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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.
| 51. |
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant `K_(c)` isA. `DeltaG=RTlnK_(c)`B. `-DeltaG=RTlnK_(c)`C. `DeltaG^(@)=RTlnK_(c)`D. `-DeltaG^(@)=RTlnK_(c)` |
| Answer» Correct Answer - D | |
| 52. |
For the reaction `A(g)+2B(g)rarr2C(g)+3D(g)`, the value of `DeltaE` at `27^(@)C` is 19.0 kcal. The value of `DeltaH` for the reaction would be `(R=2.0 cal K^(-1) mol^(-1))`A. 20.8 kcalB. 19.8 kcalC. 18.8 kcalD. 20.2 kcal |
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Answer» Correct Answer - D Use `DeltaH=DeltaE+DeltanRT` `DeltaH=19+2xx2xx10^(-3)xx300=20.2 kcal , Deltan=2`. |
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| 53. |
Given : `C_(2)H_(6(g))rarr2C_((g))+6H_((g)), DeltaH=712 kcal`. The C - C bond energy is 112kcal, what is the C-H bond energyA. 88 kcalB. 12 kcalC. 50 kcalD. 600 kcal |
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Answer» Correct Answer - D Bond energy of 6C-H bonds =712-112 = = 600 kcal. |
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| 54. |
For a gaseous reaction at 300K, `DeltaH - DeltaU = - 4.98 kJ` assuming that `R=8.3 JK^(-1) mol^(-1), Deltan_((g))` isA. 1B. 2C. `-2`D. 0 |
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Answer» Correct Answer - C `DeltaH=DeltaE+DeltanRT` `DeltanRT=DeltaH-DeltaE` `Deltanxx8.3xx300=-4.98xx10^(3)` `Deltan=-2`. |
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| 55. |
For the reaction, `A(s)+3B(s)to4C(s)+D(l)` `DeltaH` and `DeltaU` are related as-A. `DeltaH = DeltaU`B. `DeltaH = DeltaU + 3RT`C. `DeltaH=DeltaU + RT`D. `DeltaH=DeltaU-3RT` |
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Answer» Correct Answer - D `A(s)+3B(g)rarr4C(s)+D(l)` `Deltan_(g)=0-3=-3, DeltaH=DeltaE+Deltan_(g)RT [because DeltaE=DeltaU]` `thereforeDeltaH=DeltaU+(-3)(RT)` `DeltaH=DeltaU-3(RT), DeltaH=DeltaU-3RT`. |
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| 56. |
Given the bond energies `N-=N`, H-H and N-H bonds are 945,436 and 391 kJ `"mole"^(-1)` respectively, the enthalpy of the following reaction `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` isA. `-93` kJB. 102 kJC. 90 kJD. 105 kJ |
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Answer» Correct Answer - A `underset("Energy absorbed")ubrace(underset(945)(N-=)Nunderset(+3xx436)(+3H-H))rarrunderset("Energy released")undersetubrace(2xx(3xx391)=2346)(2underset(H)underset(|)overset(H)overset(|)N-H)` Net. Energy released = 2346-2253=93 kJ i.e. `DeltaH=-93 kJ`. |
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| 57. |
Which of the following is combustion reactionA. `C+O_(2)rarrCO_(2)`B. `CH_(4)+O_(2)rarrCO_(2)+H_(2)O`C. `Mg+O_(2)rarrMgO`D. All of these |
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Answer» Correct Answer - D When all substances in a compound combine with oxygen and produces `CO_(2)` and `H_(2)O`, it is commonly called burning. During combustion substance reacts with an oxidising agent such as `O_(2)` and `F_(2)`. Not all combustion reaction releases `CO_(2)` and `H_(2)O` e.g. `Mg+O_(2)rarrMgO+` Heat. `therefore 3.2"gm will liberate = "(890xx3.2)/(16)=178 kJ`. |
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| 58. |
Which is the best definition of 'heat of neutralization'A. The heat set free when one gram molecule of a base is neutralized by one gram molecule of an acid in dilute solution at a stated temperatureB. The heat absorbed when one gram molecule of an acid is neutralized by one gram molecule of a base in dilute solution at a stated temperatureC. The heat set free or absorbed when one gram atom of an acid is neutralized by one gram atom of a base at a stated temperatureD. The heat set free when one gram equivalent of an acid is neutralized by one gram equivalent of a base in dilute solution at a stated temperature |
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Answer» Correct Answer - D It is definition of heat of neutralization. |
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| 59. |
The lowest value of heat of neutralization is obtained forA. `HCl+NaOH`B. `CH_(3)COOH+NH_(4)OH`C. `NH_(4)OH+HCl`D. `NaOH+CH_(3)COOH` |
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Answer» Correct Answer - B Both `NH_(4)OH` and `CH_(3)COOH` are weak. |
