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NEET • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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481 Medium
481 of 560
Analyzing +4 | -1
Assertion :- Sum of two path functions is always path function.

Reason :- Sum of path function depends on path followed.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
Option D is correct because both the assertion and the reason are false. The sum of two path functions is not always a path function, and path functions depend on the path followed.
482 Easy
482 of 560
Remembering +4 | -1
The heat change for the reaction,H2O(l){\mathrm{H}_2}\mathrm{O}\left( l \right)→  H2O  (g)\to \;{\mathrm{H}_2}\mathrm{O}\;(g) is called as :
View answer and explanation
Correct option: A
Heat of vaporisation
Option A (Heat of vaporisation) is correct because it represents the heat change during the phase transition from liquid to gas.
483 Medium
483 of 560
Analyzing +4 | -1
Mark out the enthalphy of formation of carbon monoxide (CO)Given,C(s)+O2(g)→CO(g)  ,ΔH=−393.3  kJ/mol\mathrm{C}\left( s \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{CO}\left( \mathrm{g} \right)\;,\Delta H = - 393.3\;kJ/mol
CO(g)+12O2(g)→CO2(g),ΔH=−282.8kJ/mol\mathrm{CO}\left( \mathrm{g} \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right),\Delta H = - 282.8\mathrm{kJ}/\mathrm{mol}
View answer and explanation
Correct option: A
110.5 kJ/mol
Option A (110.5 kJ/mol) is correct because the enthalpy of formation of CO is calculated by subtracting the enthalpy of the second reaction from the first reaction.
484 Easy
484 of 560
Understanding +4 | -1
The entropy of the universe:
View answer and explanation
Correct option: A
Increasing and tending towards maximum value
Option A (Increasing and tending towards maximum value) is correct because, according to the second law of thermodynamics, the entropy of the universe always increases for spontaneous processes.
485 Medium
485 of 560
Application +4 | -1
The difference between heat of reaction at constant pressure and constant volume for the reaction given below at 25∘C25^\circ C in kJ is:
2C6H6(l)+15O2(g)→  12CO2(g)+6H2O(l) 2{C_6}{H_6}(l) + 15{O_2}(g)\xrightarrow{\,}\,12C{O_2}(g) + 6{H_2}O(l)\,
View answer and explanation
Correct option: A
– 7.43
Option A (– 7.43) is correct because it represents the difference between heat of reaction at constant pressure and constant volume, calculated using ∆H - ∆U = ∆nRT.
486 Medium
486 of 560
Application +4 | -1
The quantity of heat (in JJ ) required to raise the temperature of 1.0 kg1.0 \mathrm{~kg} of ethanol from 293.45 K293.45 \mathrm{~K} to the boiling point and then change to liquid to vapour at that temperature is closest to [Given, boiling point of ethanol 351.45 K351.45 \mathrm{~K}. Specific heat capacity of liquid ethanol 2.44 J g−1k−12.44 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{k}^{-1}. Latent heat of vaporisation of ethanol 855 J g−1855 \mathrm{~J} \mathrm{~g}^{-1} ]
View answer and explanation
Correct option: D
9.97×1059.97 \times 10^5
Option D (9.97×1059.97 \times 10^5) is correct because it accounts for both the heat required to raise the temperature to the boiling point and the heat of vaporization.
487 Easy
487 of 560
Application +4 | -1
If 'xx' mole of ideal gas at 27∘C27^\circ C expands isothermally and reversibly from a volume of 'yy' to 10y10 y, then work done is
View answer and explanation
Correct option: C
W=−300 x R ln⁡10W = - 300\,x\,R\,\ln 10
Option C (W=−300 x R ln⁡10W = - 300\,x\,R\,\ln 10) is correct because it correctly applies the formula for isothermal reversible work with the given volume ratio and temperature.
488 Medium
488 of 560
Remembering +4 | -1
Which of the following is a highly corrosive salt?
View answer and explanation
Correct option: D
HgCl2{HgCl}_2.
Option D (HgCl2{HgCl}_2) is correct because Mercuric chloride is a highly corrosive salt due to its strong oxidizing properties.
489 Medium
489 of 560
Analyzing +4 | -1
One mole of CH3COOH\mathrm{CH}_3 \mathrm{COOH} undergoes dimerization in vapour phase at 127∘C127^{\circ} \mathrm{C} as:

