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.
| 2201. |
`2NO_(2)(g)to N_(2)O_(4)(g)` `DeltaU_(f)^(@)[N_(2)O_(4)(g)]=2kcal//"moleand "DeltaU_(reaction)^(@)=-16kcal//mol,` then calculate `DeltaH_("formation" )^(@)=-16kcal//mol,` then calculate `DeltaH_("formation")^(@) of NO_(2)at 727^(@)C:`A. 9kcal/molB. 4.5kcal/molC. 8kcal/molD. 10kcal/mol |
|
Answer» Correct Answer - c |
|
| 2202. |
Which reaction has a positive `DeltaS_(reaction)^(@)? `A. `Ag^(+)(aq) +Br^(-)(aq) to AgBr(s) `B. `2C_(2)H_(6)(g) +3O_(2)(g) to 4CO_(2)(g) + 6H_(2)O(l)`C. `N_(2)(g) +2H_(2)(g) to N_(2)H_(4) (g) `D. `2H_(2)O_(2)(l) to 2H_(2)O(l)+ O_(2)(g) ` |
|
Answer» Correct Answer - D |
|
| 2203. |
when the substance `Na(s), Br_(2)(g), Br_(2)(l),NaBr(s)` are arranged in order of increasing molar entropy at `25^(@)C`, which order is correct?A. `Br_(2)(g), Br_(2)(l), ,Na(s), NaBr(s)`B. `Na(s), Br_(2)(g), Br_(2)(l),NaBr(s)`C. `Na(s), NaBr(s), Br_(2)(l),Br_(2)(g)`D. `NaBr(s), Br_(2)(g), Br_(2)(l),Na(s)` |
|
Answer» Correct Answer - C |
|
| 2204. |
Which of the following is not a state functionA. `DeltaS`B. `DeltaG`C. `DeltaH`D. `DeltaQ` |
| Answer» Correct Answer - D | |
| 2205. |
Which of the following is true for an adiabatic process:A. `DeltaH=0`B. `DeltaW=0`C. `DeltaQ=0`D. `DeltaV=0` |
| Answer» Correct Answer - C | |
| 2206. |
In view of the signs of `Delta_(r)G^(0)` for the following reactions `PbO_(2)+Pbrarr2PbO,Delta_(r)G^(0)lt0` `SnO_(2)+Snrarr2SnO,Delta_(r)G^(0)gt0` Which oxidation state are more characteristic for lead and tin?A. For lead +2 , for tin +2B. For lead +4 , for tin +4C. For lead +2 lead +2 , for tin +4D. For lead +4 , for tin +2 |
|
Answer» Correct Answer - C Negative `Delta_(r)G^(@)` value indicates that `+2` oxidation state is more stable for `Pb^(2)`. Also it is supported by inert pair effect that `+2` oxidation state is more stable for Pb and `+4` oxidation state is more stable for Sn. |
|
| 2207. |
A certain mass of gas expanded from ( 1L, 10 atm) to (4 L, 5 atm) against a constant external pressure of 1 atm. If initial temperature of gas is 300 K and the heat capacity of process is `50 J//^(@)C`. Then the enthalpy change during the process is : (`1"L" "atm" underline ~ 100 J`)A. `Delta H = 15 kJ`B. `Delta H = 15.7 kJ`C. `Delta H = 14.4 kJ`D. Delta H = 14.7 kJ` |
|
Answer» Correct Answer - B `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2)) rArr T_(2) 600 K, q = C. Delta T = 15 KJ` `w = P_("ext") Delta V = 1 (4 - 1) = 0.3KJ Delta U = 14.7 KJ` `Delta H = 14.7 + (5 xx 4 - 10 xx 1) xx 100 xx 10^(-3) = 15.7 KJ` |
|
| 2208. |
Match the column : (Given process does not include chemical reactionn and phase change) `{:(,"Column-I",,"Column-II"),((a),DeltaH=DeltaU+Delta(PV),(P),"Any matter undergong any process"),((b),DeltaH=n.C_(P)DeltaT,(q),"Isochoric process involving any substance"),((c),q=DeltaU,(r),"Ideal gas,under any process"),((d),DeltaH=DeltaU+nRDeltaT,(s),"Ideal gas under isothermal process"),(,,(t),"Anysubstance undergoing isobaric process"):}` |
|
