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JEE • CHEMISTRY • THERMODYNAMICS
Thermodynamics practice · Mixed difficulty
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Practice format and chapter quiz
These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.
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Showing 481-500 of 500 questions
481
Medium
481 of 500
Application
+4 | -1
Assign the sign of work done (based on SI convention) in
the following chemical changes taking place against
external atmospheric pressure
I.
II.
III.
IV.
the following chemical changes taking place against
external atmospheric pressure
I.
II.
III.
IV.
View answer and explanation
Correct option: A
Option A (+, -, -, +) is correct because it correctly assigns the sign of work based on the change in the number of moles of gas ().
482
Medium
482 of 500
Application
+4 | -1
At the given condition of fuel
cell the cell emf is 0.8 V and the
enthalpy of combustion of is
-772 kJ/mol. The maximum efficiency of
the given fuel cell in the given condition
is :
cell the cell emf is 0.8 V and the
enthalpy of combustion of is
-772 kJ/mol. The maximum efficiency of
the given fuel cell in the given condition
is :
View answer and explanation
Correct option: C
0.8
Option C (80%) is correct because the maximum efficiency is calculated by dividing the Gibbs free energy change by the enthalpy change of combustion.
483
Medium
483 of 500
Remembering
+4 | -1
The state of a gas can be described by quoting the relationship between
View answer and explanation
Correct option: D
Pressure, volume, temperature and amount
Option D (Pressure, volume, temperature and amount) is correct because the state of a gas is defined by the relationship between these four variables, as described by the ideal gas law.
484
Medium
484 of 500
Application
+4 | -1
One mole of a non-ideal gas undergoes a change of state (2.0 atm,3.0 L,95 K)→(4.0 atm,5.0 L,245 K) with a change in internal energy, ∆U=30.0 L atm. The change in enthalpy (∆H) of the process in L atm is:
View answer and explanation
Correct option: C
44
Option C (44) is correct because the change in enthalpy is calculated using the formula ∆H = ∆U + (P₂V₂ - P₁V₁), which gives 44 L atm.
485
Medium
485 of 500
Remembering
+4 | -1
If water is formed from ions andthe heat of formation of water is :
View answer and explanation
Correct option: A
Option A () is correct because the heat of formation of water from H+ and OH- is the heat of neutralization, which is -13.7 kcal.
486
Medium
486 of 500
Application
+4 | -1
The volume of required to completely burn at Ethyne of STP is:
View answer and explanation
Correct option: C
Option C () is correct because the balanced equation shows that 1 mole of ethyne requires 2.5 moles of oxygen for complete combustion, and 52g of ethyne is 2 moles, thus requiring 5 moles of oxygen which equals 112L at STP.
487
Medium
487 of 500
Application
+4 | -1
The entropy change in the conversion of one mole of liquid water at to vapour at the same temperature is: (Latent heat of vaporization of water, )
View answer and explanation
Correct option: C
Option C () is correct because it is calculated by dividing the molar heat of vaporization by the temperature.
488
Medium
488 of 500
Remembering
+4 | -1
Which of the following is a highly corrosive salt?
View answer and explanation
Correct option: D
.
Option D () is correct because Mercuric chloride () is known as corrosive sublimate due to its highly corrosive and poisonous nature.
489
Medium
489 of 500
Application
+4 | -1
The standard free energy change for a gaseous reaction at If equilibrium constant for a reaction is 100 and R is Then is
View answer and explanation
Correct option: A
– 2.7636
Option A (– 2.7636) is correct because it is the standard free energy change calculated using the given equilibrium constant and temperature.
490
Medium
490 of 500
Understanding
+4 | -1
To calculate the amount of work done in joules during reversible isothermal expansion of an ideal gas, the volume must be expressed in
View answer and explanation
Correct option: D
Any of these
Option D (Any of these) is correct because the work done in a reversible isothermal expansion involves a ratio of volumes, so the units cancel out, making any volume unit acceptable as long as both volumes are in the same unit.
