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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.
| 3601. |
What will happen to the space shown in the following when heated ? |
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Answer» When heat is given to the substance, the separation between the molecules of the substance increases and they set themselves by occupying the more space, hence in fig (a) distance between a and b will increase. In the fig (b) the radius of the cavity will also increase and, in fig. (c) the space between two cavities, i.e., a and b also increases. |
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| 3602. |
Assertion :- In an isolated system the entropy increases due to spontanous process. Reason :- The processes in an isolated system are adiabatic.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 - A | |
| 3603. |
If an electric fan is switched on in a closed room, will the air of the room be cooled? If not, why do we feel cold? |
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Answer» No, rather the air will be heated, because due to motion of the fan, the speed of molecules will increase. In fact, we feel cold due to evaporation of our sweat. |
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| 3604. |
Assertion : If an electric fan be switched on in a closed room, the air of the room will be cooled. Reason : Fan air decrease the temperature of the room.A. 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 - D |
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| 3605. |
If an electric fan be switched on in a closed room, will the air of the room be cooled? If not, why do we feel cold? |
| Answer» No, rather the air will be heated, because due to motion of the fan, the speed of air molecules will increase. Infact, we feel cold due to evaporation of our sweat. | |
| 3606. |
Calculate what fraction of gas molecules (a) traverses without collisions the distances exceeding the mean free path `lamda` , (b) has the free path values lying within the inerval from `lamda` to `2 lamda`. |
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Answer» (a) The fraction of gas molecules which traverses distances exceeding the mean free path without collision is just the probability to traverse the distance `s = lamda` without collision. Thus `P = e^-1 = (1)/( e) = 0.37` This probability is `P = e^-1 - e^-2 = 0.23`. |
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| 3607. |
A sample of gas goes through a process shown in graph. Arrange the volume of given points in increasing order A. `V_(1) gt V_(2) gt V_(3)`B. `V_(1) gt V_(2)=V_(3)`C. `V_(3) gt V_(2) gt V_(1)`D. `V_(3) gt V_(2)=V_(1)` |
| Answer» Correct Answer - C | |
| 3608. |
2 mole of ideal momoatmic gas was subjected to reversible adiabatic compression from initianl state of P= 1 and T = 300 K till the pressrue is `4sqrt(2)` atm and temperature is TK . The gas is then subjected to reversible isothermal expansion at T K till the internal pressure is one atm . The gas is now isobarically cooled to attain initial state. Find `W_(Net)` ( in calorie) for whole process . [Use In 2=0.7] |
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Answer» Correct Answer - `-1.2 xx 10^(03) or - 1.19 xx 10 ^(03)("taking R= 1.98")` `T_(2)= T_(1)(P_(1)/(P_(2)))^(((1-gamma)/(gamma))` ` = 600 K` `W_(I)= W_("rev")= nC_(V)[T_(2)- T_(1)] = 2 xx (3)/(2) R xx 300` `= 900 R` `W_(II)= - 2.303 nRT log .(P_(1))/(P_(2)) = - 2xx R xx 600 In ((4sqrt(2))/(1))` `W_(III) = - PDeltaV = - nRT = - 1200R xx (5)/(2) xx 0.7 = - 2100 R` `W_("Total") = 900 R - 2100 R + 600 R` ` = - 600 R = - 1200 cal` |
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| 3609. |
A monatomic ideal gas is heated at constant volume until its pressure is doubled. It is again heated at constant pressure, until its volume is doubled. Find molar specific heat for the whole process.A. `(3R)/(2)`B. `(5R)/(2)`C. `(13R)/(6)`D. `(7R)/(2)` |
| Answer» Correct Answer - C | |
| 3610. |
What are the Sign conventions uses of heat and work ? |
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Answer» Sign conventions used: (i) Heat absorbed by a system is positive. Heat given out by system is negative. (ii) Work done by a system is positive. Work done on a system is negative. (iii)The increase in internal energy of a system is positive. The decrease in internal energy of a system is negative. |