Answer» Correct Answer - `(a-p,q,r,s,t);(b-r,s,t);(c-q);(d-r,s)` |
|
| 2209. |
A piece of copper of mass `m = 90 g` at a temperature `t_1 = 90 ^@C` was placed in a calorimeter in which ice of mass `50 g` was at a temperature `-3 ^@C`. Find the entropy increment of the piece of copper by the moment the thermal equilibrium is reached. |
|
Answer» `c =` specific heat of copper `= 0.39 (J)/(g. K)` Suppose all ice does not melt, then heat rejected `= 90 xx 0.39 (90 - 0) = 3159 J` Heat gained by ice `= 50 xx 2.09 xx 3 + x xx 333` Thus `x = 8.5 gm` The hypothesis is correct and final temperature will be `T = 273 K` Hence change in entropy of copper piece =`mc 1n (273)/(363) = - 10 J//K`. |
|
| 2210. |
In the vicinity of the triple point the saturated vapour pressure `p` of carbon dioxide depends on temperature `T` as log `p = 1 - b//T`, where `a` and `b` are constants. If `p` is expressed in atomspheres, then for the sublimation process`a = 9.05` and `b = 1.80 kK`, and for the vaporization process `a = 6.78` and `b = 1.31 kK`. Find : (a) temperature and pressure at the triple point , (b) the values of the specific latent heats of sublimation, vaporization, and melting in the vicinity of the triple point. |
|
Answer» (a) The equations of the transition lines are `log p = 9.05 - (1800)/(T)` : Solid gas =`6.78 - (1310)/(T)` : Liquid gas At the triple point they intersect. Thus `2.27 = (490)/(T_(tr))` or `T_(tr) = (490)/(2.27) = 216 K` corresponding `p_(t r)` is 5.14 atmosphere In the formula `log p = a - (b)/(T)`, we compare `b` with the corresponding term in the equation in `2.210`. Then `1n p = 1 xx 2.303 - (2.303 b)/(T)` So, `2.303 = (Mq)/( R)` or `q_("sublimation") = (2.303 xx 1800 xx 8.31)/(44) = 783 J//gm` `q_("liquid-gas") = (2.303 xx 1310 xx 8.31)/(44) = 570 J//gm` Finally `q_("solid - liquid") = 213 J//gm` on subtraction. |
|
| 2211. |
An engineer claims to have made an engine delivering 10 kW power with fuel consumption of 1 g/sec . The calorific value of the fuel is 2 kcal/g . Is the claim of the engineerA. ValidB. InvaildC. Depends of engine designD. Depends on the load |
|
Answer» Correct Answer - B |
|
| 2212. |
At very low temperatures the heat capacity of crystals os equal to `C = aT^3`, where `a` is a constant. Find the entropy of a crystal as a function of temperature in this temperature interval. |
|
Answer» We can take `S rarr 0 as T rarr 0` Then `S = int_0^T C (dT)/(T) = int _0^T aT^2 dT = (1)/(3) aT^3`. |
|
| 2213. |
One mole of an ideal gas with heat capacity `C_V` goes through a process in which its entropy `S` depends on `T` as `S = alpha//T`, where `alpha` is a constant. The gas temperature varies from `T_1` to `T_2` Find : (a) the molar heat capacity of the gas as function of its temperature , (b) the amount of heat transferred to the gas , ( c) the work performed by the gas. |
|
Answer» (a) `C = T (dS)/(dT) = - (alpha)/(T)` (b) `Q = int_(T_1)^(T_2) CdT = alpha 1n (T_1)/(T_2)` ( c) `W = Delta Q - Delta U = alpha 1n (T_1)/(T_2) + C_V (T_1 - T_2)` Since for an ideal gas `C_V` is constant and `Delta U = C_v(T_2 - T_1)` (`U` does not depend on `V`). |
