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JEE • PHYSICS • THERMODYNAMICS

Thermodynamics practice · Medium

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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 271-300 of 363 questions

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271 Medium
271 of 363
Remembering +4 | -1
In thermodynamics, heat and work are:
View answer and explanation
Correct option: C
Path functions
Option C (Path functions) is correct because heat and work depend on the process or path taken, not just the initial and final states.
272 Medium
272 of 363
Understanding +4 | -1
The specific heat of an ideal gas for an isothermal process is
View answer and explanation
Correct option: D
infinite
Option D (infinite) is correct because in an isothermal process, the temperature change is zero, leading to infinite specific heat.
273 Medium
273 of 363
Application +4 | -1
An engine is supposed to operate between two reservoirs at temperature 7270C  and  2270C{727^0}C\,\,and\,\,{227^0}C . The maximum possible efficiency of such an engine is
View answer and explanation
Correct option: A
12\frac{1}{2}
The correct answer is 2 because it follows from the intensity equation in Young's double slit experiment.
274 Medium
274 of 363
Understanding +4 | -1
The internal energy of an ideal gas increases during an isothermal process when the gas is
View answer and explanation
Correct option: A
Expanded by adding more molecules to it
Option A (Expanded by adding more molecules to it) is correct because the internal energy of an ideal gas depends on the number of molecules and temperature; at constant temperature, increasing the number of molecules increases internal energy.
275 Medium
275 of 363
Remembering +4 | -1
The molar specific heat at constant pressure for a monoatomic gas is
View answer and explanation
Correct option: B
52R\frac{5}{2}R
Option B (52R\frac{{5}}{{2}}R) is correct because for a monoatomic gas, the molar specific heat at constant pressure is given by Cp=52RC_p = \frac{{5}}{{2}}R.
276 Medium
276 of 363
Application +4 | -1
nn moles of an ideal gas with constant volume heat capacity CV{C_V} undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is
View answer and explanation
Correct option: D
nRCV+nR\frac{{nR}}{{{C_V} + nR}}
Option D (nRCV+nR\frac{{nR}}{{{{C_V} + nR}}}) is correct because it represents the ratio of work done to heat supplied during an isobaric process for an ideal gas.
277 Medium
277 of 363
Analyzing +4 | -1
Assertion: A heat engine is the reverse of a refrigerator.
Reason: A refrigerator cannot work without some external work done on the system.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of assertion
Option A is correct. Heat engines and refrigerators are reverse processes, and refrigerators require external work to function.
278 Medium
278 of 363
Understanding +4 | -1
(A): An adiabatic process is an isoentropic process.
(R): Change in entropy is zero in case of adiabatic process.
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 an adiabatic process is defined as a process with no heat exchange, which implies no change in entropy, making it isoentropic.
279 Medium
279 of 363
Analyzing +4 | -1
An ideal gas in a piston cylinder arrangement is to be taken from a temperature of 500 K500 {~K} to a temperature of 600 K600 {~K}. Which of the thermodynamic process given in options should be used so that the final state of 600 K600 {~K} is achieved using a minimum energy input from outside?
View answer and explanation
Correct option: C
Isochoric
Option C (Isochoric) is correct because in an isochoric process, no work is done, so all the energy input goes into increasing the internal energy, thus minimizing the energy needed to reach 600 K.
280 Medium
280 of 363
Understanding +4 | -1
The change of entropy, when heat is absorbed by the gas is:
View answer and explanation
Correct option: A
Positive
Option A is correct because when a gas absorbs heat, its entropy increases as the molecules become more disordered.
281 Medium
281 of 363
Application +4 | -1
In a Carnot engine, when T2=0 ∘C   and    T1=200 ∘C{T_2} = 0\,^\circ C\,\,\,\mathrm{and}\;\,\,{T_1} = 200\,^\circ C its efficiency is  η1    and     when     T1=0 ∘C    and    T2=−200 ∘C  ,{\mathrm{\eta }_1}\mathrm{\;}\,\,\mathrm{and\;}\,\,\,\mathrm{when}\;\,\,\,{T_1} = 0\,^\circ C\;\,\,\mathrm{and}\,\,\;{T_2} = - 200\,^\circ C\;, its efficiency is η2{\mathrm{\eta }_2}, then what in η1/η2?{\eta _1}/{\eta _2}?