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| 3611. |
Calculate the change in internal energy when `5g` of air is heated from internal energy when `5g` of air is heated from `0^(@)C "to" 2^(@)C`. Specific heat of air at constant volume is `0.172 cal//g//^(@)C`. |
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Answer» Here, `dU=?m= 5g, DeltaT= 2^(@)C` `C_(v)= 0.172 cal//g//^(@)C` `dQ=mC_(v) DeltaT= 5xx10.172xx2 cal` `=1.72xx4.2J= 7.2J` As air is heated at constant volume, `:. dW= 0` From first law of thermodynamics, `dU+dW=dQ` `dU=dQ-dW= 7.2-0= 7.2J` |
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| 3612. |
Find the change in internal energy in joule when `20gm` of a gas is heated from `20^(@)C` to `30^(@)C` `(c_(v)=0.18 kcal//kg` `K , J=4200J//kcal`)`A. `72.8 J`B. `151.2 J`C. `302 J`D. `450 J` |
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Answer» Correct Answer - B `dU=mc_(v)dT` |
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| 3613. |
Calculate the change in internal energy when 5g of air is heated from `0^(@)` to `4^(@)C`. The specific heat of air at constant volume is `0.172 cal g^(-1) .^(@)C^(-1)`.A. 28.8 JB. 14.4 JC. 7.2 JD. 3.51 J |
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Answer» Correct Answer - B As constant volume , dW=PdV=P`xx`0=0 dQ = `mC_(V)DeltaT=5xx0.172xx4` cal `=5xx0.172xx4xx4.2J`=14.4 J dU =dQ-dW=14.4 -0=14.4 J |
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| 3614. |
The `P-V` diagram of a system undergoing thermodynnaic transformation is shown in Fig. The work done on the system in going from `A rarr B rarr C` is `50 J` and `20 cal` heat is given to the system. The change in internal erergy between `A` and `C` is A. `34J`B. `70J`C. `84J`D. `134J` |
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Answer» Correct Answer - D |
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| 3615. |
The `P - V` diagram of a system undergoing thermodynamic transformation is shown in figure. The work done by the system in going from `A rarr B rarr C is 30 J` and `40 J` heat is given to the system. The change in internal energy between `A and C` is A. 10 JB. 70 JC. 84 JD. 134 J |
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Answer» Correct Answer - a |
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| 3616. |
When a system is taken from state f along path `iaf, Q = 50 J` and `W = 20 J`. Along path `ibf, Q = 35 J`. If `W = -13 J` for the curved return path `f I, Q` for this path is A. `33j`B. `23j`C. `-7j`D. `-43j` |
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Answer» Correct Answer - D |
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| 3617. |
When a system is taken from state f along path `iaf, Q = 50 J` and `W = 20 J`. Along path `ibf, Q = 35 J`. If `W = -13 J` for the curved return path `f I, Q` for this path is A. 33 JB. 23 JC. `-7 J `D. `-43 J` |
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Answer» Correct Answer - d |
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| 3618. |
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in (figure). The change in internal energy of the gas during the transition is `(gamma=3//5)` A. `-20 KJ`B. `20 KJ`C. `-12 KJ`D. `20 KJ` |
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Answer» Correct Answer - A Here, `P_(1)= 5Kpa, P_(2)= 2Kpa` `V_(1)= 4m^(3), V_(2)= 6m^(3), r= 3//5` As change is internal energy `DeltaU= n C_(v)DeltaT= (PV)/(nR)` `DetlaT=T_(2)-T_(1)=(P_(2)V_(2)-P_(1)V_(1))/(nR), C_(v)= R/(gamma-1)` `DeltaU=(nR)/(gamma-1)((P_(2)V_(2)-P_(1)V_(1))/(nR))=(P_(2)V_(2)-P_(1)V_(1))/(gamma-1)` `=(5xx4-2xx6)/(3/5-1)=(200-12)/(-2/5)= -8/(2/5)=-20KJ` |
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| 3619. |
Hot air becomes cooler as it rises appreciably. Why? |
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Answer» As hot air rises up, it reaches region of reduced pressure. The air suffers adiabatic expansion. From first law of thermodynamics, ∆Q = ∆U + ∆W For adiabatic process, ∆Q = 0 ∆U + ∆W = 0 Since, ∆U becomes negative. Therefore, air becomes cool. |