491
Hard
491 of 500
Application
+4 | -1
Enthalpy of the reaction is Magnitude of enthalpies of formation of and is in ratio but their signs are opposite. Enthalpy of formation of is The magnitude of enthalpy of formation of is
View answer and explanation
Correct option: A
The correct answer is A () because it is derived from Hess's Law and the given enthalpy changes for the reactions, considering the ratio of enthalpies of formation of Ag+(aq) and Br-(aq).
492
Medium
492 of 500
Understanding
+4 | -1
For which reaction from the following, will be maximum ?
View answer and explanation
Correct option: B
Option B is correct because the reaction has the largest increase in the number of gaseous moles (), leading to the maximum increase in entropy ().
493
Medium
493 of 500
Understanding
+4 | -1
Which of the following is/are intensive properties?
(I) Temperature
(II) Heat
(III) Density
(IV) Refractive index
(I) Temperature
(II) Heat
(III) Density
(IV) Refractive index
View answer and explanation
Correct option: C
(I), (III) and (IV) only
Option C ((I), (III) and (IV) only) is correct because temperature, density, and refractive index are intensive properties, while heat is an extensive property.
494
Medium
494 of 500
Understanding
+4 | -1
For the reaction: , which one is true?
View answer and explanation
Correct option: B
Option B is correct because the decomposition of is endothermic () and increases the number of gas molecules, leading to an increase in entropy ().
495
Easy
495 of 500
Analyzing
+4 | -1
Enthalpy of neutralization of the reaction between and is and that of the reaction between and is The enthalpy of dissociation of is
View answer and explanation
Correct option: B
Option C ( bond will be easily broken is case of hydroxide of ' ') is correct because if X(OH) is more acidic than Y(OH), the O-H bond in X(OH) is more easily broken, indicating a stronger tendency to donate a proton.
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496
Medium
496 of 500
Analyzing
+4 | -1
One mole of undergoes dimerization in vapour phase at as:
if dimer formation is due to two -bonds involved in dimer, each of stength and the degree of dimerisation of acetic acid . The change in standard entropy during dimerization is
if dimer formation is due to two -bonds involved in dimer, each of stength and the degree of dimerisation of acetic acid . The change in standard entropy during dimerization is
View answer and explanation
Correct option: A
Option A is correct because the change in standard entropy during dimerization is calculated using the equilibrium constant and enthalpy change, resulting in approximately -104 J/K·mol.
497
Hard
497 of 500
Application
+4 | -1
2 moles of an ideal gas at temperature is expanded reversibly from 2L to 20L. Find entropy change (R=2cal/mol K).
View answer and explanation
Correct option: D
9.2
Option D (9.2) is correct because the entropy change is calculated using the formula for reversible expansion of an ideal gas.
498
Medium
498 of 500
Application
+4 | -1
Given, ;
What is the magnitude of standard enthalpy of formation of gas?
What is the magnitude of standard enthalpy of formation of gas?
View answer and explanation
Correct option: A
46.2
Option A (46.2) is correct because the standard enthalpy of formation is half of the given reaction enthalpy since 2 moles of NH3 are formed in the reaction.
499
Medium
499 of 500
Understanding
+4 | -1
Which of the following statements is correct?
View answer and explanation
Correct option: B
During complete combustion of 1 mole of butane, of heat is released
Option B (During complete combustion of 1 mole of butane, of heat is released) is correct because the combustion of butane is an exothermic reaction that releases a specific amount of heat.
500
Medium
500 of 500
Understanding
+4 | -1
Two mole of an ideal gas is expanded isothermally and reversibly from 1 L to 10 L at 300 K. The enthalpy change (in kJ) for the process is
View answer and explanation
Correct option: C
0
Option C (0) is correct because enthalpy change for an isothermal process involving an ideal gas is zero, as enthalpy depends only on temperature.
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