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.
| 1401. |
Explain the term free expansion of a gas. |
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Answer» When a balloon is ruptured suddenly, or a tyre is punctured suddenly, the air inside the balloon/ tyre rushes out rapidly to the atmosphere. This process (expansion of air inside the balloon/tyre) is so quick that there is no time for transfer of heat from the system to the surroundings or from the surroundings to the system. Such an adiabatic expansion is called free expansion. It is characterized by Q = W = 0, implying ∆ U = 0. Free expansion is an uncontrolled change and the system is not in thermodynamic equilibrium. Free expansion cannot be illustrated with a P-V diagram as only the initial state and final state are known. |
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| 1402. |
What is the critical temperature of water? |
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Answer» The critical temperature of water is 374.1°C. |
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| 1403. |
What is heat ? |
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Answer» Heat is a type of energy which travels from a hotter body to a cooler one. |
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| 1404. |
Differentiate between reversible and irreversible process. |
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Answer» Reversible Process :
Irreversible Process :
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| 1405. |
What is the principle of a pressure cooker? |
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Answer» The pressure cooker is based on the principle that the B.P. of water increase with increase in pressure. |
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| 1406. |
If two systems A and B are in thermal equilibrium with C, then will A and B also be in thermal equilibrium? |
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Answer» Yes; according to Zeroth law of Thermodynamics. |
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| 1407. |
On which law is the first law of thermodynamics based on? |
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Answer» Law of conservation of energy. |
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| 1408. |
Write the P-V relation for adiabatic expansion of ideal gas. |
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Answer» PVγ = constant, where γ = \(\frac { { C }_{ p } }{ { C }_{ v } } \) which is known as atomicity of gas |
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| 1409. |
What is the standed enthalpy of formation of MgO(s) is 300.9 Kj is evoled when 20.15 g of MgO(s) is formaed by the combustion when magnesium under standerd conditions?A. `-597.3KJ mol^(-1)`B. `-300.9KJ mol^(-1)`C. `+300.9KJ mol^(-1)`D. `+597.3KJ mol^(-1)` |
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Answer» Correct Answer - a |
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| 1410. |
What are the steps through which a refrigerant goes in one complete cycle of refrigeration ? |
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Answer» In one complete cycle of refrigeration, the refrigerant, a liquid such as fluorinated hydrocarbon, goes through the following steps : 1. The refrigerant in the closed tube passes through the nozzle and expands, into a low-pressure area. This adiabatic expansion results in reduction in pressure and temperature of the fluid and the fluid turns into a gas 2. The cold gas is in thermal contact with the inner compartment of the fridge. Here it absorbs heat at constant pressure from the contents of the fridge. 3. The gas passes to a compressor where it does work in adiabatic compression. This raises its temperature and converts it back into a liquid. 4. The hot liquid passes through the coils on the outside of the refrigerator and releases heat to the air outside in an isobaric compression process. The compressor, driven by an external source of energy, does work on the refrigerant during each cycle. |
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| 1411. |
What is refrigeration? |