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| 60. |
In order to decompose 9 g water 142.5 kJ heat is required. Hence the enthalpy of formation of water isA. `-142.5 kJ`B. `+142.5 kJ`C. `-285 kJ`D. `+285 kJ` |
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Answer» Correct Answer - C For the decomposition of 9gm of water heat required = 142.5 kJ we know `H_(2)O=2+16=18` Therefore heat required for decomposition of 18gm water `=(18)/(9)xx142.5=285 KJ` |
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| 61. |
All reaction with chemical dissociation areA. ReversibleB. Reversible and endothermicC. ExothermicD. Reversible or irriversible and endothermic or exothermic |
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Answer» Correct Answer - B Chemical dissociations are reversible and endothermic. |
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| 62. |
Which of the following values of heat of formation indicates that the product is least stableA. `-94` kcalB. `-231.6` kcalC. `+21.4` kcalD. `+64.8` kcal |
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Answer» Correct Answer - D In the formation of a compound, more the heat absorbed, less stable is the compound. |
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| 63. |
The compounds with high heat of formation are less stable becauseA. High temperature is required to synthesise themB. Molecules of such compounds are distortedC. It is difficult to synthesis themD. Energy rich state leads to instability |
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Answer» Correct Answer - D Compounds with high heat of formation are less stable because energy rich state leads to instability. |
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| 64. |
The enthalpy of water is 386 kJ. What is entropy of waterA. 0.5 kJB. 1.3 kJC. 1.5 kJD. 22.05 kJ |
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Answer» Correct Answer - B `DeltaS_("vap")=(DeltaH_("vap"))/(T)=(386)/(298)=1.3 kJ`. |
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| 65. |
If a mole of `H_(2)` molecule is heated to high temperature the following reaction takes placeA. `H_(2)(g)+436kJ =H(g)+H(g)`B. `H_(2)(g)+820 kJ=2H_(2)(g)`C. `2H_(2)(g)+436 J = 2H_(2)`D. `H_(2)+H_(2)=H^(+)+H^(+)` |
| Answer» Correct Answer - A | |
| 66. |
Heat of combustion of a substanceA. Is always positiveB. Is always negativeC. Is equal to heat of formationD. Nothing can be said without reaction |
| Answer» Correct Answer - B | |
| 67. |
One calorie is equal toA. 0.4184 JouleB. 4.184 JouleC. 41.84 JouleD. 418.4 Joule |
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Answer» Correct Answer - BOption (B) 4.184 J |
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| 68. |
If butane on combustion gives carbon monoxide. Find the number of `O_(2)` molecule requiredA. 6B. 5.5C. 4.5D. 4 |
| Answer» Correct Answer - C | |
| 69. |
A reaction that takes place with the absorption of energy isA. Burning of a candleB. Rusting of ironC. Electrolysis of waterD. Digestion of food |
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Answer» Correct Answer - C `H_(2)O_((l))overset("electrolysis")rarrH_(2(g))+(1)/(2)O_(2(g))`. |
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| 70. |
Statement 1 : There is a natural asymmetry between converting work to heat and converting heat to work. Statement 2 : No porcess is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.A. Statement 1 is true, statement-2 is true, statement 2 is a correct explanation for statement 4B. Statement 1 is true, statement 2 is true, statement 2 is not a correct explanation for statement 4C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - A Statement 2 is `II^(nd)` law of thermodynamics which concludes that total heat can never converted into equilvalent amount of work. |
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| 71. |
Statement 1 : For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero. Statemet 2 : At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy.A. Statement 1 is true, statement-2 is true, statement 2 is a correct explanation for statement 3B. Statement 1 is true, statement 2 is true, statement 2 is not a correct explanation for statement 3C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - D At equilibrium `DeltaG` (Gibbs energy) = 0 but `DeltaG^(@)` (standard Gibbs energy) `ne 0` As `DeltaG` is `II^(nd)` law of thermodynamics which concludes that total heat can never converted into equivalent amount of work. |
