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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.
| 1751. |
Molar enthalpy of vaporisation of water at `1atm, 500 K " is" 1100 " cal"//"mole"`. If 2 mole of water is vapourised in a closed rigid container at 500 K then calculate the heat required. (Assume ideal behaviour of `H_(2)O(g))` |
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Answer» Correct Answer - 200 |
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| 1752. |
Which of the following are intensive properties? Boiling point Melting point Heat capacity Volume Enthalpy Mass Internal energy Molar volume |
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Answer» Correct Answer - 3 |
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| 1753. |
Which of the following is/are intensive properties? Surface tension, mass, volume, enthalpy, density. |
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Answer» Surface tension, density. |
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| 1754. |
Which of the following is an intensive property. Surface tension, mass, volume, enthalpy, on density? |
| Answer» Surface tention and density are intensive properties. | |
| 1755. |
Give one difference between isolated systems and closed systems. |
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Answer» Isolated system can neither exchange matter nor energy with the surroundings. Closed system can exchange energy but not matter with the surroundings. |
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| 1756. |
Which of the following is an intensive property: Surface tension, mass, volume, enthalpy, density? |
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Answer» Surface tension and density are intensive properties. |
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| 1757. |
What happens to the internal energy of the system if: (a) Work is done on the system? (b) Work is done by the system? |
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Answer» (a) If work is done on the system, internal energy will increase, (b) If work is done by the system, internal energy will decrease. |
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| 1758. |
State first law of thermodynamics. |
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Answer» It states that a system absorbs heat dQ and as a result the internal energy of the system changes by dU and the system a work dW, then dQ = dU + dW But, dW = pdV Hence, dQ = dV + pdV. |
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| 1759. |
Define isolated system. |
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Answer» A system is said to be an isolated system if it can neither exchange nor matter with its surroundings. |
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| 1760. |
What do you mean by a closed system? |
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Answer» A system is said to be closed system if it can exchange only energy (not matter) with its surroundings. |
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| 1761. |
Write down the equation of state for an ideal gas. |
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Answer» PV = μRT, where R is the universal gas constant. |
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| 1762. |
In a thermodynamical process, 300 J of heat is suppliced to a gas and the 200 J of work is done by the gas. What is the change in internal energy of the system? |
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Answer» There will be a net increase in internal energy of the system by 300 -200 = 100 J. |
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| 1763. |
What is meant by equation of sate ? |
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Answer» The relation between the thermodynamic variables (p, V. T) of the system is called equation of state. |
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| 1764. |
How is efficiency of carnot engine affected by the nature of working substance? |
| Answer» The efficiency of carnot engine is not affected by the nature of working substance. | |
| 1765. |
Define the internal energy. |
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Answer» The internal energy of a system is the sum of kinetic and potential energies of the molecules of the system i.e., U = K.E + P. E. |
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| 1766. |
Can whole of the heat be converted into work ? |
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Answer» No, the second law of thermodynamics does not allow it. |
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| 1767. |
Is it possible to construct a heat engine, which is free from thermal pollution? |