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| 3620. |
If hot air rises , why is it cooler at the top of mountain than near the sea level? |
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Answer» Since atmospheric pressure decreases with height, pressure at the top of the mountain is lesser. When the hot air rises up,it suffer adiabatic expansion at the top of the mountain.For an adiabatic change,first law of thermodynamics may be express as dU + dW =0 (dQ = 0) dW = -dU Therefore work done by the air in rising up (dW =+ve ) result in decrease in the internal Energy of the air (dU = -ve) and hence a fall in the temperature. |
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| 3621. |
Which of the following are state function?A. Internal energyB. Irreversible expansion workC. Reversible expansion workD. Molar enthalpy |
| Answer» Correct Answer - A::D | |
| 3622. |
During the isothermal expansion of an ideal gas:A. The internal energy remains unaffactedB. The temperature remains constantC. The enthalpy remains unaffectedD. the enthalpy becomes zero |
| Answer» Correct Answer - A::B::C::D | |
| 3623. |
In which of the following cases is the reaction spontaneous at all temperatures?A. `DeltaH gt 0,DeltaS gt 0`B. `DeltaH lt 0,DeltaS gt 0`C. `DeltaH lt 0,DeltaS lt 0`D. `DeltaH lt 0,DeltaS = 0` |
| Answer» Correct Answer - B::D | |
| 3624. |
In which of the following cases entropy decreases?A. Solid changing to liquidB. Expansion of gasC. Crystal dissolvesD. Polymerisation |
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Answer» Correct Answer - D Polymerisation leads to more ordered structure. |
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| 3625. |
Which of the following conditions is (are) favourable for the feasibility of a reaction?A. `DeltaH=-ve, TDeltaS=+ve`B. `DeltaH=-ve, TDeltaS=-ve, TDeltaS lt DeltaH`C. `DeltaH=+ve,TDeltaS=+ve,TDeltaS lt DeltaH`D. `DeltaH=+ve,TDeltaS=+ve,TDeltaS gt DeltaH` |
| Answer» Correct Answer - A::B::D | |
| 3626. |
In which of the following entropy increases?A. Rusting of ironB. Melting of iceC. Crysttallization of sugar from solutionD. Vaporization of champhor |
| Answer» Correct Answer - A::B::D | |
| 3627. |
Which of the following state function is not zero at standard state :-A. EnthalpyB. EntropyC. Free energyD. None |
| Answer» Correct Answer - B | |
| 3628. |
In whiich of the following cases, do you consider the increase in entropy take `(s)` place ?A. Pure liquid or liquid solutions are formed from solidsB. Gases are formed, either from solids or liquidsC. the number of molecules of gases increase in the course of a chemical reactionD. The temperature of a substance is increased |
| Answer» Correct Answer - A::B::C::D | |
| 3629. |
In which of the following cases entropy decreases?A. Solid changing to liquidB. Expansion of a gasC. Crystals dissolveD. Polymerisation |
| Answer» Correct Answer - D | |
| 3630. |
For the reaction, `N_(2)(g)+O_(2)(g)hArr2NO(g), " " Delta G^(@)=18.6 " In " 10 "kcal at " 300 K`. If initially 1 bar of `N_(2) " and " 10 " bar of " O_(2)` is taken in a vessel at 300 K, then calculate the value of abcd. [Use : `R= 2 "cal"//"mol-K"`] `"where ab " = "two digit number such that " K_(eq) ("equilibrium constant of above reaction" ) = 10^(-(ab))` and `cd= "two digit number such tht " P_(NO)=10^(-(cd)) "bar"`. [For example if `K_(eq)("equilibrium constant")= 10^(-10)` `:. ab=10 and if P_(NO)=10^(-12)" bar"` `:.cd=12 " hence abcd"=1012]` |
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Answer» Correct Answer - 3115 |
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| 3631. |
The `P-V` diagram of a gas undergoing a cyclic process ABCDA is shown in (figure). Where `P` is in `N//m^(2)` and `V` is in `cm^(3)`. Identify the incorrect statement A. `0.4J` of work is done by the gas from A to BB. `0.2J` of work is done on the gas from C to DC. No work is done by the gas in going from B to CD. Work done by the gas in going from B to C and on the gas from D to A |
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Answer» Correct Answer - D The incorrect statement is (d).This is because, in going from B to C or D to A, `dV=0` `:. dW= P.dV=0` |
