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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.
| 2351. |
Assertion (A): Internal energy change in a cyclic process is zero. Reason (R ) : Internal energy is a state funciton.A. If both (A) and (R ) are correct, and (R ) is the correct explanation for (A).B. If the both (A) and (R ) are correct, but(R ) is not a correct explanation for (A).C. If (A) is correct, but (R ) is incorrect.D. If (A) is incorrect, but (R ) is correct. |
| Answer» For cyclic process, `DeltaU =0`. Energy is state funciton. | |
| 2352. |
Assertion (A): Internal energy change in a cyclic process is zero. Reason (R ) : Internal energy is a state funciton.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 2353. |
Assertion : The difference between `DeltaH and DeltaU` is not significant for systems consisting of only solids and/or liquids. Reason : Solids and liquids do not suffer any significant volume changes upon heating.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 2354. |
Assertion : The solubility of most salts in water increases with rise of temperature. Reason : For most of the ionic compounds, `Delta_("sol")H^(@)` is positive and the dissociation process is endothermic.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 2355. |
Assertion. Heat of neutralisation for both `HNO_(3)` and HCl with NaOH is53.7 kJ per mole. Reason. NaOH is a strong electrolyte`//` baseA. 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 - B | |
| 2356. |
Assertion : Heat of neutralisation of `HNO_(3)` and `NaOH` is same as that of HCl and KOH. Reason : Both `HNO_(3)` and HCl are storng acids and NaOH and KOH are strong bases.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 2357. |
Enthyalpy change during a reaction does not depend uponA. Conditions of a reactionB. Initial and final concentrationC. Physical states of reactants and productsD. Number of steps in the reaction |
| Answer» Correct Answer - D | |
| 2358. |
Correct expression among the following:A. `E = H +PV`B. `H = E+PV`C. `H =E-PV`D. `P=E+HV` |
| Answer» Correct Answer - B | |
| 2359. |
Which of the following reaction has `Delta H = Delta U`?A. `2 SO_(2) (g) + O_(2) (g) rarr 2 SO_(2) (g)`B. `H_(2) (g) + Cl_(2) (g) rarr 2 HCl (g)`C. `2 NH_(4) NO_(3) (s) rarr 2 N_(2) (g) + 4 H_(2) O (g) + O_(2) (g)`D. `C_(2) H_(5) OH (l) + 3 O_(2) (g) rarr 2 CO_(2) (g) + 3 H_(2) O (l)` |
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Answer» Correct Answer - B For any reaction. `Delta H = Delta U + Delta n_(g) Rt` `Delta H = Delta U` only when `Delta n_(g) = 0`, this is true for the second reaction. `Delta n = [ 2 mol HCl (G)] - [1 mol H_(2) (g) + 1 mol Cl_(2) (g)]` ` 2 mol - 2 mol = 0 mol` |
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| 2360. |
the change in internal energy of a ayatem depends onA. Initial and final states of the systemB. The path if reversibleC. The path if irreversibleD. Initial, final states and also on the path |
| Answer» Correct Answer - A | |
| 2361. |
For a reaction, `P+Q rarr R+S`. The value of `DeltaH^(@)` is `-"30 kJ mol"^(-1) and DeltaS" is "-"100 J K"^(-1)"mol"^(-1)`. At what temperature the reaction will be at equilibrium?A. `27^(@)C`B. `52^(@)C`C. `30^(@)C`D. `45^(@)C` |
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Answer» Correct Answer - A `DeltaG=DeltaH-TDeltaS,` At equilibrium, `DeltaG=0, DeltaH=TDeltaS` `30000=Txx100` `T="300 K or "27^(@)C` |
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| 2362. |
An adiabatic expansion of an ideal gas always hasA. decrease in temperatureB. `q = 0`C. `W = 0`D. `Delta H = 0` |
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Answer» Correct Answer - B A process is called adiabatic if the system does not exchange heat with the surrounding, i.e., `q = 0`. An adiabatic process may involve increase of decrease in temperature of the system. In an adiabatic expansion of an ideal gas, there is no change of temperature because there is no intermolecular interaction between its particles. |