| Answer» No, as in accordance with second law of thermodynamics, whole of heat cannot be converted into work. | |
| 1768. |
Is it possible to construct a heat engine, which is free from thermal pollution? |
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Answer» No, as in accordance with second law of thermodynamics, whole of heat cannot be converted into work. |
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| 1769. |
What do you mean by open system? |
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Answer» A system is said to be an open system,if it can exchange both energy and matter with its surroundings. |
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| 1770. |
One mole of a substance is cooled at the rate of `0.4 kJ min^(-1)` as shown in the graph. Curve `AB`, points `B` and `C` and curve `CD` represent respectively, the cooling of the liquid, start of freezing, completion of freezing and cooling of the solid based on this data. The entropy of fusion in `J molK^(-1)` is : A. `10`B. `20`C. `30`D. `40` |
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Answer» Correct Answer - C The cooling shows fusion process for `30` minute `(40-10)` at `400K` `DeltaS=q/T=(0.4xx10^(3)xx30)/400=30 J` |
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| 1771. |
Bond energies can be obtained by using the following relation: `DeltaH (reaction) = sum`Bond energy of bonds, broken in the reactants `-sum` Bond energy fo bonds, formed in the products Bond enegry depends on three factors: a. Greater is the bond length, lesser is the bond enegry. b. Bond energy increases with the bond multiplicity. c. Bond enegry increases with electronegativity difference between the bonding atoms. The heat of formation of `NO` from its elements is `+90 kJ mol^(-1)`, What is the approximate bond dissociation enegry of the bond in `NO`? `BE_(N=N) = 941 kJ mol^(-1) BE_(O=O) = 499 kJ mol^(-1)`A. `630 kJ mol^(-1)`B. `700 kJ mol^(-1)`C. `860 kJ mol^(-1)`D. `810 kJ mol^(-1)` |
| Answer» `DeltaH = H_(P) -H_(R)` | |
| 1772. |
Calculate the work done during the process, when one mole of gas is allowed to expand freely into vacuum. |
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Answer» Gas expands in vacuum i.e., `P_(ext)=0` and thus, irreversible process. Therefore, `W=-P_(ext)(V_(2)-V_(1))=0` |
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| 1773. |
Bond energies can be obtained by using the following relation: `DeltaH (reaction) = sum`Bond energy of bonds, broken in the reactants `-sum` Bond energy fo bonds, formed in the products Bond enegry depends on three factors: a. Greater is the bond length, lesser is the bond enegry. b. Bond energy increases with the bond multiplicity. c. Bond enegry increases with electronegativity difference between the bonding atoms. In `CH_(4)` molecule, which of the following statement is correct about the `C -H` bond enegry?A. All `C-H` bonds of methane have same enegry.B. Average of all `C -H` bond energies is considered.C. Fourth `C -H` bond required highest enegry to break.D. None of the above |
| Answer» Average bond energy is considered. | |
| 1774. |
Bond energies can be obtained by using the following relation: `DeltaH (reaction) = sum`Bond energy of bonds, broken in the reactants `-sum` Bond energy fo bonds, formed in the products Bond enegry depends on three factors: a. Greater is the bond length, lesser is the bond enegry. b. Bond energy increases with the bond multiplicity. c. Bond enegry increases with electronegativity difference between the bonding atoms. Use the bond enegries to estimate `DeltaH` for this reaction: `H_(2)(g) +O_(2)(g) rarr H_(2)O_(2)(g)` `{:("Bond","Bond energy"),(H-H,436 kJ mol^(-1)),(O-O,142 kJ mol^(-1)),(O=O, 499kJ mol^(-1)),(H-O,460kJ mol^(-1)):}`A. `-127 kJ`B. `-109 kJ`C. `-400 kJ`D. `-800 kJ` |
| Answer» `DeltaH = H_(P) -H_(R)` | |
| 1775. |
A change in the free energy of a system at constant temperature and pressure will be: `Delta_(sys)G = Delta_(sys)H -T Delta_(sys)S` At constant temperature and pressure `Delta_(sys) G lt 0` (spontaneous) `Delta_(sys)G = 0` (equilibrium) `Delta_(sys)G gt 0` (non-spontaneous) For a spontaneous reaction `DeltaG`, equilibrium `K` and `E_(cell)^(Theta)` will be, respectivelyA. `-ve, gt 1, +ve`B. `-ve, lt1, -ve`C. `+ve, gt1, -ve`D. `-ve, gt1, -ve` |
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Answer» `DeltaG =- nRT In K` `DeltaG =- nFE^(Theta)` |
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| 1776. |
A change in the free energy of a system at constant temperature and pressure will be: `Delta_(sys)G = Delta_(sys)H -T Delta_(sys)S` At constant temperature and pressure `Delta_(sys) G lt 0` (spontaneous) `Delta_(sys)G = 0` (equilibrium) `Delta_(sys)G gt 0` (non-spontaneous) For a system in equilibrium, `DeltaG = 0`, under conditions of constantA. Temperature and pressureB. Pressure and volumeC. Temperature and volumeD. Energy and volume |
