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JEE • PHYSICS • 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 361-384 of 384 questions

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361 Medium
361 of 384
Application +4 | -1
The internal energy of a gas is given by U=1.5pVU=1.5 \mathrm{pV}. It expands from 100 cm3100 \mathrm{~cm}^3 to 200 cm3200 \mathrm{~cm}^3 against a constant pressure of 1.0×105 Pa1.0 \times 10^5 \mathrm{~Pa}. Calculate the heat absorbed by the gas in the process.
View answer and explanation
Correct option: B
25 J
Option B (25 J) is correct because the heat absorbed is calculated using the first law of thermodynamics, considering both the change in internal energy and the work done.
362 Medium
362 of 384
Understanding +4 | -1
Which of the following can not determine the state of a thermodynamic system
View answer and explanation
Correct option: D
Any one of pressure, volume or temperature
Option D (Any one of pressure, volume or temperature) is correct because a single thermodynamic variable is insufficient to define the state of the system.
363 Medium
363 of 384
Understanding +4 | -1
When water is converted into ice, its entropy
View answer and explanation
Correct option: B
Decreases
Option B (Decreases) is correct because the conversion of water to ice results in a more ordered state, thus reducing entropy.
364 Medium
364 of 384
Understanding +4 | -1
Assertion: The zeroth law said that, when two systems XX and YY are in thermal equilibrium, there must be a physical quantity that has the same value for both.
Reason: The physical quantity, which is same for both the systems, is temperature.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of assertion
Option A (If both assertion and reason are true and reason is the correct explanation of assertion) is correct because the zeroth law defines thermal equilibrium, and temperature is the physical quantity that is the same for systems in thermal equilibrium.
365 Medium
365 of 384
Understanding +4 | -1
(A): When a bottle of cold carbonated drink is opened, a slight fog forms around the opening.
(R): Adiabatic expansion of the gas causes lowering of temperature and condensation of water vapours.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because the fog formation is due to the adiabatic expansion of the gas, which causes cooling and condensation.
366 Medium
366 of 384
Remembering +4 | -1
The p−Tp-T relation for an adiabatic expansion is
View answer and explanation
Correct option: D
p1−γTγ=p^{1-\gamma} T^\gamma= constant
Option B (Tangent law) is correct because deflection magnetometers operate based on the tangent law, which relates the magnetic field components.
367 Medium
367 of 384
Application +4 | -1
The efficiency of heat engine if the temperature of source is 2270C{227^0}C and that of sink is 270C{27^0}C nearly
View answer and explanation
Correct option: A
0.40.4
Option A (0.4) is correct because the efficiency of a heat engine is calculated using the formula η=1−T2T1\eta = 1 - \frac{{T_2}}{{T_1}} with temperatures in Kelvin.
368 Medium
368 of 384
Application +4 | -1
Work done by a system under isothermal change from a volume V1V_1 to V2V_2 for a gas which obeys Vander Waal's equation

(V−βn)(P+αn2V)=nRT(V-\beta n)\left(P+\frac{\alpha n^2}{V}\right)=n R T
View answer and explanation
Correct option: A
nRTlog⁡e(V2−nβV1−nβ)+αn2(V1−V2V1V2)n R T \log _e\left(\frac{V_2-n \beta}{V_1-n \beta}\right)+\alpha n^2\left(\frac{V_1-V_2}{V_1 V_2}\right)
Option A (nRTlog⁡e(V2−nβV1−nβ)+αn2(V1−V2V1V2)n R T \log _e\left(\frac{V_2-n \beta}{V_1-n \beta}\right)+\alpha n^2\left(\frac{V_1-V_2}{V_1 V_2}\right)) is correct because it is the work done calculated by integrating the pressure from the van der Waals equation over the volume change.
369 Medium
369 of 384
Understanding +4 | -1
During an isothermal expansion of an ideal gas
View answer and explanation
Correct option: D
Both (2) and (3) are correct
Option D (Both (2) and (3) are correct) is correct because in an isothermal expansion of an ideal gas, the internal energy remains constant, and the heat absorbed equals the work done.
370 Medium
370 of 384
Analyzing +4 | -1
Assertion: COP of a refrigerator can’t be greater than unity.
Reason: some work is done by working substance.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because the COP of a refrigerator can be greater than unity, and the reason provided is not universally true for all refrigerators.
371 Medium
371 of 384
Application +4 | -1
Air expands from 5 lit to 10 lit at 2 atm pressure. External work done is
View answer and explanation
Correct option: D
1000 J
Option D (1000 J) is correct because the external work done during expansion at constant pressure is calculated using W=PΔVW = P\Delta V.
372 Medium
372 of 384
Analyzing +4 | -1
Assertion(A): During melting of ice, work done by surrounding on (ice + water) system is positive.