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| 72. |
For a spontaneous process the correct statement isA. Entropy of the system always increasesB. Free energy of the system always increasesC. Total entropy change is always negativeD. Total entropy change is always positive |
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Answer» Correct Answer - D Definition based. |
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| 73. |
Match the transformation in Column I with appropriate options is Column II `{:(,"Column I",,"Column II"),((A),CO_(2)(s)rarrCO_(2)(g),(p),"Phase transition"),((B),CaCO_(3)(s)rarrCaO(s)+CO_(2),(q),"Allotropic change"),((C),2H• rarrH_(2)(g),(r),DeltaH" is positive"),((D),P_(("white,solid"))rarrP_(("red,solid")),(s),DeltaS" is positive"),(,,(t),DeltaS " is negative"):}` |
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Answer» Correct Answer - A `rarr` p, r, s; B `rarr` r, s; C `rarr` t ; D `rarr`p, q, t (A) `CO_(2)(s)rarrCO_(2)(g)` It is phase transition. The process is endothermic (sublimation). Gas is produced, so entropy increases. (B) On heating `CaCO_(3)` decomposes, So, process is endothermic. The entropy increases as gaseous product is formed. (C) `2H•rarrH_(2)(g)` Entropy decreases as number of gaseous particles decreases. (D) It is phase transition White and red P are allotropes Red P is more stable than white ,brgt So `DeltaG` is -ve |
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| 74. |
Following data is known about melting of a compound AB. DeltaH=9.2 kJ mol^(-1), DeltaS=0.008 kJ K^(-1) mol^(-1)`. Its melting point isA. 736 KB. 1050 KC. 1150 KD. `1150^(@)C` |
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Answer» Correct Answer - C `T_(m)=(DeltaH_("fusion"))/(DeltaS_("fusion"))=(9.2)/(0.008)=1150 K`. |
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| 75. |
In the process : `H_(2)O(s, -10^(@)C, 1atm)rarrH_(2)O(1, 10^(@)C, 1 atm)` `C_(p)` for ice = 9 cal `deg^(-1) mol^(-1), C_(p)` for `H_(2)O=18` cal `deg^(-1)mol^(-1)`. Latent heat of fusion of ice = 1440 cal `mol^(-1)` at `0^(@)C`. The entropy change for the above process is 6.258 cal. `deg^(-1)mol^(-1)` Give the total number of steps in which the third law of thermodynamics is used |
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Answer» Correct Answer - B Step 1. (using the third law of thermodynamics) : (For changing) `H_(2)O(s)(-10^(@)C, 1 atm)rarrH_(2)O(s, 0^(@)C1 atm)` `DeltaS_(1)=underset(-10)overset(0)intn(C_(p))/(T)dT=1xx9xx2.3xxlog.(273)/(263)=0.336" cal deg"^(-1) mol^(-1)` Step 2 (using the second law of thermodynamics) : `H_(2)O(s)(0^(@)C, 1 atm)rarrH_(2)O(l)(0^(@)C, 1 atm)` `DeltaS_(2)=(q_(rev))/(T)=(1440)/(273)=5.258" cal deg"^(-1)mol^(-1)` Step 3 (using the third law of thermodynamics) : `H_(2)O(l)(0^(@)C, 1 atm)rarrH_(2)O(l)(10^(@)C, 1 atm)` `DeltaS_(3)=underset(0)overset(10)intn(C_(p))/(T)dT=1xx18xx2.3xxlog.(283)/(273)=0.647 " cal deg"^(-1)mol^(-1)` `DeltaS=DeltaS_(1)+DeltaS_(2)+DeltaS_(3)=0.336+5.258+0.647 = 6.258 " cal deg"^(-1) mol^(-1)`. |
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| 76. |
If for a given substance melting point is `T_(B)` and freezing point is `T_(A)`, then correct variation shown by graph between entropy change and temperature isA. B. C. D. |
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Answer» Correct Answer - A For a pure substance `T_(A)` and `T_(B)` represent the same temperature. Hence A is a correct choice. |
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| 77. |
Which of the following statements is trueA. The total entropy of the universe remains constantB. The total entropy of the universe is continuously decreasingC. The total entropy of the universe is continuously decreasingD. The total energy of the universe remains constant |
| Answer» Correct Answer - D | |
| 78. |
Mixing of two different ideal gases under isothermal reversible condition will lead toA. Increase of Gibbs free energy of the systemB. No change of entropy of the systemC. Increase of entropy of the systemD. Increase of enthalpy of the system |