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| 3632. |
The standared enthalpies of formation at `298K` for `C C1(g), H_(2)O(g), CO_(2)(g)` and `HC1(g)` are `-106.7, -241.8, -393.7`, and `-92.5kJ mol^(-1)`, respectively. Calculate `DeltaH^(Theta)underset(298K)` for the reaction `C C1_(4)(g) +2H_(2)O(g) rarr CO_(2)(g) +4HCI_(g)` |
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Answer» Correct Answer - `-41.4 kcal` |
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| 3633. |
A cylinderical vessel made of thermally isulating material is divided into two equal parts by an insulating (moveable & smooth) piston. Both parts hae same ideal gas `((V_(p))/(C_(v))=(3)/(2))`. The gas in left part is supplied heat such that volume of right part becomes one fourth of initial value. `(PVT,PVT)` (1) In right part, the pressure becomes eight times of its inital value (2) In right part, the temperature becomes two times (3) In right part, work done by gas =3PV (4) In right part, work done by gas=-2PV |
| Answer» Correct Answer - 1,2,4 | |
| 3634. |
A reaction will never the spontaneous at any temperature and pressure ifA. `DeltaS =+ve, DeltaH =+ve`B. `DeltaS =+ve, DeltaH =- ve`C. `DeltaS =- ve, DeltaH =+ve`D. `DeltaS =-ve, DeltaH =- ve` |
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Answer» `DeltaG = DeltaH - T DeltaS` When `DeltaS =-ve` and `DeltaH =+ve. DeltaG` will always be `+ve`. Hence, reaction will be non-spontaneous. |
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| 3635. |
Which of the following reaction is expected never to be spontaneous :-A. `2O_(3)rarr 3O_(2)` `Delta H=-Ve, Delta S = +Ve`B. `Mg+H_(2)rarr MgH_(2)` `Delta H =-Ve, Delta S=-Ve`C. `Br_(2)(I)rarr Br_(2)(g)` `Delta H=-Ve, Delta S=-Ve`D. `2Ag+3N_(2)rarr2Ag N_(3)` `Delta H=+Ve, Delta S=-Ve` |
| Answer» Correct Answer - D | |
| 3636. |
In case of an ideal gas, Joule Thomson coefficient is -A. ZeroB. PositiveC. NegativeD. Infinite |
| Answer» Correct Answer - A | |
| 3637. |
The Joule. Thomson expansion of a gas is anA. Isothermal processB. Isochoric processC. Isoenthalpic processD. Isobaric process |
| Answer» Correct Answer - C | |
| 3638. |
`NH_(3)(g) + 3Cl_(2)(g) rarr NCl_(3)(g) + 3HCl(g), " "DeltaH_(1)` `N_(2)(g)+3H_(2)(g)rarr 2NH_(3)(g), " "DeltaH_(2)` `H_(2)(g)+ Cl_(2)(g) rarr 2HCl(g) , " " DeltaH_(3)` The heat of formation of `NCl_(3)` in the terms of `DeltaH_(1), DeltaH_(2) "and" DeltaH_(3)` isA. `DeltaH_(f)=-DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2) DeltaH_(3)`B. `DeltaH_(f)=DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2) DeltaH_(3)`C. `DeltaH_(f)=-DeltaH_(1)-(DeltaH_(2))/(2)-(3)/(2) DeltaH_(3)`D. None |
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Answer» Correct Answer - A `(1)/(2)N_(2)(g)+(3)/(2)Cl_(2)(g)rarr NCl_(3)(g)` `DeltaH=-DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)` |
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| 3639. |
Heat of neutralization of strong acid by a strong base is a constant value due to:A. salt formed does not hydrolysisB. only `H^(+)` and `OH^(-)` ions reactC. The strong base and strong acid react completelyD. The strong base and strong acid react in aqueous solution |
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Answer» Correct Answer - B In neutralization of strong acid and base only `H^(+)` and `OH^(-)` ions react in every case. So, heat of neutralization is remain constant and equal to `-13.7 kcal` (approximate). |
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| 3640. |
Equal volume of HCOOH and NaOH are mixed. If x is the heat of formation of water, then heat avolved of neutralization is:A. more than xB. equal to xC. twice of xD. less than x |
| Answer» Correct Answer - D | |
| 3641. |
A system is given 300 calories of heat and it does 600 joules of work. How much does the internal energy of the system change in this process (j=4.18joules/cal)A. `654`jouleB. `156.5`jouleC. `-300`jouleD. `-528.2`joule |
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Answer» Correct Answer - A |
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| 3642. |
At what temperature does the reduction of lead oxide to lead by carbon becomes spontaneous? `PbO +C rarr Pb(s) +CO(g)` For the reacion, `DeltaH` and `DeltaS` at `25^(@)C` are `108.4 kJ mol^(-1)` and `190 J K^(-1) mol^(-1)` respectively. |