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Answer» Refrigeration is artificial cooling of a space or substance of a system and/or maintaining its temperature below the ambient temperature. |
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| 1412. |
The thermal efficiency of a heat engine is 25%. If in one cycle the heat absorbed from the hot reservoir is 50000 J, what is the heat rejected to the cold reservoir in one cycle ? |
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Answer» Data : η = 25% = 0.25, QH = 50000 J W η = \(\frac{W}{Q_H}\) ∴ W = ηQH = (0.25)(50000)J = 12500J Now, W = QH – |QC| ∴ |QC| = QH – W = (50000 – 12500) J = 37500J This is the heat rejected to the cold reservoir in one cycle. [Notes : QC = – 37500 J] |
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| 1413. |
Which Statement is always true for an exothermic reaction ?A. The enthalpy change is negative .B. The entropy change is negative .C. The reaction absorbs heat from the surroundings.D. The reaction is spontaneous. |
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Answer» Correct Answer - a |
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| 1414. |
When acids an bases react they liberate some amount of energy which is represented as `DeltaH` neutralization if one equivalent of acid and base reacts. The energy liberated depends on the concentration of acid and base and also on whether the acid is strong or weak. For strong acids and bases, the value is constant and for weaker acids as bases the magnitude is generally lesser. Using this info and the date below answer the question that follow : [Given data : `Delta_("neutralization")HCl//NaOH = - 14 Kcal//eq` (at infinite dil.) `DeltaH_("ionisation")CH_(3)COOH = 2kcal//mol` `DeltaH_("ionisation")NH_(4)OH = 3kcal//eq]` In which of the following will heat released be same ?A. `HCl (0.2 M, 500 ml)`, mixed with `NaOH (0.3 M, 500 ml)`B. `HNO_(3)(0.5M, 200 ml)` mixed with `KOH (0.4 M, 280 ml)`C. `H_(2)SO_(4)(0.1 M, 200 ml)` mixed with `NaOH (0.3 M, 1000 ml)`D. `CH_(3)COOH (0.1 M, 1000 ml)` mixed with `NaOH (0.3 M, 500 ml)` |
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Answer» Correct Answer - a,b |
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| 1415. |
It is observed that when acid and bases react, some energy is released due to neutralization reaction. While in case of strong acid and bases, the energy released in greater, in case of weaker acids of bases, energy released will be lesser. The difference is attributed to energy required for ionization. From the above information and data below answer the questions that follow: `Delta H_("neutrization SA/SB") `=-57.5 kJ/equivalent, `DeltaH_("ionization") of NH_(4)OH`=10kJ/mole, `DeltaH_("ionization") " of " H_(2)C_(2)O_(4)`=20kJ/ mole. What will be the energy released when 500ml of0.1 M `H_(2)SO_(4)` solution reacts with 1L of 0.1 M `NH_(4)OH` solutioni if the base is 30% ionized in the given solution?A. 5.75kJB. 4.75kJC. 5.05kJD. 5.45kJ |
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Answer» Correct Answer - c |
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| 1416. |
It is observed that when acid and bases react, some energy is released due to neutralization reaction. While in case of strong acid and bases, the energy released in greater, in case of weaker acids of bases, energy released will be lesser. The difference is attributed to energy required for ionization. From the above information and data below answer the questions that follow: `Delta H_("neutrization SA/SB") `=-57.5 kJ/equivalent, `DeltaH_("ionization") of NH_(4)OH`=10kJ/mole, `DeltaH_("ionization") " of " H_(2)C_(2)O_(4)`=20kJ/ mole. What will be final temperature attained if all the heat release in neutralization of 1 L of 0.2 M `NH_(4)OH` with 2 L of 0.1 M HCl increase the temperature of the final solution having density 0.95gm/ ml and specific heat capacity=`(1)/(3) J//gm^(@)"C"` if original temperature was `27^(@)"C"`? Assume weak base to be completely unionized.A. `310^(@)C`B. 300KC. 310KD. 290K |
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Answer» Correct Answer - c |
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| 1417. |