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| 2363. |
Internal energy does not includeA. Vibrational energyB. Rotational energyC. Energy arising by gravitional pullD. Nuclear energy |
| Answer» Correct Answer - C | |
| 2364. |
(a) An endothermicreaction `A rarr B` proceeds to completion . Predict the sign of `DeltaS`. (b) What will be the sign of `DeltaS` for the reaction `: N_(2)(g) + O_(2)(g) rarr 2NO(g)` ? Given reasons in support of your answer. |
| Answer» Correct Answer - (a)Since it is feasible , so `DeltaG` is-ve. As it is endothermic `DeltaH ` is `+ve`.Hence for `DeltaG`to be `-ve, DeltaS` must be`+ve` . (b) `DeltaS`is `-ve` because there are two different reactant while there is only one product. | |
| 2365. |
One mole of a real gas is subjective to a process form `(2"bar",30lit,300k)` to `(2"bar",50lit,400k)` Given `C_(v)=40J//mol//K,C_(P)=50J//mol//K` Calculate `triangleU`.A. `4000J`B. `2000J`C. `1000J`D. `5000J` |
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Answer» Correct Answer - C |
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| 2366. |
An adiabatic expansion of an Ideal gas always hasA. Constant TemperatureB. `q=0`C. `w=0`D. `Delta H=0` |
| Answer» Correct Answer - B | |
| 2367. |
Which of the following reactions will have the value of `DeltaS` with a negative sign?A. `H_(2)O_((l))rarrH_(2)O_((g))`B. `2SO_(2(g))+O_(2(g))rarr2SO_(3(g))`C. `Cl_(2(g))rarr2Cl_((g))`D. `CaCO_(3(s))rarr CaO_((s))+CO_(2(g))` |
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Answer» Correct Answer - B `2SO_(2(g))+O_(2(g)) rarr 2SO_(3(g))` `Deltan_(g)=2-3=-1, DeltaS=-ve` |
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| 2368. |
A reaction is at equilibrium at `100^(@)C` and the enthalpy change for the reaction is `"42.6 kJ mol"^(-1)`. What will be the value of `DeltaS` in `"J K"^(-1)"mol"^(-1)`?A. 120B. 426.2C. 373.1D. 114.2 |
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Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS` At equilibiurm, `DeltaS=(DeltaH)/(T)=("42600 J mol"^(-1))/("373 K")="114.2 J K"^(-1)"mol"^(-1)` |
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| 2369. |
An ideal gaseous sample at intial state `i(P_(0),V_(0),T_(0))` is allowed to expand to volume `2 V_(0)` using two different process, in the first process, thje equation of process is `2PV^(2)=K_(1)` and in the second process the equation of the process is `PV=K_(2)`. Then:A. Work done in the first process will be greater than work in second process (magnitude wise).B. The order of value of work done cannot be compared unless we know the value of `K_(1)` and `K_(2)`C. Value of work done (magnitude) in second process is greater in above expansion irrespective value of `K_(1) and K_(2)`D. Ist process is not possible. |
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Answer» Correct Answer - C |
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| 2370. |
For a cyclic process, the condition isA. `DeltaU=0`B. `Delta H=0`C. `DeltaU gt 0` and `Delta H gt 0`D. both `Delta U=0` and `Delta H=0` |
| Answer» Correct Answer - D | |
| 2371. |
A gaseous system changes from state `A(P_(1), V_(1), T_(1))` to `B(P_(2), V_(2) , T_(2))`, B to C `(P_(3), V_(3) T_(3))` and finally from C to A. The whole process may be calledA. Cyclic processB. Reversible processC. Isobaric processD. Spontaneous process |
| Answer» Correct Answer - A | |
| 2372. |
A process in which no heat change takes place is calledA. An isothermal processB. An adiabatic processC. An isobaric processD. An isochoric process |
| Answer» Correct Answer - B | |
| 2373. |
Which of the following statement is correct?A. Only internal energy is a state function but not workB. Only work is a state function but not internal energyC. Both internal energy and work are state functionsD. Neither internal energy nor work is a state function |
| Answer» Correct Answer - A | |
| 2374. |
Which one fo the following statement 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 |
| Answer» Correct Answer - A | |
| 2375. |