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Answer» Correct Answer - C During mixing, `DeltaS_(mix)` is always positive. |
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| 79. |
The condition of spontaneity of process isA. Lowering of entropy at constant temperature and pressureB. Lowering of Gibbs free energy of system at constant temperature and pressureC. Increase of entropy of system at constant temperature and pressureD. Increase of Gibbs free energy of the universe at constant temperature and pressure |
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Answer» Correct Answer - B `dG_(P,T)=-ve` is the criterion for spontaneity. |
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| 80. |
Among the following for spontaneity of chemical reaction there should beA. Decrease in entropy and increase in free energyB. Decrease in entropy and free energy bothC. Increase in entropy and decrease in free energyD. Increase in entropy and free energy both |
| Answer» Correct Answer - C | |
| 81. |
For spontaneity of a cell, which is correctA. `DeltaG=0, E^(@)=0`B. `DeltaG=-ve, E^(@)=0`C. `DeltaG=+ve, E^(@)=+ve`D. `DeltaG=-ve, E^(@)=+ve` |
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Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS, DeltaG=DeltaE+PDeltaV-TDeltaS` For spontaneity `(DeltaG=-ve)` `becauseDeltaG=-ve, DeltaE=+ve`. |
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| 82. |
The essential condition for the feasibility of a reaction is thatA. The reaction should be exothermicB. The entropy of products must be larger than that of reactantsC. The reaction is to be accompanied with free energy decreaseD. The reaction has to possess high activation energy |
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Answer» Correct Answer - C If `DeltaG=-ve, reaction is spontaneous. |
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| 83. |
Which of the following is true for an adiabatic processA. `DeltaH = 0`B. `DeltaW = 0`C. `DeltaQ = 0`D. `DeltaV = 0` |
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Answer» Correct Answer - C For adiabatic process `DeltaQ=0`. |
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| 84. |
Which of the following is not a state functionA. Internal energyB. EnthalpyC. WorkD. Entropy |
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Answer» Correct Answer - C Work is not a state function as during a process its value depends on the path followed. The value of enthalpy, internal energy entropy depends on the state and not on the path followed to get that state, hence these are state functions. |
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| 85. |
Under equilibrium conditions, which of the following is trueA. `DeltaS=0`B. `DeltaH gt 0`C. `DeltaS gt 0`D. `DeltaH lt 0` |
| Answer» Correct Answer - A | |
| 86. |
Internal energy does not includeA. Nuclear energyB. Rotational energyC. Vibrational energyD. Energy arising by gravitational pull |
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Answer» Correct Answer - D `E=E_(el e).+E_("nucl")+E_("chemical")+E_("potential")+E_("kinetic")(E_(t)+E_(v)+E_(r))`. |
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| 87. |
A gaseous system during a thermodynamic process does not undertake any volume change. It is calledA. Isochoric processB. Isobaric processC. Isothermal processD. Isoentropic process |
| Answer» Correct Answer - A | |
| 88. |
A well stoppered thermos flask contains some ice cubes. This is an example of aA. Closed systemB. Open systemC. Isolated systemD. Non-thermodynamic system |
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Answer» Correct Answer - C In isolated system neither exchange of matter nor exchange of energy is possible with surroundings. |
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| 89. |
The work done by a system is 8 joule, when 40 joule heat is supplied to it. What is the increase in internal energy of systemA. 25 JB. 30 JC. 32 JD. 28 J |
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Answer» Correct Answer - C `q=40 J` `w=8 J` (Work done by the system) `DeltaE=q+w=40-8=32 J` |
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| 90. |
If `Klt1.0`, what will be the value of `DeltaG^(@)` of the followingA. `1.0`B. ZeroC. NegativeD. Positive |
