm )n= 100 J]A. `-24 J`B. ` -84 J`C. `-164 J`D. `-248 J` |
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Answer» Correct Answer - B `C_(4)H_(4)+HCl to C_(2) H_(5)Cl` `Deltan_(g) = - 1` HCl is limiting reagent `rArr DeltaU= DeltaH - Deltan_(g)RT` ` = - 185 +(1 xx 8.314 xx 10 ^(-3) xx 300) = - 69.8 kJ` For two mole of HCl `= - 69.8 xx2 = - 139.6 kJ` |
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| 3584. |
What is the change internal energy when a gas contracts from 377 mL to 177 mL under a constant pressure of 1520 torr, while at the same time being cooled by removing 124 J heat ? `[Take :( 1 L atm ) =100 J)]`A. 40.52 JB. `-83.48 J`C. `-248 J`D. None of these |
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Answer» Correct Answer - B `DeltaH=DeltaU+P.DeltaV` `implies -124 = DeltaU+(1520)/(760)xx(177-377)xx10^(-3)xx101.3` `DeltaU=-83.48 J` |
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| 3585. |
Staetement -1: The magniyude of the work involed in an isothermal expansion is greater than that involved in an adiabatic expansion. Statement -2: P-v cure (pon y -axas and V on X-axas) decrease more repidly for reversible asiabatic expansion compared to reversible isothermal expansion starting from same initial state.A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 3586. |
At `25^(@)C`, a 0.01 mole sample of a gas is compressed from 4.0 L to 1.0 L at constant temperature. What is the work done for this process if the external pressure is 4.0 bar?A. `1.6 xx 10^(3) J`B. `8.0 xx 10^(2) J`C. `4.0 xx 10^(2) J`D. `1.2 xx 10^(3) J` |
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Answer» Correct Answer - D `w=-4underset(1)overset(4)intdV=+4xx3xx100, w=+1.2 xx10^(3)J` |
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| 3587. |
If the boundary of system moves by an infinitesimal amount, the work involved is given by `dw=-P_("ext")dV` for irreversible process `w=-P_("ext")DeltaV " "( "where "DeltaV=V_(f)-V_(i))` for reversible process `P_("ext")=P_("int")pmdP~=P_("int")` so for reversible isothermal process `w = -nRTln.(V_(f))/(V_(i))` 2mole of an ideal gas undergoes isothermal compression along three different plaths : (i) reversible compression from `P_(i)=2` bar and `V_(i) = 8L` to `P_(f) = 20` bar (ii) a single stage compression against a constant external pressure of 20 bar, and (iii) a two stage compression consisting initially of compression against a constant external pressure of 10 bar until `P_("gas")=P_("ext")`, followed by compression against a constant pressure of 20 bar until `P_("gas") = P_("ext")` Work done (in bar-L) on the gas in reversible isothermal compression is :A. 9.212B. 36.848C. 18.424D. None of these |
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Answer» Correct Answer - B `DeltaU=-300+(-50-150)=-500` `T= 100 K` `w_("rev")=-2.303xxnxxRxxTlog.(P_(1))/(P_(2))` `=-2.303xx2xx0.08xx100xxlog((1)/(10))` `=36.848 " bar-L"` |
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| 3588. |
Statement-1 : Work done in isothermal reversible process is more than irreversible process. And Statement-2 : Irreversible process is an infinitesimally slow process.A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 3589. |
What would be the sign convention for work done on a system for a cylinder containing one mole of an ideal gas where `P_(ext) = P_(int)` ? |
| Answer» Correct Answer - W = 0 | |
| 3590. |
In previous problem if expansion is carried out freely `(P_("ext")=0)`, then:A. `W=0`B. `W="RT In"3`C. `DeltaS="R In" 3`D. `Q="Rt In" 3` |
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Answer» Correct Answer - A::C `DeltaS` will be same because entropy is a state function and in free expansion `P_("external") =0` . So , work done is zero. |
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| 3591. |
If the boundary of system moves by an infinitesimal amount, the work involved is given by `dw=-P_("ext")dV` for irreversible process `w=-P_("ext")DeltaV " "( "where "DeltaV=V_(f)-V_(i))` for reversible process `P_("ext")=P_("int")pmdP~=P_("int")` so for reversible isothermal process `w = -nRTln.(V_(f))/(V_(i))` 2mole of an ideal gas undergoes isothermal compression along three different plaths : (i) reversible compression from `P_(i)=2` bar and `V_(i) = 8L` to `P_(f) = 20` bar (ii) a single stage compression against a constant external pressure of 20 bar, and (iii) a two stage compression consisting initially of compression against a constant external pressure of 10 bar until `P_("gas")=P_("ext")`, followed by compression against a constant pressure of 20 bar until `P_("gas") = P_("ext")` Total work done on the gas in two stage compression is :A. 40B. 80C. 160D. None of these |
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Answer» Correct Answer - B `w_("irr (total)")=w_(1)+w_(2)` `=-10((nRT)/(10)-(nRT)/(2))` `=-20((nRT)/(20)-(nRT)/(10))` `=5xxnRT=80 " bar-L"` |