It is observed that when acid and bases react, some energy is released due to neutralization reaction. While in case of strong acid and bases, the energy released in greater, in case of weaker acids of bases, energy released will be lesser. The difference is attributed to energy required for ionization. From the above information and data below answer the questions that follow: `Delta H_("neutrization SA/SB") `=-57.5 kJ/equivalent, `DeltaH_("ionization") of NH_(4)OH`=10kJ/mole, `DeltaH_("ionization") " of " H_(2)C_(2)O_(4)`=20kJ/ mole. What will be `DeltaH_("neutralization") " of " H_(2)C_(2)O_(4)` with NaOH per mole of `H_(2)C_(2)O_(4)` if acid is assumed to be completely unionized in the following?A. 37.5kJ/moleB. `-75kJ//"mole"`C. `95kJ//"mole"`D. `-3507kJ//"mole"` |
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Answer» Correct Answer - c |
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| 1418. |
What is the relationship among the magnitudes of the enthalpies of combustion `(DeltaH_(comb))`, fusion `(DeltaH_(fus))` and vaporization `(DeltaH_(vap))` for a hydrocarbon such as hexane,`C_(6)H_(14)`?A. `DeltaH_(fus)ltDeltaH_(fus)lt DeltaH_(vap)`B. `DeltaH_(vap)ltDeltaH_(fus)lt DeltaH_(comb)`C. `DeltaH_(comb)ltDeltaH_(vap)lt DeltaH_(fus)`D. `DeltaH_(fus)ltDeltaH_(vap)lt DeltaH_(comb)` |
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Answer» Correct Answer - d |
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| 1419. |
For a process to be spontaneous, `DeltaG` must be………… .A. `(DeltaG)_("system")` must be negativeB. `(DeltaG)_("system")` must be positiveC. `(DeltaS)_("system")` must be positiveD. `(DeltaS)_("system")` must be negative |
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Answer» Correct Answer - A For spontaneous reaction `(DeltaG_("system"))_(T,P)=-ve` |
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| 1420. |
The reaction of cyanamide, `NH_(2)CN(s)`, with dioxygen was carried out in a bomb calorimeter, and `DeltaU` was found to be `-742.7 kJ mol^(-1)` at `298 K`. Calculate enthalpy change for the reaction at `298 K`. `NH_(2)CN(g)+3/2O_(2)(g) rarr N_(2)(g)+CO_(2)(g)+H_(2)O(l)` |
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Answer» Correct Answer - `-743.939 kJ` Enthalyp change for a raction `(DeltaH)` is given by the expression, `DeltaH=DeltaU+Deltan_(g)RT` Where, `DeltaU`= change in internal energy `Deltan_(g)` = change in number of moles For the given reaction, `Deltan_(g)=sumn_(g) ("products")-sumn_(g)("reactants")` `=(2-2.5)` moles `Deltan_(g)=-0.5` moles And, `DeltaU=-742.7 kJ"mol"^(-1)` T=298 K `R=8.314xx10^(-3)kJ "mol"^(-1)` Substituting the values in the expression of `DeltaH :` `DeltaH=(-742.7 kJ "mol"^(-1))+(-0.5 "mol")(298 K) (8.314xx10^(-3)KJ "mol"^(1)K^(-1))` `=-742.7-1.2` `DeltaH=-743.9 kJ "mol"^(-1)` |
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| 1421. |
Which of the following is correct option for free expansion of an ideal gas under adiabatic condition?A. `q=0,DeltaT lt0,w cancel(=) 0`B. `q=0, DeltaTcancel(=) 0,w=0`C. `qcancel(=) 0,DeltaT , w=0`D. `q=0,DeltaT=0,w=0` |
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Answer» Correct Answer - D For adiabatic expansion, `q=0` For free expansion, `P_(ext) = 0`. Hence,`w-0` `DeltaU = q+w=0+0=0`. As U is a function of temperature `( U = (3)/(2) nRT)` and as `DeltaU = 0, :.DeltaT =0` |
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| 1422. |
The reaction of cynamide, `NH_(2)CN(s)`, with dioxygen was carried out in a bomb calorimeter, and `DeltaU` was found to be `- 742.7 k J mol^(-1)` at 298 K. Calculate the enthalpy change of the reaction at 298 K `NH_(2)CN(s) + (3)/(2)O_(2)(g)rarrN_(2)(g)+CO_(2)(g) +H_(2)O(l)` |
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Answer» `Deltan_(g)=(n_(p) -n_(p))_(g)= 2-(3)/(2) = +(1)/(2)` mol `DeltaH =DeltaU +Deltan_(g) RT = - 742.7 kJ mol^(-1) +(+(1)/(2)mol)(8.314xx10^(-3)kJ K^(-1) mol^(-1))(298K) ` `= - 742.7 +1.2kJ mol^(-1) = - 741.5 kJ mol^(-1)` |
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| 1423. |
Standard heat of formation at `298K` is arbitraily taken to be zero forA. Liquid bromineB. Gaseous bromine atomsC. Gaseous bromine moleculesD. Solid bromine |
| Answer» Liquid bromine `[Br_(2)(l)]` is standard state of bromine. | |
| 1424. |