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Answer» Correct Answer - D `DeltaG^(@)=-RT lnK`. When `K lt 1.0, DeltaG^(@)=+ve`. |
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| 91. |
When the value of entropy is greater, then the ability of workA. Is maximumB. Is minimumC. Is mediumD. None of these |
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Answer» Correct Answer - A When the value of entropy is greater, then ability of work is maximum. |
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| 92. |
Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is given out by the system, isA. `+1 kJ`B. `-5 kJ`C. `+5 kJ`D. `+3 kJ` |
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Answer» Correct Answer - D Change in internal energy is given by the relation, `DeltaE=q+w` `DeltaE=-q+w` (as heat is given out by the system) `DeltaE=-1+4` `DeltaE=+3kJ`. |
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| 93. |
The cooling in refrigerator is due toA. Reaction of the refrigerator gasB. Expansion of iceC. The expansion of the gas in the refrigeratorD. The work of the compressor |
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Answer» Correct Answer - C This is based on Joule-Thomson effect. |
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| 94. |
According to the first law of thermodynamics which of the following quantities represents change in a state functionA. `q_(rev)`B. `q_(rev)-w_(rev)`C. `q_(rev)//w_(rev)`D. `q_(rev) + w_(rev)` |
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Answer» Correct Answer - D `DeltaE=q+w, DeltaE` is a state function. |
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| 95. |
Which one of the following statements is falseA. Work is a state functionB. Temperature is a state functionC. Change in the state is completely defined when the initial and final states are specifiedD. Work appears at the boundary of the system |
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Answer» Correct Answer - A Work is not a state function because it depends upon the path followed. |
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| 96. |
Among the following, the state function(s) is(are)A. Internal energyB. Irreversible expansion workC. Reversible expansion workD. Molar enthalpy |
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Answer» Correct Answer - A::D State function are internal energy and molar enthalpy. Work is path function whether it is reversible or irreversible. |
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| 97. |
In thermodynamics, a process is called reversible whenA. Surroundings and system change into each otherB. There is no boundary between system and surroundingsC. The surroundings are always in equilibrium with the systemD. The system changes into the surroundings spontaneously |
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Answer» Correct Answer - C In a reversible process, the driving and the opposite forces are nearly equal, hence the system and the surroundings are nearly equal, hence the system and the surroundings always remain in equilibrium with each other. |
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| 98. |
Among the following the intensive property is (properties are)A. Molar conductivityB. Electromotive forceC. ResistanceD. Heat capacity |
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Answer» Correct Answer - A::B Molar conductivity and electromotive force (emf) are intensive properties as these are size independent. |
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| 99. |
The following is (are) endothermic reaction(s)A. Combustion of methaneB. Decomposition of waterC. Dehydrogenation of ethane to ethyleneD. Conversion of graphite to diamond |
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Answer» Correct Answer - B::C::D All combustion reactions are exothermic in nature. |
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| 100. |
It is general principle that the less energy a system contains, it isA. Less stableB. More stableC. UnstableD. More unstable |
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Answer» Correct Answer - B The less energy of a system and more is its stability. |
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