View answer and explanation
Correct option: A
0.577
Option A (0.577) is correct because it represents the ratio of the efficiencies calculated using the Carnot engine efficiency formula with the given temperatures converted to Kelvin.
282 Medium
282 of 363
Understanding +4 | -1
The theory of refrigerator is based on
View answer and explanation
Correct option: D
None of the above
Option A (Joule –Thomson effect) is incorrect, Option B (Newton’s particle theory) is incorrect, Option C (Joule’s effect) is incorrect, therefore Option D (None of the above) is the correct answer because the refrigerator's operation is based on the second law of thermodynamics.
283 Medium
283 of 363
Application +4 | -1
An engine takes in compressed steam at 127∘C{127^ \circ }C and rejects it at 47∘C{47^ \circ }C . Efficiency of the engine is
View answer and explanation
Correct option: C
0.2
Option C (20%) is correct because the efficiency is calculated by using the formula η=1−T2T1\eta = 1 - \frac{{{T_2}}}{{{T_1}}} with temperatures in Kelvin.
284 Medium
284 of 363
Remembering +4 | -1
In which of the processes, does the internal energy of the system remain constant?
View answer and explanation
Correct option: D
Isothermal
Option D (Isothermal) is correct because an isothermal process occurs at a constant temperature, meaning the internal energy remains constant.
285 Medium
285 of 363
Application +4 | -1
The ΔHf∘\Delta H_f^{\circ} for CO2( g),CO(g)\mathrm{CO}_2(\mathrm{~g}), \mathrm{CO}(\mathrm{g}) and H2O(g)\mathrm{H}_2 \mathrm{O}(\mathrm{g}) are −393.5,−110.5-393.5,-110.5 and −241.8 kJ mol−1-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively. The standard enthalpy change (in kJ\mathrm{kJ} ) for the reaction,
CO2( g)+H2( g)⟶CO(g)+H2O(g)\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) is
View answer and explanation
Correct option: B
41.2
Option B (25 mm) is correct because the separation is calculated by determining the positions of the proton and alpha particle when their velocities are perpendicular, considering their circular paths in the magnetic field.
286 Medium
286 of 363
Understanding +4 | -1
In an isobaric process, ΔQ=γKγ−1\Delta Q=\frac{\gamma K}{\gamma-1}, where γ\gamma is Cp/Cv\boldsymbol{C}_{\boldsymbol{p}} / \boldsymbol{C}_{\boldsymbol{v}} What is KK ?
View answer and explanation
Correct option: D
Work done
Option D is correct because in an isobaric process, the heat added is related to the change in internal energy and the work done, and K represents the work done.
287 Medium
287 of 363
Application +4 | -1
An ideal gas at pressure 2.5×105 Pa2.5 \times 10^5 \mathrm{~Pa} and temperature 300 K300 \mathrm{~K} occupies 100cc100 \mathrm{cc}. It is adiabatically compressed to half its original volume. Calculate (a) the final temperature and (b) the work done by the gas in the process. Take γ=1.5\gamma=1.5.
View answer and explanation
Correct option: A
424 K; 21 J
Option A (424 K; 21 J) is correct because it accurately calculates the final temperature using the adiabatic process formula and the work done using the appropriate formula.
288 Medium
288 of 363
Understanding +4 | -1
The work done, W during an isothermal process in which 1 mole of the gas expands from an initial volume V1{V_1}to a final volume V2{V_2} is given by (R=gas constant, T=temperature)
View answer and explanation
Correct option: C
RTlog⁡e(V2V1)RT{\log _e}\left( {\frac{{{V_2}}}{{{V_1}}}} \right)
Option C is correct. The work done during an isothermal process is given by the formula W=RTlog⁡e(V2V1)W = RT \log_e(\frac{V_2}{V_1}).
289 Medium
289 of 363
Understanding +4 | -1
In an isothermal expansion
View answer and explanation
Correct option: C
Internal energy remains unchanged
Option C is correct because in an isothermal process, the temperature remains constant, and therefore the internal energy, which depends on temperature, also remains unchanged.
