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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 301-330 of 500 questions
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301
Medium
301 of 500
Application
+4 | -1
During isothermal expansion of one mole of an ideal gas from to at , thework done is (gas constant ) :
View answer and explanation
Correct option: C
Option C (-1257.43 cal) is correct because it's the work done calculated using the formula for isothermal expansion of an ideal gas.
302
Medium
302 of 500
Application
+4 | -1
A gas expands against a variable pressure given by (where in atm and in ). During expansion from volume of 1 litre to 10 litre, the gas undergoes a change in internal energy of . How much heat is absorbed by the gas during expansion?
View answer and explanation
Correct option: C
Option C (5065.8 J) is correct because the heat absorbed is calculated using the first law of thermodynamics, where q = ΔU - w, and the work done is found by integrating P dV.
303
Easy
303 of 500
Understanding
+4 | -1
The heat of neutralisation for strong acid and strong base forming 2 moles of water is
View answer and explanation
Correct option: A
Option A is correct because the heat of neutralization for a strong acid and strong base is -57.1 kJ per mole of water formed, so for 2 moles, it's -2 * 57.1 kJ.
304
Easy
304 of 500
Remembering
+4 | -1
The law of Lavoisier and Laplace is based on :
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Correct option: A
The principle of conservation of energy
Option A (The principle of conservation of energy) is correct because the law of Lavoisier and Laplace is based on the conservation of energy.
305
Medium
305 of 500
Application
+4 | -1
At what will be the change in standard internal energy change for the given reaction
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Correct option: C
Option C (-310 kJ) is correct because it is calculated using the formula ΔU = ΔH - Δn_gRT, accounting for the change in the number of moles of gas.
306
Easy
306 of 500
Understanding
+4 | -1
The gas in a refrigerator causes cooling on expansion because:
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Correct option: B
Heat of the gas is lost as work is done by the gas
Option B is correct. When a gas expands in a refrigerator, it does work, which causes it to lose heat and cool down.
307
Easy
307 of 500
Understanding
+4 | -1
Energy equivalent to one erg, one joule and one calorie are in order:
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Correct option: C
Option C () is correct because 1 calorie is greater than 1 joule, which is greater than 1 erg.
308
Medium
308 of 500
Remembering
+4 | -1
Work done in reversible adiabatic process is given by:
View answer and explanation
Correct option: B
Option B () is correct because it represents the work done in a reversible adiabatic process in terms of the number of moles, gas constant, adiabatic index, and temperature change.
309
Medium
309 of 500
Understanding
+4 | -1
For an isothermal reversible expansion process, the value of can be calculated by the expression
View answer and explanation
Correct option: A
Option A () is correct because it represents the heat absorbed during an isothermal reversible expansion.
310
Hard
310 of 500
Application
+4 | -1
The heat of combustion of ethanol into carbon dioxides and water is at constant pressure. The heat evolved (in cal) at constant volume and (if all gases behave ideally) is ( cal )……
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Correct option: A
-326400
The correct answer is A (-326400) because it is calculated by applying the formula relating heat of combustion at constant pressure and constant volume, accounting for the change in the number of moles of gas.
311
Medium
311 of 500
Understanding
+4 | -1
At the reaction, proceeds spontaneously because the magnitude of
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Correct option: B
Option B () is correct because for a spontaneous reaction, the change in Gibbs free energy () must be negative, and . Therefore, must be less than for the reaction to be spontaneous.
312
Medium
312 of 500
Understanding
+4 | -1
In an isobaric process, when temperature change from to is equal to:
View answer and explanation
Correct option: A
Option A () is correct because it represents the entropy change in an isobaric process derived from the general entropy change formula.
313
Medium
313 of 500
Understanding
+4 | -1
An ideal monoatomic gas initially at expands adiabatically into vacuum to double its volume. The final temperature of gas is
View answer and explanation
Correct option: A
Option A () is correct because adiabatic expansion into a vacuum is a free expansion, which is an isothermal process for an ideal gas.
314
Medium
314 of 500
Application
+4 | -1
Choose the reaction in which is not equal to ?
View answer and explanation
Correct option: B
Option B () is correct because the change in the number of gaseous moles () is not zero, making not equal to .
315
Medium
315 of 500
Remembering
+4 | -1
Which of the following represents the first law
of thermodynamics
of thermodynamics
View answer and explanation
Correct option: A
Option A () is correct because it represents the first law of thermodynamics, stating that the change in internal energy equals heat plus work.
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316
Medium
316 of 500
Understanding
+4 | -1
For the reaction the signs of and respectively are
View answer and explanation
Correct option: C
Option C () is correct because bond formation releases energy (negative Δ H) and decreases randomness (negative Δ S).