The molar heat of formation of `NH_(4)NO_(3)(s)` is `-367.54 kJ ` and those of `N_(2)O(g)`and `H_(2)O(l) ` are `+ 81.46kJ ` and `- 285.78 kJ` respectively at `25^(@)C`and 1.0 atmospheric pressure. Calculate `DeltaH ` and `Delta U` for the reaction `NH_(4)NO_(3)(s) rarr N_(2)O(g) + 2H_(2)O(l)` |
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Answer» Correct Answer - `DeltaH = -122.56;Delta U = - 125. -4 kJ ` `DeltaH_("reaction") =Sigma Delta_(f) H (` Products ) `- Sigma Delta_(f) H(` Reactants) `Deltan_(g) = n_(p) - n_(r) = 1-0 = 1 ` ` Delta U = Delta H - Delta n_(g)RT` |
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| 1425. |
The molar heat of formation of `NH_(4)NO_(3)(g)` is -367.54kJ and those of `N_(2)O(g)` and `H_(2)O(l)` are `+81.46kJ` and -285.78 kJ respectively at `25^(@)C` and 1.0 atmospheric pressure. Calculate `Delta H` and `DeltaU` for the reaction. |
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Answer» Correct Answer - `DeltaH = - 122. 56 kJ mol^(-1) , DeltaU = - 125. 04kJ mol^(-1)` `NH_(4)NO_(3)(s) rarrN_(2)O(g) + 2H_(2)O(l), DeltaH _("reaction") = SigmaDelta_(f) H (` Products `) -SigmaDelta_(f) H (` Reactants) `Deltan_(g) =n_(p)- n_(r) = 1-0 = 1` `DeltaU = DeltaH - Delta n_(g) RT` |
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| 1426. |
The factor of `DeltaG` values is importantin metallurgy.The `DeltaG` values for the following reactions at `800^(@)C` are given as `S_(2)(g) + 2O_(2)(g) rarr 2SO_(2)(g), DeltaG = - 544KkJ` `2Zn(s) + S_(2)(g) rarr2ZnS(s),DeltaG= - 293kJ` `2Zn(s) +O_(2) (g) rarr2ZnO(s),DeltaG = - 480 kJ ` The `DeltaG` for the reaction `2ZnS(s) + 3O_(2)(g) rarr2ZnO9g) +2SO_(2)(g)` will beA. `-731kJ`B. `- 787 kJ`C. `- 534kJ`D. `- 554kJ` |
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Answer» Correct Answer - a From the given reactions, `Delta_(f)G^(@) (SO_(2)) = - ( 544)?( 2) = - 272 kJ mol^(-1)` `Delta_(f)G^(@) ( ZnS) = - ( 293)/( 2) = - 146 .5kJmol^(-1)` `Delta_(f)G^(@) (ZnO)= - ( 480)/( 2) = - 240 kJ mol^(-1)` For the required reaction `Delta_("reaction") G^(@) = SigmaDelta_(f)G^(@) (`Products) - `Delta_(f)G^(@) (` Reactants) ` [2 Delta_(f)G^(@) (ZnO) + 2Delta_(f)G^(@) ( SO_(2))]- [2Delta_(f)G^(@) ( ZnS) + 3 xx 0]` `= - 480 + ( - 544) - ( - 293) = - 731kJ` |
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| 1427. |
Calculate heat of combusion of ethene: |
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Answer» `DeltaH^(Theta) =` Sum of bond energies of reactants -Sum of bond energies of products `= [DeltaH_((C=C))+4DeltaH_((C-H)) +3xxDeltaH_((O=O))]` `-[4 xx DeltaH_((C=O)) +4 xx DeltaH_((O-H))]` `=[619 +4 xx 414 +3 xx 499] -[4 xx 724 +4 xx 460]` `=- 964 kJ mol^(-1)` |
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| 1428. |
Distinguish between exothermic and endothermic reactions. |
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Answer»
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| 1429. |
Define ‘standard enthalpy of formation’. |
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Answer» Standard enthalpy of formation of a compound is the change in enthalpy. when one mole of the compound is formed from its elements in their standard states under standard conditions i.e., at 298 K and 101.3 kPa pressure. |
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| 1430. |
Heat of reaction depond upon:A. Physical state of reactants and productsB. Whether the reaction is carried out at constant pressure or at constant volumeC. Method by which the final products are obtained from the reactantsD. Temperature of the reaction |
| Answer» Correct Answer - A::B::D | |
| 1431. |
Select the correct statenents :A. All conbustion reactuions are exothermicB. heat of combustion are always exothermicC. `N_(2) + O_(2) rarr 2NO , DeltaH=+ve`D. `F_(2) +(1)/(2) O_(2) rarr F_(2)O , DeltaH =+ve` |
| Answer» Correct Answer - B::C::D | |
| 1432. |
For which of these is `DeltaH_(f)^(@)` not equal to zone?A. `Br_(2)(l)`B. `Fe(S)`C. `I_(2)(s)`D. `O_(3)(g)` |
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Answer» Correct Answer - d |
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| 1433. |