"Closed system" isA. Perfectly sealedB. Perfectly insulatedC. Both 1 & 2D. Neither insulated nor sealed |
| Answer» Correct Answer - A | |
| 2376. |
In a closed systemA. Energy is not exchangedB. Matter is exchangedC. Energy is only exchangedD. Energy and matter are exchanged |
| Answer» Correct Answer - C | |
| 2377. |
In open system, system and surroundings exchangeA. Energy onlyB. Matter onlyC. Energy and matterD. Neither energy nor matter |
| Answer» Correct Answer - C | |
| 2378. |
Hot water in a thermos flask is an example forA. Isolated systemB. Open systemC. Closed systemD. Adiabatic system |
| Answer» Correct Answer - A | |
| 2379. |
500 mL of0.1 M`H_(2)SO_(4)` was added into 1 L of 0.1 M NaOH solution. The heat evolved was x calories. If further 500 mL of `H_(2)SO_(4)` is added into the solution, now heat evolved will beA. `x cal`B. 2 x calC. zeroD. `x// 2 cal` |
| Answer» Correct Answer - c | |
| 2380. |
Find (in terms of a ) the amount of energy required to raise the temperature of a substgance from, 3 K to 5 K at constant pressure. At low temperatures, `C_(p)=aT^(3).` Express your answer after dividing by a. |
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Answer» Correct Answer - 136 |
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| 2381. |
Use the following data to answer the question below : Calculate the resonance energy of anthracene, s |
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Answer» Correct Answer - 84 |
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| 2382. |
What is the `S^(@)(OH^-)` (in cal/mol K) at `300 K` ? Given : `Komega(H_(2)O)=10^(14),` `DeltaH_("neut")(H^(+)+OH^(-))=-13.5"kcal"` Fill OMR excluiding decimaol places. |
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Answer» Correct Answer - 19 |
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| 2383. |
`C_(6)H_(6) rarr C_(6)H_(12)Delta H = -204 KJ` heat of hydrognentaion of each C=C bond in benzeneA. `-68 kJ`B. `-93kJ`C. `-56 kJ`D. `-88 kJ` |
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Answer» Correct Answer - A for these C=C bonds in benzene is -204 KJ for one C=C bond is ............? |
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| 2384. |
The molar specific heat of a gas at constant volume is 20.8 J/mol.k. Two moles of the gas are heated at constant volume so that the rise in temperature is 10 K. Find the heat supplied to the gas. |
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Answer» Data : Cv = 20.8 J/mol.K, n = 2, Tf – Ti = 10 K. The heat supplied to the gas, Q = nCv(Tf – Ti) = (2) (20.8) (10) J = 416J |
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| 2385. |
The molar specific heat of a gas at constant pressure is 29.11 J/mol.k. Two moles of the gas are heated at constant pressure so that the rise in temperature is 40 K. Find the heat supplied to the gas. |
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Answer» Data : Cp = 29.11 J/mol.K, n = 2, Tf – Ti = 40 K. The heat supplied to the gas, Q = nCp (Tf – Ti) = (2) (29.11) (40) J = 2329 J |
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| 2386. |
The molar specific heat of Ar at constant volume is 12.47 J/mol.K. Two moles of Ar are heated at constant pressure so that the rise in temperature is 20 K. Find the work done by the gas on its surroundings and the heat supplied to the gas. Take R = 8.314 J/mol.K. |
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Answer» Data : Cv = 12.47 j/mol.K, n = 2,Tf – Ti = 20 K, R = 8.314 J/mol.K 1. W = nR (Tf – Ti) = (2) (8.314) (20) J = 332.6 J This is the work done by the gas on its surroundings. 2. Q = nCv (Tf – Ti) + W = (2) (12.47) (20) + 332.6 = 498.8 + 332.6 = 831.4 J This is the heat supplied to the gas. |
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| 2387. |
For the reaction `A(g)+2B(s)toC(l)+D(g)," "Delta H=-40 kJ//"mole"` 10 mole of each A and B are mixed in a closed rigid container and allowed to react at 300 K. Calculate the heat transfer (in kJ) occurs in the process. [Use : `R=8.3 "J"//"K-mole"`] |