| Answer» `(DeltaG)_(T,P) = 0` | |
| 1777. |
`C_(P) -C_(V)` for an ideal gas is………….. .A. `R^(2)`B. `sqrt(R )`C. `R`D. `R//2` |
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Answer» Correct Answer - C For an ideal gas, `Delta H = Delta U + Delta (pv)` `=Delta U + Delta (nRT)` `= Delta U + nR Delta T` `:. Delta H = Delta U + nR Delta T` `C_(P) Delta T = C_(V) Delta T + nR Delta T` `C_(P) = C_(V) + nR` `C_(P) - C_(V) = nR` |
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| 1778. |
Calculate the work done in an open vessel at `300K,` when `112g` ion reacts with dil. `HCl`. |
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Answer» `Fe+2HCl rarr FeCl_(2)+H_(2)` Work done `=-Pxx(V_(2)-V_(1))` Since, mole of `Fe` used `=112/56=2` `:.` Mole of `H_(2)` formed `=2` Work is done by giving out `2`moles of `H_(2)` `=-Pxx[V_(H_(2))-V_(i)] i.e., =-PxxV_(H_(2))` [`V_(i)=0` for initial condition] For `H_(2) PxxV_(H_(2))=nRT` `:. V_(H_(2))=(nRT)/P` `:.` Work done `=-Pxx(nRT)/P=-nRT` `=-2xx2xx300=-1200 cal` |
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| 1779. |
The open system (`s`) is (are) which :A. can exchange matter with the surroundingsB. can exchange energy with the surroundingsC. can exchange both matter and energy with the surroundingsD. cannot exchange either matter or energy with the surroundings |
| Answer» Correct Answer - A::B::C | |
| 1780. |
The work done when 6.5 g of zinc reacts with dil `HCl` is an open beaker at 298 K isA. `-495.52 J`B. `247.76 J`C. `-247.76 J`D. `-123.88 J` |
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Answer» Correct Answer - C `Zn_((s))+2HCl_((aq)) rarr ZnCl_((aq))+H_(2(g))` |
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| 1781. |
The enthalpy change of a reaction does not depend uponA. state of reactants bad productsB. nature of reactants and productsC. different intermediate reactionD. initial and final enthalpy change of a reaction. |
| Answer» Correct Answer - C | |
| 1782. |
Statement: The change in internal energy and change in heat enthalpy does not depend upon the path by which change are brought in. Explanation: Both `DeltaE` and `DeltaH` are path independent as `E` and `H` are state functions.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` are correct and `E`C. Both `S` and `E` are correct but `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
| Answer» Correct Answer - c | |
| 1783. |
Constant- volume calorimeter measuresA. `Delta H`B. `Delta U`C. `H`D. `U` |
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Answer» Correct Answer - B `Delta U = q_(v)`, it is the amount of heat exchanged by the system with its surrounding at constant volume. |
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| 1784. |
The work done is heating one mole of an ideal gas at constant pressure from `15^(@)C` to `25^(@)C` isA. `+19.87` calB. `-198.7` calC. `+198.7` calD. `-19.87` cal |
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Answer» Correct Answer - D Rise of `10^(@)C rarr 1.987xx10=19.87 cal` |
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| 1785. |
The enthalpy of reaction, `Delta_(1) H`, isA. `Delta_(r) = H` (product) `+ H` (reactants)B. `Delta_(r) = H` (reactants) `- H` (products)C. `Delta_(r) =H` (product) `- H` (reactants)D. `Delta_(r) = H` (reactants) `+ H` (products) |
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Answer» Correct Answer - C The enthalpy of reaction is the difference between the ethalpies of the products and the enthalpies of the reactants. |
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| 1786. |
During a process, a system abdorbs `710J` of heat and does work. The change in `DeltaU` for the process is `460J`. What is the work done by the system? |
| Answer» `DeltaU = q +w or 460 = 710 +w rArr w =- 250J` | |
| 1787. |
A gas expends from `2 L` to `6L` against a constant pressure of `0.5` atm on absobing `200 J` of heat. Calculate the change in internal energy. |
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Answer» `Delta U = q+w` `= 200 J-0.5 atm (62-22)` `= 200 J-2 L atm` `= 200 J - 2 xx 101. 3 J ("1 L atm = 101.3 J")` `= - 2.6 J` |
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| 1788. |
The heat of neutralization of a strong acid by a strong base is a constant because `:` |
| Answer» It is because strong acid and strong base are completely ionised in aqueous solution. | |
| 1789. |
A gas expends from 1.5 to 6.5 L against a constant pressure of 0.5 atm and during this process the gas also absorbs 100 J of heat. The change in the internal energy of the gas isA. `153.3 J`B. `353.3 J`C. `-153.3 J`D. `-353.3 J` |