|
| 2214. |
For one complete cycle of a thermodynamic process gas as shown in the P-V diagram, which of following correct? A. `DeltaE_("int")=0,Qlt0`B. `DeltaE_("int")=0,Qgt0`C. `DeltaE_("int")gt0, Qlt0`D. `DeltaE_("int")lt0, Qgt0` |
|
Answer» Correct Answer - A |
|
| 2215. |
For one complete cycle of a thermodynamic process gas as shown in the P-V diagram, which of following correct? A. `DeltaE_("int")=0,QgtO`B. `DeltaE_("int")=0,QltO`C. `DeltaE_("int")gt0,QltO`D. `DeltaE_("int")lt0,QgtO` |
|
Answer» Correct Answer - A |
|
| 2216. |
Select the correct statements for the equilibrium under standard conditions. `H_(2)O(s) hArr H_(2)O(l),DeltaS_(1)^(Theta)` `H_(2)O(l) hArr H_(2)O(v), Delta_(2)^(Theta)` `H_(2)O(s) hArr H_(2)O(v), DeltaS_(3)^(Theta)`A. `DeltaS_(1)^(Theta) gt DeltaS_(2)^(Theta)`B. `DeltaS_(2)^(Theta) gt gt gt DeltaS_(1)^(Theta)`C. `DeltaS_(3)^(Theta) gt DeltaS_(2)^(Theta)`D. `DeltaS_(3)^(Theta) gt DeltaS_(1)^(Theta)` |
|
Answer» `H_(2)O(s)` has more ordered arrangement. Also, `DeltaS^(Theta)_(2)` `= S_(H_(2)O)^(Theta)(l) - S_(H_(2)O)^(Theta)(l)` `DeltaS_(1)^(Theta) = S_(H_(2)O)^(Theta)(l) -S_(H_(2)O)^(Theta)(s)` `:. S_(H_(2)O(v))^(Theta)` is maximum and thus `DeltaS_(2)^(Theta) gt gt gt DeltaS_(1)^(Theta)`. |
|
| 2217. |
The equilibrium constant for the reaction given below is `2.0xx10^(-7)` at `300K.` Calculate the standard free energy change for the reaction,`PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`. Also, calculate the standard entropy change if `DeltaH^@=28.40kJmol^(-1)`. |
|
Answer» `DeltaG^(@)=-2.303xx8.314xx300xxlog[2.0xx10^(-7)]` `=+38479.8 J mol^(-1)` `+38.48 k J mol^(-1)` Also, `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `:. DeltaS^(@)=(DeltaH^(@)-DeltaG^(@))/T` `=(28.40-38.48)/300` `=-0.03356 k J=-33.6 J K mol^(-1)` |
|
| 2218. |
If `x` and `y` are arbitrary extensive variables, thenA. `(x +y)` is an extensive variable.B. `x//y` is an intensive variable.C. `dx//dy` is an intensive variableD. Both (b) and (c ). |
| Answer» The sum of two extensive properties is always extensive while the ratio of extensive properties is intensive and the derivation is also intensive. | |
| 2219. |
Calculate the Gibbs energy change for the reaction given below:`Zn_((s))+Cu_((s))^(2+)toCu_((s))+Zn_((aq.))^(2+)` Given that standard Gibbs energy (`G^@`) for `Zn_((aq.))^(2+),Cu_((aq.))^(2+)` are `-147.2 kJ mol^(-1)` and `65.0 kJ mol^(-1)` respectively. |
|
Answer» `DeltaG^(@)=SigmaG_("products")^(@)-SigmaG_("reaction")^(@)` `=[0+(-147.2)-(0+65.0)]` `=-212.2 kJ` |
|
| 2220. |
Which of the following reactions is associated with the most negative change in entropy?A. `2SO_(2)(g) + O_(2)(g) rarr 2SO_(3)(g)`B. `C_(2)H_(4)(g) + H_(2)(g) rarr C_(2)H_(6)(g)`C. `C("s, graphite") + O_(2) rarr CO_(2) (g)`D. `3C_(2)H_(3)(g) rarr C_(6)H_(6)(l)` |
| Answer» Correct Answer - D | |
| 2221. |
For the gas - phase decomposition , `PCI_(5)(g)overset(Delta)hArr PCI_(3)(g) + CI_(2)(g):`A. `DeltaH lt 0, DeltaS lt 0`B. `DeltaH gt 0, DeltaS gt 0`C. `DeltaH gt 0, DeltaS lt 0`D. `DeltaH lt 0, DeltaS gt 0` |