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Answer» Correct Answer - 200 |
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| 2388. |
One mole of an ideal gas is heated from 300 K to 700 K at constant pressure. The change ininternal energy of the gas for this process is 8 kJ. What would be the change in enthalpy (in kJ) for the same process? (`R=8 " J"//"mole" -K`) |
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Answer» Correct Answer - 2 |
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| 2389. |
`Delta H ` for the reaction `H-C-= N(g) + 2H_(2)(g) rarr H - underset(H) underset(|)overset(H ) overset(|)(C) - overset(H) overset(|)(N) - H (g)` is `- 150 kJ `. Calculate the bond energy of`C -= N`bond. [ Given bond energies of `C-H= 414 kJ mol^(-1), H-H= 435kJ mol^(-1), C-N = 293kJ mol^(-1), N-H =396 kJ mol^(-1)`] |
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Answer» Correct Answer - `893 kJ mol^(-1)` `Delta_(r)H^(@) = Sigma ` B.E. ( Reactants ) `-Sigma ` B.E. ( Produts) `- 150 = [ B.E. ( C-H) + B.E.(C-=N) + 2B.E.( H-H) ]-[3 xx B.E. (C-H) + B.E.(C-N) + 2 xx B.E. (N-H)]` |
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| 2390. |
Calculate the C-C bond energy from the following data `:` (i) 2 C ( graphite )`+3H_(2)(g) rarr C_(2)H_(6)(g), DeltaH =- 84.67 kJ ` (ii) C ( graphite ) `rarr C(g), Delta H = 716 .7 kJ ` (iii) `H_(2)(g) rarr 2H(g), Delta H = 435.9 kJ` Assume the C-H bond energy as 416 kJ |
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Answer» Correct Answer - `329.77kJ` Let us calculate `DeltaH` for the reaction, `C_(2)H_(6)(g),i.e., H - underset(H) underset(|) overset(H) overset(|) (C)-underset(H) underset(|) overset(H) overset(|) (C) -H(g) rarr 2C(g) +6H(g)` ` 2 xx `Eqn. (ii) `+3 xx ` Eqn. (iii) -Eqn. (i) gives `Delta _(r)H= 2825.77 kJ` If x is the bond energy of C-C bond,then `x+ 6 B.E. ( C-H)= 2825.77` `x +6 ( 416) = 2825.77` or `x=329.77kJ` |
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| 2391. |
For a process to be spontaneous :A. `(DeltaG_("system"))_(T,P)=0`B. `DeltaS_("system")+DeltaS_("surrounding")gt0`C. `DeltaS_("system")+DeltaS_("surrounding")lt0`D. `(DeltaG_("system"))_(T,P)lt0` |
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Answer» Correct Answer - B::D For spontaneous process `DeltaS_("system")+DeltaS_("Surrounding")gt0` `(DeltaG_("system"))_(T,P)lt0` |
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| 2392. |
`1`mole each of `CaCl_2, Al_4C_3, Mg_2C_3` reacts with excess water in separate open flasks work done during the dissolution shows the order:A. `CaC_(2)=Mg_(2)C_(3) lt Al_(4)C_(3)`B. `CaC_(2)=Mg_(2)C_(3)=Al_(4)C_(3)`C. `Mg_(2)C_(3) lt CaC_(2) lt Al_(4)C_(3)`D. `Mg_(2)C_(3) lt Al_(4)C_(3) lt CaC_(2)` |
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Answer» Correct Answer - A In `Al_(4)C_(3)`, there moles of `CH_(4)` are formed. |
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| 2393. |
`H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(l)` `BE (H-H) = x_(1), BE (O=O)=x_(2)` `BE(O=H)=x_(3)` Latent heat of vaporisation of water liquid into water vapour `=x_(4)`, then `Delta_(f)H` (heat of formation of liquid water) isA. `x_(1) +(x_(2))/(2) -x_(3) +x_(4)`B. `2x_(3)-x_(1) -(x_(2))/(2)-x_(4)`C. `x_(1)+(x_(2))/(2)-2x_(3)-x_(4)`D. `x_(1)+(x_(2))/(2)-2x_(3)+x_(4)` |
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Answer» `DeltaH = (BE)_("reactant") -(BE)_("products")` [But all the species must be in gaseous state. In product, `[H_(2)O(l) rarr H_(2)O(g)] DeltaH` must be added. Hence, `H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(l)` `DeltaH = [(BE)_(H-H)+(1)/(2)(BE)_(O=O)]` `= [(DeltaH)_(vap) +2(BE)_(O-H)]` `= x_(1) +(x_(2))/(2) -[x_(4)+2x_(3)]` `= x_(1) +(x_(2))/(2) -x_(4) - 2x_(3)` |
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| 2394. |
The change of energy on freezing `1.00 kg` of liquid water at `0^(@)C` and1 atmisA. `236 .7 kJ kg^(-1)`B. ` 333.4kJ kg^(-1)`C. ` - 333.4kJ kg^(-1)`D. `- 236 .7 kJ kg^(-1)` |