290 Medium
290 of 363
Application +4 | -1
Temperature of 100gm100 \mathrm{gm} water is changed from 0∘C0^{\circ} \mathrm{C} to 3∘C3^{\circ} \mathrm{C} in this process, heat supplied to water will be (specified heat of water =1 cal g−1∘C−1=1 \mathrm{~cal} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}
View answer and explanation
Correct option: C
less than 300 cal300 \mathrm{~cal}
Option C (less than 300 cal300 \mathrm{~cal}) is correct because the density of water increases from 0∘C0^{\circ} \mathrm{C} to 4∘C4^{\circ} \mathrm{C}, resulting in negative work done by the system.
291 Medium
291 of 363
Remembering +4 | -1
The specific heat of an ideal gas varies with temperature TT as
View answer and explanation
Correct option: D
T0{T^0}
Option D (T0{T^0}) is correct because the specific heat of an ideal gas is independent of temperature.
292 Medium
292 of 363
Understanding +4 | -1
If a system undergoes contraction of volume, then the work done by the system is
View answer and explanation
Correct option: B
negative
Option B is correct because when a system undergoes contraction of volume, the work done *by* the system is negative.
293 Medium
293 of 363
Understanding +4 | -1
(A): CpC_p is always greater than CVC_V in gases.
(R): Work done at constant pressure is more than at constant volume.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A is correct because CpC_p is always greater than CVC_V for gases, and the reason correctly explains this: work done at constant pressure is more than at constant volume, requiring more heat input.
294 Medium
294 of 363
Analyzing +4 | -1
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A: Efficiency of a reversible heat engine will be highest at −273∘C- 273^\circ C temperature of cold reservoir.

Reason R: The efficiency of Carnot’s engine depends not only on temperature of cold reservoir but it depends on the temperature of hot reservoir too and is given as η=(1−T2T1)\eta = \left( {1 - \frac{{{T_2}}}{{{T_1}}}} \right).
In the light of the above statements, choose the correct answer from the options given below
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 efficiency of a reversible heat engine is highest at 0K (-273°C) and the Carnot efficiency equation explains this dependence on both hot and cold reservoir temperatures.
295 Medium
295 of 363
Application +4 | -1
A given mass of a gas is compressed isothermally until its pressure is doubled. It is then allowed to expand adiabatically until its original volume is restrored and is pressure is then found to be 0.75 of its initial pressure. The ratio of the specific heats of the gas is approximately
View answer and explanation
Correct option: B
1.41
Option B (1.41) is correct because the ratio of specific heats (γ) is determined by analyzing the isothermal and adiabatic processes, resulting in approximately 1.41.
296 Medium
296 of 363
Application +4 | -1
A thermally insulated vessel contains an ideal gas of molecular mass MM and ratio of specific heats 1.4. Vessel is moving with speed vv and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by (RR = Universal gas constant)
View answer and explanation
Correct option: B
Mv25R\frac{{M{v^2}}}{{5R}}
Option B (Mv25R\frac{{M{v^2}}}{{5R}}) is correct because the increase in temperature is due to the conversion of kinetic energy into internal energy of the gas.
297 Medium
297 of 363
Application +4 | -1
A carnot engine, having efficiency of η=120\eta = \frac{1}{{20}} as heat engine, is used as a refrigerator. If the work done on the system is 20 J, the amount of energy absorbed from the reservoir at lower temperature is
View answer and explanation
Correct option: D
380 J
Option D (380 J) is correct because it represents the heat extracted from the cold reservoir when the Carnot engine operates as a refrigerator.
298 Medium
298 of 363
Understanding +4 | -1
A sample of gas expands from volume V1{V_1}to V2{V_2} . The amount of work done by the gas is maximum when the expansion is
View answer and explanation
Correct option: A
Isothermal
Option A (Isothermal) is correct because in an isothermal process, the gas expands at a constant temperature, resulting in maximum work done compared to other processes.
299 Medium
299 of 363
Understanding +4 | -1
The volume of an ideal diatomic gas is doubled isothermally, the internal energy
View answer and explanation
Correct option: D
Remains unchanged
Option D (Remains unchanged) is correct because the internal energy of an ideal gas depends only on its temperature, and the process is isothermal.
300 Medium
300 of 363
Remembering +4 | -1
The first law of thermodynamics is based on the law of conservation of
View answer and explanation
Correct option: A
Energy
Option A (Energy) is correct because the first law of thermodynamics is a statement of the conservation of energy.
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