|
Answer» Correct Answer - B Reaction is endothermic and no. of moles increases on decomposition. |
|
| 2222. |
Choose the correct statement(s).A. During adiabatic expansion of an ideal gs ,magnitude of work obatained id equal to `DeltaH` of gas.B. For same change in temperature of ideal gas, more work done by system through adiabatic than irreversible process.C. During boiling of `H_(2)O` at 1 atm , `100^(@)` C kineticx energy of `H_(2)O` molecules increase .D. For same change in volume of ideal gas,decrease in internal energy is more in adiabatic expansion than in isobaric expnsion if expansion starts with same intial state. |
|
Answer» Correct Answer - b,d |
|
| 2223. |
Which of the following statement are false ?A. For every change in an isolated system `DeltaT=0`B. For evergy change in an isolated system `W=0`C. For evergy change in an isolated system that has no macroscopic change in kinetic or potential energy ,`DeltaV=0`D. the magnitude of the slpoe of an adiabatic curve is greater than the magnitude of the slpoe of an isothermal curve for the same values of pressue and volume |
|
Answer» Correct Answer - a,d |
|
| 2224. |
The combustion of 0.200 mol of liquid carbon disulphide, `CS_(2)` to give `CO_(2)(g) and SO_(2)(g)` releases 215 KJ of heat . What is `DeltaH_(F)^(@)` for `CS_(2)(l)` in `Kjxxmol^(-1)`? , `{:(DeltaH_(f)^(@),kj.mol^(-1)),(CO_(2)(g),-393.5),(SO_(2)(g),-296.8):}`A. 385B. 87.5C. `-385`D. `-475` |
|
Answer» Correct Answer - b |
|
| 2225. |
Given the thermochemical equations: `Br_(2)(l)+Fe(g)to2BrF(g) DeltaH^(@)=-188KJ` `Br_(2)(l)+3F_(2)(g)to2BrF_(3)(g) DeltaH^(@)=-768KJ` detemine `DeltaH^(@)` for the reaction `BrF(g)+F_(2)(g)toBrF_(3)(g)`A. `-956KJ`B. `-478KJ`C. `-5820KJ`D. `-290KJ` |
|
Answer» Correct Answer - d |
|
| 2226. |
Identify the correct statement for change of Gibbs energy for a system `(Delta_(sys)G)` at constant temperature and pressure:A. if `Delta_(sys)G =0`, the system is still moving in a particular direction.B. if `Delta_(sys)G lt 0`, the process is not spontaneousC. if `DeltaG gt 0`, the process is spontaneousD. if `DeltaG = 0`, the system has attained equilibrium |
| Answer» At equilibrium, `DeltaG = 0`. | |
| 2227. |
Select the correct statementA. There is a natural asymmetry between converting work to heat and converting heat to work.B. No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.C. For energy chemical reaction at equilibrium, standard change in Gibbs free energy is zero.D. At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy.. |
| Answer» `DeltaG =- ve` for spontaneous process and `DeltaG^(Theta) != 0` at equilibrium. Statement (b) represents Carnot theorem. | |
| 2228. |
For a reaction `2A(g)+2B(g) rarr 2C(g)+D(g)`. If heat of reaction at constant pressure is `-28` K.cal. The heat of reaction at constant volume at `27^(@)C` isA. `-27.4` K. calB. `27.4` K. calC. `28.4` K. calD. `28` K. cal |
|
Answer» Correct Answer - A `DeltaH=DeltaU+Delta nRT` |