2CH3COOH(g)⇌(CH3COOH)2( g)2 \mathrm{CH}_3 \mathrm{COOH}(\mathrm{g}) \rightleftharpoons\left(\mathrm{CH}_3 \mathrm{COOH}\right)_2(\mathrm{~g})

if dimer formation is due to two H\mathrm{H}-bonds involved in dimer, each of 33 kJ33 \mathrm{~kJ} stength and the degree of dimerisation of acetic acid 98.2%98.2 \%. The change in standard entropy during dimerization is
View answer and explanation
Correct option: A
−104JK−1 mol−1-104 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
Option A (−104JK−1 mol−1-104 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}) is correct because it represents the change in standard entropy during dimerization, calculated using the Gibbs-Helmholtz equation and the given enthalpy and temperature.
490 Hard
490 of 560
Application +4 | -1
Enthalpies of formation of CCl4(g),H2O(g)\mathrm{CCl}_{4}(\mathrm{g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), CO2(g)\mathrm{CO}_{2}(\mathrm{g}) and HCl(g)\mathrm{HCl}(\mathrm{g}) are −105,−242,−349-105,-242,-349 and -92 kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} respectively. The magnitude of enthalpy of the reaction given below is _____ kJ\mathrm{kJ} mol−1\mathrm{mol}^{-1}. (Nearest integer)
CCl4(g)+2H2O(g)→CO2(g)+4HCl(g)\mathrm{CCl}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{HCl}(\mathrm{g})
View answer and explanation
Correct option: A
173
The correct answer is 173 because it is calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products.
491 Easy
491 of 560
Understanding +4 | -1
Which of the following is an intensive property?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because temperature, viscosity, and surface tension are all intensive properties, meaning they do not depend on the amount of substance.
492 Easy
492 of 560
Analyzing +4 | -1
Two bars of different metals are heated to 60∘C60^{\circ} \mathrm{C} and then immersed in identical insulated containers each containing 200 g200 \mathrm{~g} of water at 20∘C.20^{\circ} \mathrm{C}. Will the metal with higher or lower atomic mass cause a greater temperature rise in water?
View answer and explanation
Correct option: A
Lower atomic mass
Option A (Lower atomic mass) is correct because metals with lower atomic mass will experience a larger temperature change for the same amount of heat transfer.
493 Easy
493 of 560
Understanding +4 | -1
In a irreversible process, the value of ΔSSystem+ΔSSurroundings\Delta {S_{\mathrm{System}}} + \Delta {S_{\mathrm{Surroundings}}} is
View answer and explanation
Correct option: A
+ ve+ \,ve
Option A (+ ve+ \,ve) is correct because in an irreversible process, the total entropy change (system + surroundings) is always positive according to the second law of thermodynamics.
494 Easy
494 of 560
Remembering +4 | -1
Hess’s law is used to calculate
View answer and explanation
Correct option: A
Enthalpy of reaction
Option A is correct. Hess's Law is a statement of the conservation of energy, and it is primarily used to calculate the enthalpy change of a reaction.
495 Medium
495 of 560
Remembering +4 | -1
Work done by an ideal gas at a constant
volume is .........
View answer and explanation
Correct option: C
0
The correct answer is C (0) because at constant volume, no work is done by the gas.
496 Easy
496 of 560
Remembering +4 | -1
Select the path function among the following.
View answer and explanation
Correct option: D
W\mathrm{W}
Option D (W\mathrm{W}) is correct because work is a path function, meaning its value depends on the path taken during a process.
497 Easy
497 of 560
Analyzing +4 | -1
Assertion : Molar enthalpy of vaporisation of water is different from ethanol.