317
Hard
317 of 500
Application
+4 | -1
Calculate the standard internal energy change for the following reaction at .
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Correct option: A
Option A () is correct because it is calculated by using the given enthalpies of formation and the formula relating Δ H and Δ E.
318
Medium
318 of 500
Application
+4 | -1
gas at S.T.P. is expanded so that volume is doubled. Find the work done in Joules. (1 litre atm = )
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Correct option: A
Option A (1134.56 J) is correct because the work done during the expansion is calculated using the formula W = P(V2 - V1), where P is the pressure and V is the volume.
319
Easy
319 of 500
Remembering
+4 | -1
The enthalpies of elements under the following conditions are assumed to be
View answer and explanation
Correct option: A
Zero at 298 K and 1 atm
Option A (Zero at 298 K and 1 atm) is correct because, by convention, the standard enthalpy of formation of an element in its standard state is defined as zero at 298 K and 1 atm.
320
Medium
320 of 500
Understanding
+4 | -1
Considering entropy(s) as the thermodynamic parameter, the criterion for the spontaneity of any process is:
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Correct option: A
Option A () is correct because the spontaneity of a process is determined by the total entropy change of the system and surroundings, which must be greater than zero for a spontaneous process.
321
Easy
321 of 500
Understanding
+4 | -1
For the dissociation reaction, , heat of atomization of is
View answer and explanation
Correct option: A
Option A (81 Kcal) is correct because the heat of atomization of H is half of the dissociation energy of H2.
322
Medium
322 of 500
Analyzing
+4 | -1
The standard molar enthalpies of formations of and are and respectively. Molar enthalpies of vaporization of and are and respectively. The bond dissociation enthalpy of is Calculate the bond dissociation enthalpy of bond in assuming that the bond dissociation enthalpy of bond is same in both and
View answer and explanation
Correct option: A
Option A (142 kJ/mol) is correct because by using Hess's Law and the given enthalpies of formation and vaporization, along with the bond dissociation enthalpy of O2, we can calculate the O-O bond dissociation enthalpy.
323
Hard
323 of 500
Application
+4 | -1
Determine for the following reaction
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Correct option: C
Option C () is correct because it is calculated by using the Gibbs-Helmholtz equation, considering the change in enthalpy and entropy for the given reaction.
324
Medium
324 of 500
Application
+4 | -1
A gas, while expanding absorbs of heat and does of work. Thus, (change in internal energy) for the gas is
View answer and explanation
Correct option: D
Option D () is correct because the change in internal energy is calculated using the first law of thermodynamics, considering the signs of heat absorbed and work done.
325
Medium
325 of 500
Application
+4 | -1
For a perfectly crystalline solid , where is constant. If is , molar entropy at is :
View answer and explanation
Correct option: C
Option C () is correct because it is calculated by integrating from 0 to 20 K, using the given relationship .
326
Medium
326 of 500
Analyzing
+4 | -1
Consider the following conversions:
(i)
(ii)
(iii)
(iv)
That according to given information the incorrect statements is:
(i)
(ii)
(iii)
(iv)
That according to given information the incorrect statements is:
View answer and explanation
Correct option: D
and are negative whereas and are positive
Option D is correct because the first electron gain enthalpy for O, F, and Cl are negative, but the second electron gain enthalpy for oxygen is positive due to electron-electron repulsion.
327
Easy
327 of 500
Application
+4 | -1
A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If ∆H is the enthalpy change and ∆E is the change in internal energy, then
View answer and explanation
Correct option: B
∆H<∆E
Option B (∆H<∆E) is correct because the number of moles of gas decreases during the reaction, leading to a negative ∆n and thus ∆H < ∆E.
328
Hard
328 of 500
Application
+4 | -1
Standard entropy of and are 30, 60 and respectively. For the reaction . To be at equilibrium, the temperature will be:
View answer and explanation
Correct option: B
The correct answer is because at equilibrium, , and the temperature is calculated using .
329
Medium
329 of 500
Analyzing
+4 | -1
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If is the initial temperature and is the final temperature, which of the following statements is correct?
View answer and explanation
Correct option: A
Option A ((Tf)irrev > (Tf)rev) is correct because in an irreversible adiabatic expansion, less work is done, resulting in a smaller decrease in internal energy and a higher final temperature compared to a reversible process.
330
Easy
330 of 500
Understanding
+4 | -1
Which is an extensive property of the system?
View answer and explanation
Correct option: B
Volume
Option B (Volume) is correct because volume is an extensive property, meaning it depends on the amount of substance present.
Revision notes and key takeaways
Review THERMODYNAMICS in your textbook or class notes, then use this set to check your understanding.
Showing 301-330 of 500 questions