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Answer» Correct Answer - C Enthalpy of fusion of ice `= 6.03 kJ mol^(-1)` at `0^(@)C` `:.`Enthalpy of freezing of water `= - 6.03 kJmol^(-1) `at`0^(@)C` i.e.,.Enthalpy change on freezing of 18 g ofwater at `0^(@)C =- 6.03 kJ` `:.`Enthalpy change on freezing1000 g ofwter. `= - ( 6.03)/( 18) xx 1000 =- 335 kJ ` |
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| 2395. |
A boy after swimming comes out from a pool covered with a film of water weighing 80 g .How much heat must be supplied to evaporate this water ? `(Delta_(v) H^(@)=40.79 kJmol^(-1))`A. `1.61 xx10^(2) kJ`B. `1.71 xx 10^(2)kJ`C. ` 1.81xx10^(2) kJ `D. ` 1.91 xx 10^(2) kJ ` |
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Answer» Correct Answer - C Given `: Delta_("vap") H^(@) `for`H_(2)O= 40.79 kJ mol^(-1)` Water to be evaporated `= 80 g ( 80)/( 18) mole` `=4.44` mole Heat required for evaporated of 1 moleof `H_(2)O =40.79kJ` `:. `Heat required for evaporated of 4.44 moleof`H_(2)O=40.79 xx 4.44 = 1.81 xx 10^(2) kJ` |
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| 2396. |
The standard molar entropy of `H_(2)O(l) ` is `70 JK^(-1) mol^(-1)`. Will the standard molar entropy of `H_(2)O(s)` be more, or less than `70 JK^(-1) mol^(-1)` ? |
| Answer» In ice, molecules of`H_(2)O` are less random than in liquid water. Hence , molar entropy of ice, `H_(2)O(s)`, will be less than that of liquid water, `H_(2)O(l)`. | |
| 2397. |
The entropy change for cooling 1.6 g of an organic compound (mol. Wt. 32, molar heat capacity at constant pressure, `59 JK^(-) mol^(-)`) from 1000 K to 800 K is :A. `-0.659 JK^(-1)`B. `+0.659 JK^(-1)`C. `-1.318 JK^(-1)`D. `+1.318 JK^(-1)` |
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Answer» Correct Answer - A `DeltaS=2.303 nC_(p)"log" T_(2)/T_(1)` |
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| 2398. |
Assertion (A): Enthalpy of graphite is lower than that of diamond. Reason (R ) : Entropy of graphite is lower than that of diamond.A. If both (A) and (R ) are correct, and (R ) is the correct explanation for (A).B. If the both (A) and (R ) are correct, but(R ) is not a correct explanation for (A).C. If (A) is correct, but (R ) is incorrect.D. If (A) is incorrect, but (R ) is correct. |
| Answer» Graphite is most stable of carbon, hence its energy is lower than that of diamond and entropy of graphite is also lower than that of diamond. | |
| 2399. |
A cylinder contains 0.50 mol of an ideal gas at temperature of 310 K. as the gas expands isothermally from an initial volume of 0.31 `m^(3)` to a final volume of 0.45`m^(3)`, find the amount of heat that must be added to the gas in order to maintain a constant temperature. |
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Answer» Since the process is isothermal, the internal energy of an ideal gas does not change, i.e., `DeltaU=0`. The relation `DeltaU=Q-W` gives `Q=W`, where W is positive (work is done by the system). Now, we know that the gas gains heat from the surrounding (positive Q). as the work done W is given by `int_(V_(i))^(V_(f))PdV`, we obtain W as `W=int_(V_(i))^(V_(f))PdV=int_(V_(i))^(V_(f))(nRT)/(V)dV=nRT int_(V_(i))^(V_(f))(dV)/(V)=nRT" In "V|_(V_(i))^(V_(f))=nRT" In "((V_(f))/(V_(i)))` Substituting `n=0.50mol,R=8.31J//(mol" "K),V_(f)=0.45m^(3) and V_(i)=0.31m^(3)`, we obatain Q=W=480J. |
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| 2400. |
An ideal gas is made to undergo a termodynamic process given by `V prop T^(2)` , find the molar heat capacity of the gas for the above process.A. `(R)/((gamma-1))`B. `(gammaR)/((gamma-1))`C. `((2gamma-1)/(gamma-1))R`D. `((2gamma-1)/(gamma+1))R` |
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Answer» Correct Answer - C Given: `VpropT^(2)` `V=KT^(2)` where `k` is constant. Differentating both sides, of the above equation, we have `dV=2kTdT or P=(nR)/(kT) dW=PdV` `=(nR)/(kT)2kTdT=2nRdT(dW)/(ndT)=2R` The molar heat capacity of the gas for the process will be `C=C_(v)+W=(R)/(gamma-1)+2R=((2gamma-1)/(gamma-1))R` |
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