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Answer» Correct Answer - C `Q=DeltaU-W` |
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| 1790. |
In a certain process, `400J` of work is done on a system which gives off `200J` of heat. What is `DeltaU` for the process? |
| Answer» `DeltaU = q+w =- 200 +400 = 200J` | |
| 1791. |
Reaction enthalpy does not depend uponA. amounts states reactions involvedB. physical states of the reactants and productsC. allotropic molificationsD. pathway of reaction |
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Answer» Correct Answer - D Because enthalpy is a state function. The magnitude of `Delta_(1) H` depends only upon reactants (initial state) and products (final state). |
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| 1792. |
`H_(2)(g) +CI_(2)(g) rarr 2HCI(g) +185 kJ`. State whether this reaction is exo or endothermic and why? |
| Answer» It is an expthermic reaction because heat is being evolved. | |
| 1793. |
10 g of argon gas is compressed isothermally and reversibly at a temperature of `27^(@)C` from 10 L to 5 L. Calculate q and `Delta H` for this process. Atomic wt. of `Ar=40`.A. `+ 103.99` cal, 0B. `+39.99` cal, 3 JC. `-39.99` cal, 3 JD. `-103.99` cal, 0 |
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Answer» Correct Answer - D `W= -P(V_(2)-V_(1))" "DeltaH=DeltaU+P Delta V` |
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| 1794. |
Which of the following equations refers to standard enthalpy of reaction?A. `CH_(4) (g) + 2 O_(2) (g) rarr CO_(2) (g) + 2 H_(2) O (l)`B. `CH_(4) (g) + 2 O_(2) (g) rarr CO_(2) (g) + 2 H_(2) (g)`C. `CH_(4) (g) + 2 O_(2) (g) rarr CO_(2) (s) + 2 H_(2) O (l)`D. `CH_(4) (g) + 2 O_(2) (l) rarr CO_(2) (s) + 2 H_(2) O (l)` |
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Answer» Correct Answer - A The standar enthalpy of reaction `(Delta_(r) H^(@))` refers to the enthalpy change `(Delta H)` when the specified numbe of moles of reactants all at standard states are converted completely states. |
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| 1795. |
Caclulate `DeltaU`, internal enegry change of a system, if it absorbs `25 kJ` of heat and does `5kJ` of work. |
| Answer» `DeltaU =q +w = 25 - 15 = 10 kJ` | |
| 1796. |
A system absorbs `20kJ` heat and also does `10kJ` of work. The net internal enegry of the systemA. Increases by `10kJ`B. Decreases by `10kJ`C. Increases by `30 kJ`D. Decreases by `30kJ` |
| Answer» `DeltaU = q +w = 20 - 10 = 10 kJ` | |
| 1797. |
In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas releases 30 joules of heat and 10 joules of work was done on the gas. If the initial internal energy of the gas was 30 joules , then the final internal energy will beA. 2 jB. `-18` jC. `10j`D. `58j` |
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Answer» Correct Answer - C |
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| 1798. |
A system absorbs `20kJ` heat and also does `10kJ` of work. The net internal enegry of the systemA. Increases by 10 kJB. Decreases by 10 kJC. increases by 30 kJD. Decreases by 30 kJ |
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Answer» Correct Answer - A `DeltaU=Q+W` |
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| 1799. |
Standard thermodynamic conditions chosen for substance when listing or comparing thermodynamic data refer toA. one atmosphere pressure and `273 K`B. one bar pressure and any specified temperatureC. one atmosphere pressure and `298 K`D. one bar pressure and 298 K |
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Answer» Correct Answer - B For the ease of comparison and tabulation, we refer to thermochemical or thermodynamic changes under standard conditions. To indicate a change at standard pressure (1 bar pressure), we add a superscript zero. If some temperature other than `25^(@)C` (298 K) is specified, we indicate it with a subscript, e.g., `Delta H^(@)` (300 K). If no subscript appears, a temperature of `25^(@)C` (298 K) is limplied. |
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| 1800. |
Which of the following describes the thermodynamic standard state of carbon?A. GraphiteB. DiamondC. BuckminsterfullerenceD. Charcoal |
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Answer» Correct Answer - A The thermodynamic standard state of a substance is its most stable, pure form (physical state and allotrope) under standard pressure and at some specific temperature `(25^(@)C` or 298 K unless otheriwse specified). Under these conditions, the most stable, pure form of carbon is graphite. |
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