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| 3592. |
Gibbs Helmholtz equation relates the enthalpy, entropy and free energy change of the process at constant pressure and temperature as `DeltaG=DeltaH-TDeltaS " (at constant P, T)"` In General the magnitude of `DeltaH` does not change much with the change in temperature but the terms `TDeltaS` changes appreciably. Hence in some process spontaneity is very much dependent on temperature and such processes are generally known as entropy driven process. The Dissolution of `CaCl_(2).6H_(2)O` in a large volume of water is endothermic to the extent of 3.5 kcal `"mol"^(-1)` and `DeltaH` for the reaction is -23.2 kcal `"mol"^(-1)`. `CaCl_(2)(s)+6H_(2)O(l)rarrCaCl_(2).6H_(2)O(s)` Select the correct statement :A. `DeltaH_("solution")` for anhydrous `CaCl_(2)` is - 19.7 kcal/mol and the process is enthalpy drivenB. `DeltaH_("solution")` for anhydrous `CaCl_(2)` is - 19.7 kcal/mol and the process is entropy drivenC. Dissolution of `CaCl_(2).6H_(2)O` in water is enthalpy driven processD. The `Delta_(r )S` the reaction `CaCl_(2)(s)+6H_(2)O(l)rarrCaCl_(2).6H_(2)O(s)` is negative |
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Answer» Correct Answer - A `CaCl_(2)(s)+aq+6H_(2)OrarrCaCl_(2)(aq)+x` `CaCl_(2)(aq)rarrCaCl_(2).6H_(2)O(s)+aq+3.5` `CaCl_(2)(s)+6H_(2)OrarrCaCl_(2).6H_(2)O(s)+x+3.5` `x+3.5 =23.2` `x = 19.7` `DeltaH=-x=-19.7` enthalpy driven |
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| 3593. |
Gibbs Helmholtz equation relates the enthalpy, entropy and free energy change of the process at constant pressure and temperature as `DeltaG=DeltaH-TDeltaS " (at constant P, T)"` In General the magnitude of `DeltaH` does not change much with the change in temperature but the terms `TDeltaS` changes appreciably. Hence in some process spontaneity is very much dependent on temperature and such processes are generally known as entropy driven process. Fro the reaction at 298 K, `A_(2)B_(4)rarr2AB_(2)` `DeltaH=2" kJ"` and `DeltaS` = 20 J/K at constant P and T, the reaction will beA. spontaneous and entropy drivenB. spontaneous and enthalpy drivenC. non-spontaneousD. at equilibrium |
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Answer» Correct Answer - A `(DeltaG)_(PT)=2000-(20xx298)` `=-3960 " J"//"mol"` |
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| 3594. |
Statement-1 : The work done in an open container at 300 K, when 112 g of iron reacts with dil. `HCl` is 200 cal. statement-2 : Work done `= P_(ext)(V_(2)-V_(1))`, where `V_(1)` and `V_(2)` are initial and final volume of reaction mixture & `P_(ext)` is external applied pressure.A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - A | |
| 3595. |
Assertion. Molar entropy of vaporisation of water is different from ethanol. Reason. Water is more solar than ethanol.A. If both A and R are true,andR is the true explanation of A.B. If both A and R are true,but R is not the true explanation of A.C. If A is true, but R is falseD. If both A and R are false. |
| Answer» Correct Answer - C | |
| 3596. |
Statement-1 : Many endothermic reactions that are not spontaneous at room temperature become spontaneous at high temperatures. Statement-2: Energy of the system increases with increase in temperature.A. Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - C | |
| 3597. |
Assertion. Water is liquid state is more stable than ice at room temperature. Reason. Water in liquid form has higher entropy than ice.A. If both A and R are true,andR is the true explanation of A.B. If both A and R are true,but R is not the true explanation of A.C. If A is true, but R is falseD. If both A and R are false. |
| Answer» Correct Answer - a | |
| 3598. |
Assertion :- Water in liquid state is more stable than ice at room temperature. Reason :- Water in liquid from has higher entropy than ice.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false. |
| Answer» Correct Answer - B | |
| 3599. |
Assertion :- Entropy is always constant for a closed system. Reason :- Closed system is always reversible.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are false. |
| Answer» Correct Answer - D | |
| 3600. |
Assertion : In an isolated system the entropy increases. Reason : The processes in an isolated system are adiabatic.A. If both assetion and reason are true and reaasons is the correct expanation of assetionB. If both assetion and reason are tur but reason is not the correct explanation of assetionC. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - B The entropy of an isolated system increases. In an isolated system, there is no exchange of heat , therefore, process is adiabatic. |
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