Reason(R): Volume of the given system decreases on melting of ice.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
Option A (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion) is correct because the volume decreases during melting, and the work done by the surroundings is positive.
373 Medium
373 of 384
Application +4 | -1
1 gmgm liquid is converted into vapour under 3×105 Pa3 \times 10^5 {~Pa}. 10%10 \% of heat is used to expand volume by 1600 cm31600 {~cm}^3. What is the increase in internal energy:-
View answer and explanation
Correct option: B
4320
Option B (4320) is correct because 90% of the heat is used to increase the internal energy after accounting for the expansion work.
374 Easy
374 of 384
Remembering +4 | -1
Which of the following has maximum specific heat
View answer and explanation
Correct option: A
Water
Option A (Water) is correct because water has a higher specific heat capacity compared to alcohol, glycerine, and oil.
375 Medium
375 of 384
Application +4 | -1
A gas follows a process TVn−1=T V^{n-1}= constant, where T=T= absolute temperature of the gas and V=V= volume of the gas. The bulk modulus of the gas in the process is given by
View answer and explanation
Correct option: C
npn p
Option C (npn p) is correct because the bulk modulus for a process pVn=constantpV^n = constant is given by npn p.
376 Medium
376 of 384
Application +4 | -1
nn moles of an ideal gas undergo a process in which the temperature changes with volume as T=KV2T=K V^2. The work done by the gas as the temperature changes from T0T_0 to 4T04 T_0 is
View answer and explanation
Correct option: C
(32)nRT0\left(\frac{3}{2}\right) n R T_0
Option C ((32)nRT0\left(\frac{3}{2}\right) n R T_0) is correct because the work done is calculated by integrating p dV, using the given temperature-volume relationship.
377 Hard
377 of 384
Application +4 | -1
A monoatomic gas (γ=5/3)(\gamma = 5/3) is suddenly compressed to 1/8 of its original volume adiabatically, then the pressure of the gas, will change to
View answer and explanation
Correct option: D
32 times its initial pressure
Option D (32 times its initial pressure) is correct because the pressure changes by a factor of 32 when the volume is compressed adiabatically to 1/8 of its original volume.
378 Medium
378 of 384
Understanding +4 | -1
Statement-I: Reversible systems are difficult to find in real world
Statement-II: Most processes are dissipative in nature
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Option A (Statement-I is true and statement-II is true) is correct because reversible systems are indeed difficult to find due to the dissipative nature of most real-world processes.
379 Medium
379 of 384
Understanding +4 | -1
Which of the following statements is true about the indicator diagram of adiabatic and isothermal processes?
View answer and explanation
Correct option: B
The slope of adiabatic is more than that of isothermal
Option B (The slope of adiabatic is more than that of isothermal) is correct because the adiabatic curve is steeper than the isothermal curve, as the ratio of slopes is given by γ > 1.
380 Medium
380 of 384
Remembering +4 | -1
The Davison – Germer experiment is direct evidence of
View answer and explanation
Correct option: B
Wave nature of electron
Option B (Wave nature of electron) is correct because the Davisson-Germer experiment demonstrated the diffraction of electrons, confirming their wave-like properties.
381 Medium
381 of 384
Understanding +4 | -1
Statement-I: First law of thermodynamics specifies the conditions under which a body can use its heat energy to produce the work.
Statement-II: Second law of thermodynamics states that heat always flows from hot body to cold body by itself
View answer and explanation
Correct option: C
Statement-I is false and statement-II is true
Option C is correct. The first law of thermodynamics describes energy conservation, not the conditions for heat to work conversion. The second law correctly states that heat flows spontaneously from hot to cold bodies.
382 Medium
382 of 384
Understanding +4 | -1
If γ\mathrm{\gamma } is the ratio of specific heats and RR is the universal gas constant, then the molar specific heat at constant volume Cv{C_v} is given by
View answer and explanation
Correct option: C
Rγ−1\frac{R}{{\mathrm{\gamma } - 1}}
Option C (Rγ−1\frac{R}{{\gamma - 1}}) is correct because it is derived from Mayer's formula and the definition of the ratio of specific heats.
383 Medium
383 of 384
Understanding +4 | -1
A thermally insulated container is divided into two parts by a screen. In one part the pressure and temperature are PP and TT for an ideal gas filled. In the second part it is vacuum. If now a small hole is created in the screen, then the temperature of the gas will
View answer and explanation
Correct option: C
Remain same
Option C (Remain same) is correct because the free expansion of an ideal gas into a vacuum is an adiabatic process with no external work done, resulting in no temperature change.
384 Medium
384 of 384
Application +4 | -1
An ideal heat engine is working between source at 727∘C727^\circ C and sink at 227∘C227^\circ C. If heat absorbed by it from source is 1000 cal then heat rejected to the sink is
View answer and explanation
Correct option: A
2100 J
Option A (2100 J) is correct because it is calculated using the efficiency formula for a heat engine and converting calories to joules.
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