Thermodynamics mainly deals with:A. interrelation of various forms of energy and their transformation from one form to anotherB. energy changes in the processes which depends ony on initial and final states of the microscopoic systems containing a few moleculesC. how and ay what rate these energy transformation are carried outD. the system in equilibrium state or moving from one equilibrium state to another equilibrium state |
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Answer» Correct Answer - A::D Thermodyanmics deals with interrelation of various from of energy and their transformation into each other. It also deals with thermal or mechanical equilibrium. However, it does not tell any thing about the rate of reaction |
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| 1434. |
The second law of thermodynamics deals with transfer of: (A) work done (B) energy (C) momentum (D) mass |
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Answer» Correct answer is: (B) energy |
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| 1435. |
In the following questions morethan one correlation is possible between options of both columns. Match the following `:` `{:(,"Reaction",,"Entropy change"),((i) ,"A liquid vaporises",(a),DeltaS=0),((ii),"Reaction is non-spontaneous at all temperatures and " DeltaH " is positive",(b),DeltaS="positive"),((iii),"Reversible expansion of an ideal gas ",(c ) , DeltaS = "negative"):}` |
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Answer» Correct Answer - `(i)-(b),(ii)-(c ),(iii)-(a)` (i) When a liquid vaporises, i.e.,liquid `rarr` vapour, entropy increases, i.e.,`DeltaS = + ve`. Hence, `(i) rarr(b)`. (ii) When `DeltaH = +ve. i.e.,` energy factor oppose . The process is non-spontaneous at all temperature if entropy factor also opposes the process, i.e., `DeltaS = - ve` . Hence,`(ii) rarr (c )` (iii) In the reversible expansion of an ideal gas, the system remains in equilibrium at every stage. Hence, `DeltaS =0, i.e., (iii) rarr (a)`. |
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| 1436. |
The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of`10 dm^(3)` to a volume of 100 `dm^(3)` at `27^(@)` is `:`A. `35.8 J mol^(-1)K^(-1)`B. ` 32.3J mol^(-1) K^(-1)`C. ` 42.3 J mol^(-1) K^(-1)`D. `38.3 J mol^(-1) K^(-1)` |
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Answer» Correct Answer - C For isothermal reversible expansion of an ideal gas, `DeltaS = nR ln. (V_(2))/(V_(1))=2 xx 8.314 xx 2.303 log. ( 100)/(10)` `= 38.3 J mol^(-1) K^(-1)` |
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| 1437. |
In the following questions morethan one correlation is possible between options of both columns. Match the following `:` `Delta`( Parameters) Description `{:(,Delta_(r)H^(c-),Delta_(r)S^(c-),Delta_(r)G^(c-),,),((i),+,-,+,(a),"None-spontaneous at high temperature"),((ii),-,-,+ "at high T",(b),"Spontaneous at all temperatures"),((iii),-,+,-,(c ),"Non-spontaneous at all temperatures"):}` |
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Answer» Correct Answer - `(i)-(c ),(ii),-(a),(iii)-(b)` (i) When `Delta_(r)G^(@)` is `+ve` reaction is non-spontaneous at all temperature, i.e., `(i) rarr (a)` (ii) `Delta_(r)G^(@) = + ve` at high T means the reaction is non-spontaneous at high temperature, i.e., (ii) `rarr` (a) . (iii) `Delta_(r) H^(@) = -ve` means it favours ,`Delta_(r)S^(@) = + ve`means it also favours. `Delta_(r)G^(@) = -ve` means reaction is spontaneous at all temperatures, i.e, (iii) `rarr`(a). |
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| 1438. |
If water kept in an insultated vessel at `- 10^(@)C` suddenly freezes, the entropy change of the systemA. decreasesB. increasesC. is zeroD. equals to that of the surroundings. |
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Answer» Correct Answer - C As the system is insultatedfrom the surroundings, no heat enters or leaves the system. Hence,`DeltaS= q//T =0`. |
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| 1439. |
Predict which of the following reactions(s) has a positive entropy change?A. `N_(2)(g)+3H_(2)(g)to2NH_(3)(g)`B. `HCl(g)+NH_(3)(g)to2NH_(4)Cl(g)`C. `"Graphite(s)" to "diamonds(s)"`D. `F_(2)(g) to2F(g)` |
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Answer» Correct Answer - D |
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| 1440. |