Reason : Water is more polar than ethanol.
View answer and explanation
Correct option: B
If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
Option B is correct. Both the assertion and the reason are true, but the reason doesn't fully explain the assertion. The difference in molar enthalpy of vaporization is due to a combination of factors, including hydrogen bonding and other intermolecular forces, not just polarity.
498 Hard
498 of 560
Understanding +4 | -1
For adiabatic expansion of a perfect gas dPP\frac{d P}{P} is :
View answer and explanation
Correct option: C
−γdVV-\gamma \frac{d V}{V}
Option C (−γdVV-\gamma \frac{d V}{V}) is correct because for an adiabatic expansion of a perfect gas, the relationship between pressure and volume changes is given by PVγ=constantPV^\gamma = constant, which leads to the given expression.
499 Medium
499 of 560
Understanding +4 | -1
An exothermic reaction is one in which the reacting substances
View answer and explanation
Correct option: A
Have more energy than the products
Option A (Have more energy than the products) is correct because an exothermic reaction releases energy, meaning the reactants have higher energy than the products.
500 Medium
500 of 560
Remembering +4 | -1
Heat of reaction at constant volume is equal to :
View answer and explanation
Correct option: A
∑UP−∑UR\sum {U_P} - \sum {U_R}
Option A, ∑UP−∑UR\sum {U_P} - \sum {U_R}, is correct because the heat of reaction at constant volume is equal to the change in internal energy, which is the difference between the internal energy of products and reactants.
501 Medium
501 of 560
Application +4 | -1
A gas is allowed to expand in an insulated container against a constant external pressure of 4 atm from the initial volume of 1.8 L to a final volume of 5.2 L. The change in internal energy of the gas is
View answer and explanation
Correct option: C
–1.38 kJ
Option C (–1.38 kJ) is correct because it represents the change in internal energy, which is equal to the work done in an adiabatic process against constant external pressure.
502 Easy
502 of 560
Understanding +4 | -1
The heat of formation of water is exothermic in nature because :
View answer and explanation
Correct option: C
H2(g)and  O2(g){\mathrm{H}_2}\left( \mathrm{g} \right)and\;{O_2}(g)have higher internal energy than water
Option C (H2(g)and  O2(g){\mathrm{H}_2}\left( \mathrm{g} \right)and\;{O_2}(g)have higher internal energy than water) is correct because the formation of water releases energy, meaning the reactants have higher internal energy than the product.
503 Easy
503 of 560
Understanding +4 | -1
From the reaction, P(white)→P(red);{\mathrm{P}_{\left( {\mathrm{white}} \right)}} \to {\mathrm{P}_{\left( {\mathrm{red}} \right)}}; ∆H= -18.4 kJ it following that :
View answer and explanation
Correct option: D
White P can be converted into red P and red P is more stable
Option D (White P can be converted into red P and red P is more stable) is correct because the negative ∆H indicates that the conversion of white phosphorus to red phosphorus is exothermic, meaning red phosphorus is more stable.
504 Hard
504 of 560
Application +4 | -1
The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is (Specific heat of water liquid and water vapour are 4.2 kJ K−1 kg−14.2\,kJ\,{K^{ - 1}}\,k{g^{ - 1}} and 20 kJ K−1 kg−120\,kJ\,{K^{ - 1}}\,k{g^{ - 1}}; heat of liquid fusion and vapourisation of water are 334 kJ kg−1334\,kJ\,k{g^{ - 1}}\mathrm{ } and 2491 kJ kg−12491\,kJ\,k{g^{ - 1}} respectively).
(log⁡273=2.436,  log⁡373=2.572,  log⁡383=2.583)(\log 273 = 2.436,\,\,\log 373 = 2.572,\,\,\log 383 = 2.583)
View answer and explanation
Correct option: A
9.26 kJ  kg−1 K−19.26\,kJ\,\mathrm{ }\,k{g^{ - 1}}\,{K^{ - 1}}
Option A (9.26 kJ kg−1 K−19.26\,kJ\,k{g^{ - 1}}\,{K^{ - 1}}) is correct because it represents the total entropy change calculated by summing the entropy changes for each phase transition and heating step.
505 Hard
505 of 560
Application +4 | -1
When an electric current of 0.50 A0.50 \mathrm{~A} from a 12 V12 \mathrm{~V} supply is passed for 18056sec\frac{1805}{6} \mathrm{sec} through a resistance in thermal contact with water maintained at 1 atm and 373 K,373 \mathrm{~K}, it is found that 0.9 g0.9 \mathrm{~g} of water is vaporized. The molar internal energy change of water is (Take 8.314×373=31008.314 \times 373=3100)
View answer and explanation
Correct option: A
33 kJ33 \mathrm{~kJ}
Option A (33 kJ) is correct because it involves calculating the heat supplied, then accounting for the work done during vaporization to find the change in internal energy.
506 Hard
506 of 560
Application +4 | -1
50 mL of water takes 5 min to evaporate from a vessel on a heater connected to an electric source which delivers 400 W. The enthalpy of vaporisation of water is
View answer and explanation
Correct option: B
43.2 kJ per mol
Option B (43.2 kJ per mol) is correct because it is calculated by finding the total energy used for vaporization and then converting it to per mole.
507 Medium
507 of 560
Understanding +4 | -1
The Gibbs energy change for a reversible reaction at equilibrium is:
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because at equilibrium, the forward and reverse reaction rates are equal, resulting in no net change in Gibbs free energy (∆G = 0).
508 Medium
508 of 560
Remembering +4 | -1
The bond energies of F2{\mathrm{F}_2},Cl2,Br2\mathrm{C}{\mathrm{l}_2},\mathrm{B}{\mathrm{r}_2} and I2{\mathrm{I}_2} are 155.4, 243.6, 193.2 and 151.2 kJ mol−1\mathrm{mo}{\mathrm{l}^{ - 1}} respectively. The strongest bond is :
View answer and explanation
Correct option: B
Cl-Cl
Option B (Cl-Cl) is correct because it has the highest bond energy among the given options, indicating the strongest bond.
509 Medium
509 of 560
Analyzing +4 | -1
Assertion: In a process, if work =0=0 then ΔU=\Delta U= q.

Reason: qq is difference between initial state and final state of a system.
View answer and explanation
Correct option: C
If Assertion is correct but Reason is incorrect.
Option C is correct. If work = 0, then ∆U = q is correct based on the first law of thermodynamics, but the reason is incorrect because q is not the difference between initial and final states.
510 Easy
510 of 560
Remembering +4 | -1
For a spontaneous process, entropy :
View answer and explanation
Correct option: A
Increases
Option A is correct. For a spontaneous process, the total entropy of the system and surroundings increases, according to the Second Law of Thermodynamics.
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