|
| 2229. |
If `K lt 1` then the value of `Delta G^(@)` will beA. positiveB. negativeC. zeroD. 1 |
|
Answer» Correct Answer - A According to thermodynamics, `Delta G^(@) = - 2.303 RT log K` If `K lt 1.0`, then `log K` will be negative. Since `R` and `T` are positve quantities, `Delta_(r) G^(@)` will be positive. |
|
| 2230. |
If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride vapor, thenA. `Delta H gt Delta U`B. `Delta H = Delta U`C. `Delta H lt Delta U`D. No relationship |
| Answer» Correct Answer - C | |
| 2231. |
The molar heat capacity of water isA. `4.184 JK^(-1)`B. `75.3 JK^(-1)`C. `85 JK^(-1)`D. `1 JK^(-1)` |
|
Answer» Correct Answer - B `C_(P, m)=4.18xx18` |
|
| 2232. |
Enthalpies of formation of `K^(+)(aq), Cl^(-)(aq)` and `KCl(s)` are `- 60.0 ,- 40.0 ` and `- 104.0` kcal` mol^(-1)` .If one mole of KCl(s) is dissolved in large excess of water at `25^(@)C`, the enthalpy of solution will beA. `-4.0 kcal mol^(-1)`B. `+4.0 kcalmol^(-1)`C. `- 20.0 kcal mol^(-10`D. `-20.0 kcal mol^(-1)` |
| Answer» Correct Answer - b | |
| 2233. |
In which of the following entropy decreases? (a) melting of ice (b) evaporation of water (c) crystallization of sugar (d) dissolution of salt |
|
Answer» (c) crystallization of sugar |
|
| 2234. |
The SI unit of entropy is ……(a) Nm (b) Cal mol-1 (c) KJ mol-1 (d) JK-1 |
|
Answer» Answer: (d) JK-1 |
|
| 2235. |
Find the final value of a gram molecule of a gas after an isothermal expansion at `127^(@)C`, if the original volume is `400 c.c` Given amount of work done by a gram molecule of a gas during expansion is `2302.6 "joule", R= 8.3 "joule mole"^(-1)K^(-1)` |
|
Answer» Correct Answer - `800 c.c` `V_(2)= ?, T_(1)= 127+273= 400K, V_(1)= 400 c.c` `W=2302.6J , R= 8.3 "joule mole"^(-1)K^(-1)` As `W= 2.3026 RT "log"_(10)((V_(2))/(V_(1)))` `2302.6= 2.3026xx8.3xx400 "log"_(10)((V_(2))/(V_(1)))` `"log"_(10)((V_(2))/(V_(1)))= (2302.6)/(2.3026xx8.3xx400)` `=0.3012= log_(10)2` `:. (V_(2))/(V_(1))= 2, V_(2)= 2V_(1)= 2xx400= 800 c.c` |
|
| 2236. |
The process depicted by the equation. H2O(s) → H2O(I) ∆H = + 1.43 kcal represents(a) fusion (b) melting (c) evaporation (d) boiling |
|
Answer» Answer: (a) fusion |
|
| 2237. |
A thermodynamic state function is a quantity ……(a) used to determine heat changes (b) whose value is independent of path (c) used to determine pressure volume work (d) whose value depends on temperature only |
|
Answer» (b) whose value is independent of path |
|
| 2238. |
A heat engine absorbs 760 kJ heat from a source at 380kK . It rejects (1) 650 kJ, (ii) 560 kJ, (iii) 504 kJ of heat to sink at 280 K . State which of these represent a reversible , an irreversible and an impossible cycle . |
|
Answer» Correct Answer - (i) irreversible, (ii) reversible (iii) impossible `eta= 1 - (T_(L))/(T_(H)) = 26.31 %` `eta_("rev")gteta_("irr")` (i) `eta_(1)=(q_(H)-q_(L))/(q_(H))=14%" "("Irreversible")` (ii) `underset("Reversible")(eta_(2)= 26.31 %)` (iii) `underset("impossible")(eta_(3)= 33.68 %)` |
|
| 2239. |