Select correct statements:A. `int(dq)/(T) "is always independent of path followed between two points".`B. Entropy of universe is conserved.C. For a process in isolated system, entropy either increases or remains contantD. Enrtropy decreases with increase in temperature at constant V. |
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Answer» Correct Answer - C |
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| 1441. |
An athelete is given 100 g of glucose `(C_(6)H_(12)O_(6))` of energy equivalent to 1560 kJ. He utilizes `50%` of this gained energy in the event. In order to avoid storage of energy in body , the weight of water he would need to perspire is : (The enthalpy of evaporation of water is 44 kJ`//`mole)A. 319 gB. 422 gC. 293 gD. 378 g |
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Answer» Correct Answer - A |
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| 1442. |
What is bond energy ? Why is it called enthalpy of atomisation ? |
| Answer» Bond energy is the amount of energy required to dissociate one moleof bonds present between the atoms in the gaesous molecules. As the molecules dissociate completely into atoms in the gaseous state, therefore, bond energy of a diatomic molecule is called enthalpy of atomisation. | |
| 1443. |
Given that `C+O_(2) rarr CO_(2), Delta H^(@) = - x kJ` `2CO+ O_(2) rarr 2CO_(2) ,Delta H ^(@) = - y kJ` The enthalpy of formation of carbon monoxide will beA. `y - 2x`B. `( 2x-y)/(2)`C. `(y-2x)/(2)`D. `2x-y` |
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Answer» Correct Answer - c Aim`: C+ (1)/(2)O_(2) rarrCO,`Eqn,`(i) - (1)/(2) Eqn. (ii). ` |
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| 1444. |
Two reactions are given below`:` (i) `H_(2)O(g)+(1)/(2)O_(2)(g) rarrH_(2)O(l)` (ii) `H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)O(g)` Which one will have greater value of `DeltaH` and why ? |
| Answer» Reaction(i) will have greater value of `DeltaH`because when `H_(2)O(g)` condenses to form`H_(2)O(l)` , some heatis evolved ( called heat of condensation). | |
| 1445. |
Write an expression in the form of a chemical equation for standard enthalpy of formation,`Delta_(f)H^(@)` for CO. |
| Answer» `C(s)+(1)/(2)O_(2)(g)rarrCO(g) , Delta_(r)H=Delta_(f)H^(@)` | |
| 1446. |
`14 g` oxygen at `0^(@)C` and `10 atm` is subjected to reversible adiabatic expasnion to a pressure of `1atm`. Calculate the work done in a. Litre atomsphere. b. Caloride (given, `C_(P)//C_(V) = 1.4)`. |
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Answer» `P_(1) = 10atm` at `T = 273 K` for `(14)/(32) mol O_(2)`. `P_(2) atm` at `T = T_(2)` for `(14)/(32) mol O^(2)`. For adiabatic expansion, we have `T^(gamma).P^(1-gamma) =` constant `:. ((T_(1))/(T_(2)))^(gamma) = ((P_(2))/(P_(1)))^(1-gamma)` or `gamma "log" (T_(1))/(T_(2)) - (1-gamma) "log"(P_(2))/(P_(1))` or `1.4 "log"(273)/(T_(2)) = (1-1.4)"log" (1)/(10) rArr T_(2) = 141.14K` Work done in adiabatic expansion `= (nR)/((gamma -1)) (T_(2) - T_(1))` In `L-atm = (14)/(32) xx (0.0821(141.4-273))/((1.4-1))` `W_(rev) = 11.82L-atm` In `cal = (14)/(32) xx (2xx(141.4-273))/((1.4-1))` `W_(rev) =- 287.88cal` |
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| 1447. |
Find the work done when one mole of the gas is expanded reversibly po and isothermally from 5 atm to 1e atm at `25^(@) C`. |
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Answer» Correct Answer - w =- 3.988 kJ `W = - 2.303 nRT log (P_(1))/(P_(2)) = - 3988 J` |
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| 1448. |
Explain: “The enthalpy of formation of nitric oxide is +90.4 kJ”. |
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Answer» 1/2 N2(g) + 1/2 O2(g) → ANO(g)ΔH = +90.4kJ When 1 mole of nitric oxide is formed from its elements 90.4 kJ of heat is absorbed. |
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| 1449. |
Write Gibb’s Helmholtz Equation. |
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Answer» Gibb’s Helmholtz Equation can be given as : |
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| 1450. |
What is the heat of neutralisation of NH4OH and HCl? |
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Answer» Enthalpy of neutralization of ammonium hydroxide (weak base) and hydrochloric acid (strong acid) is -51.5 kJ. |
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