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Answer» `DeltaG = DeltaH - T DeltaS = 0` or `T = (DeltaH)/(DeltaS) = (108.4)/(190xx10^(-3) = 570.5K` Reaction will be spontaneous at temperature higher than `570.5K`, |
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| 3643. |
Which of the following is true fo the reacion `H_(2)O(l)hArr H_(2)O(g)` at `100^(@)C` and 1 atmosphereA. `Delta S = 0`B. `Delta H = 0`C. `Delta H = Delta E`D. `Delta H = T Delta S` |
| Answer» Correct Answer - D | |
| 3644. |
The enthalpy change for a given reaction at `298 K` is `-x cal mol^(-1)`. If the reaction occurs spontaneously at `298 K`, the entropy change at that temperatureA. can be negative but numerically larger than `X//298`B. can be negative bu numerically smaller than `X//298`C. cannot be negativeD. cannot be positive |
| Answer» Correct Answer - B | |
| 3645. |
A reaction occurs spontaneously in forward direction ifA. `T Delta S gt Delta H` where `Delta H` is positive and `Delta S` is negative.B. `T Delta S = Delta H` where both `Delta H and Delta S` are positiveC. `T Delta S lt Delta H` where both `Delta H and Delta S` are positiveD. `T Delta S gt Delta H` where both `Delta H and Delta S` are positive |
| Answer» Correct Answer - D | |
| 3646. |
The enthalpy change for a given reaction at `298 K` is `-x cal mol^(-1)`. If the reaction occurs spontaneously at `298 K`, the entropy change at that temperatureA. Can be negative but numerically larger than `x//298 cal K^(-1)mol^(-1)`B. Can be negative but numerically smaller than `x//298 cal K^(-1)mol^(-1)`C. Cannot be negativeD. Cannot be positive |
| Answer» Correct Answer - B | |
| 3647. |
If an eddothermic reaction occurs spontaneously at constant temperature and pressure, then which of the following is true?A. `Delta S lt 0`B. `Delta S gt 0`C. `Delta H lt 0`D. `Delta G gt 0` |
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Answer» Correct Answer - B According to thermodynamics (at constant `T` and `P`) `Delta G = Delta H - T Delta S` For reaction to be spontaneous, `Delta G` should be negative `(lt 0)`. Since the reaction is endothermic, `Delta H` is positive `(gt 0)`. Thus, `T Delta S` should be positive and greater than `Delta S` must be positive `(gt o)`, |
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| 3648. |
The standard enthalpy of formation of `NH_(3)` is -46.0 kJ `mol^(-1)`. If the enthalpy of formation of `H_(2)` from its atoms is -436 kJ `mol^(-1)` and that of `N_(2)` is -712kJ `mol^(-1)`, the average bond enthalpy of N-H bond in `NH_(3)` is :-A. `-964 KJ mol^(-1)`B. `+352 KJ mol^(-1)`C. `+1056 KJ mol^(-1)`D. `-1102 KJ mol^(-1)` |
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Answer» Correct Answer - B `({:(N_(2)(g) + (3)/(2) H_(2)(g)rarr NH_(3),:,DeltaH_(f)^(@)=-46.0 "KJ mol"^(-1)),(2H(g)rarrH_(2)(g),,,Delta_(f)^(@)=-436 "KJ mol"^(-1)),(2N(g)rarrN_(2)(g),,,DeltaH_(f)^(@)=-712"KJ mol^(-1)),(NH_(3)(g) rarr (1)/(2)N_(2)(g)+(3)/(2)H_(2)(g),,,DeltaH =+46),((3)/(2)H_(2)rarr3H,,,DeltaH=+436xx(3)/(2)),((1)/(2)N_(2)rarrN,,,DeltaH=+712xx(1)/(2)):})/` `(NH_(3)(g) rarr N(g)+ 3H(g) " "," "DeltaH =+ 1056 "KJ mol^(-1)` Average bond enthaply of N-H bond `=(1056)/(3) =+352 "KJ mol^(-1)` |
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| 3649. |
Which of the following is an exothermic reaction?A. `N_(2(g))+O_(2(g))+180.8 kJ rarr 2NO_((g))`B. `N_(2(g))+3H_(2(g))-92kJ rarr 2NH_(3(g))`C. `C_("(graphite)")+H_(2)O_((g)) rarr CO_((g))+H_(2(g))-131.3 kJ`D. `C_("(graphite)")+2S_((s)) rarr CS_(2(l))-91.9 kJ` |
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Answer» Correct Answer - B `DeltaH=-ve`, exothermic |
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| 3650. |
Which of the following is an exothermic reaction?A. `H_(2(g))+Cl_(2(g)) rarr 2HCl_((g)), DeltaH=-184.6 kJ`B. `N_(2(g))+O_(2(g)) rarr 2NO_((g)), DeltaH=+180.8 kJ`C. `C_("(graphite)")+H_(2)O_((g)) rarr CO_(2)+H_(2(g))-131.4 kJ`D. `C_("(graphite)")+2S_((g))+91.9 kJ rarr CS_(2(l))` |
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Answer» Correct Answer - A `DeltaH=-ve`, exothermic |
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