The efficiency of a Carnot cycle is `1//6`. By lowering the temperature of sink by `65 K`, it increases to `1//3`. The initial and final temperature of the sink are |
| Answer» Correct Answer - `325K, 260K` | |
| 2240. |
`A` gas for which `gamma=1.5` is suddenly compressed to `1//4` th of the initial volume. Then the ratio of the final to initial pressure isA. `1:16`B. `1:8`C. `1:4`D. `8:1` |
|
Answer» Correct Answer - D |
|
| 2241. |
If a triatomic gas is heated at constant pressure, precentage of the heat energy which is used to increase the internal energy isA. `75%`B. `14%`C. `60%`D. `100%` |
|
Answer» Correct Answer - A `(dU)/(dQ)xx100=(1)/(gamma)xx100` |
|
| 2242. |
`0.1` moles of diatomic gas at `27^(@)C` is heated at constant pressure, so that the volume is doubled. If `R=2cal.mol^(-1)k^(-1)` , the work done isA. `150cal`B. `60cal`C. `40cal`D. `30cal` |
|
Answer» Correct Answer - B `(V_(1))/(V_(2))=(T_(1))/(T_(2)),dW=pdV=RdT` |
|
| 2243. |
Which relation is correct for isometric processA. `DeltaQ=DeltaU`B. `DeltaW=DeltaU`C. `DeltaQ=DeltaW`D. None of these |
|
Answer» Correct Answer - D |
|
| 2244. |
Which is incorrectA. In an isobaric process `Deltap=0`B. In an isochoric process,`DeltaW=0`C. In an isothermal process `DeltaT=0`D. In an isothermal process `DeltaQ=0` |
|
Answer» Correct Answer - D |
|
| 2245. |
What is meant by the free energy of a system? What will be the direction of chemical reaction when (i) ΔG = 0 (ii) ΔG > 0 (iii) ΔG < 0 |
|
Answer» Free energy is defined as energy which can be converted into useful work (i) ΔG = 0, the reaction will be in equilibrium (ii) ΔG > 0, the reaction will not take place (iii) ΔG < 0, the reaction will be spontaneous |
|
| 2246. |
What are the State of system & state function? |
|
Answer» State of system is described in terms of T, P, V etc. The property which depends only on state of system not upon path is called state function. e.g., P, V, T, E, H, S etc. |
|
| 2247. |
Define the terms Extensive & Intensive Properties. Give an examples. |
|
Answer» Properties which depends on quantity of matter called extensive properties. eg. mass, volume, heat capacity, enthalpy, entropy etc. The properties which do not depends on matter present depends upon nature of substance called Intensive properties. eg. T,P, density, refractive index, viscosity, bp, pH, mole fraction etc. |
|
| 2248. |
Explain the term Internal energy. |
|
Answer» The total energy with a system. i.e. U = Ee + En + Ec + Ep + Ek + ------ ∆U = U2 – U1 or UP – UR & U is state function and extensive properly. If U1 > U2 energy is released. |
|
| 2249. |
Of the following a correct statement isA. `Delta H` is positive for exothermic reactionB. `Delta H` is negative for endothermic reactionC. The heat of neutralization of strong acid with strong base is always the sameD. The enthalpy of fusion is negative |
| Answer» Correct Answer - C | |
| 2250. |
`DeltaH` is negative for the reactionA. Neutralization of `HCl` with `NaOH`B. Formation of `NO`C. Boiling of waterD. Dissociation of `NaCl` |
| Answer» Correct Answer - A | |