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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 241-270 of 363 questions

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241 Medium
241 of 363
Application +4 | -1
An ideal heat engine exhausting heat at 27∘C{27^ \circ }C is to have 25% efficiency. It must take heat at
View answer and explanation
Correct option: A
127∘C{127^ \circ }C
Option A (127∘C{127^ \circ }C) is correct because it is calculated using the efficiency formula for a heat engine and converting temperatures to Kelvin.
242 Medium
242 of 363
Understanding +4 | -1
Which is the correct statement
View answer and explanation
Correct option: A
For an isothermal change PV=PV =Constant
The correct answer is A (For an isothermal change PV=PV =Constant) because this is a direct consequence of Boyle's law, which applies to isothermal processes.
243 Medium
243 of 363
Understanding +4 | -1
If the door of a refrigerator is kept open, then which of the following is true
View answer and explanation
Correct option: B
Room is heated
Option B (Room is heated) is correct because a refrigerator transfers heat from its inside to the surrounding environment, and keeping the door open will eventually heat the room.
244 Medium
244 of 363
Remembering +4 | -1
Which of the following the scalar quantity is
View answer and explanation
Correct option: A
Temperature
The correct answer is A (Temperature) because temperature is a scalar quantity, meaning it only has magnitude and no direction.
245 Medium
245 of 363
Understanding +4 | -1
Which of the following process is reversible?
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because all the listed processes are irreversible due to entropy increase.
246 Medium
246 of 363
Understanding +4 | -1
The quantity of heat required to raise one mole through one degree kelvin for a monoatomic gas at constant volume is
View answer and explanation
Correct option: A
32R\frac{3}{2}R
Option A (32R\frac{3}{2}R) is correct because the molar specific heat at constant volume for a monoatomic gas is 32R\frac{3}{2}R.
247 Medium
247 of 363
Application +4 | -1
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx=T{V^x} = constant, then xx is
View answer and explanation
Correct option: A
25\frac{2}{5}
Option A (25\frac{2}{5}) is correct because it is the value of x derived from the adiabatic process equation for a diatomic gas.
248 Medium
248 of 363
Application +4 | -1
A certain mass of gas at 273K273K is expanded to 81 times its volume under adiabatic condition. If γ=1.25\gamma = 1.25 for the gas, then its final temperature is
View answer and explanation
Correct option: B
−182∘C- {182^ \circ }C
Option B (−182∘C- {182^ \circ }C) is correct because it is calculated using the adiabatic process equation TVγ−1=constantT{V^{\gamma - 1}} = constant.
249 Medium
249 of 363
Application +4 | -1
A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 litres. If the ratio of final pressure of the gas to initial pressure of the gas is 1681\frac{16}{81}. Then the ratio of CpCv\frac{C_{p}}{C_{v}} will be
View answer and explanation
Correct option: A
43\frac{4}{3}
Option A (43\frac{4}{3}) is correct because it is calculated using the adiabatic process equation PVγ=constantP V^{\gamma} = constant and solving for γ=CpCv\gamma = \frac{C_p}{C_v}.
250 Medium
250 of 363
Understanding +4 | -1
(A): Heat flows always from a hotter body to a colder body by itself.
(R): Without an external aid heat cannot flow from colder body to a hotter body.
View answer and explanation
Correct option: B
Both (A) and (R) are true and (R) is not the correct explanation of (A)
Option B is correct because heat naturally flows from a hotter to a colder body, and external assistance is required for the reverse process.
251 Medium
251 of 363
Analyzing +4 | -1
An engineer claims to have made an engine delivering 10kW10kW power with fuel consumption of 1g/s1g/s. The calorific value of the fuel is 2kcal/g2kcal/g. Is the claim of the engineer
View answer and explanation
Correct option: B
Invalid
Option B (Invalid) is correct because the engine's claimed efficiency exceeds 100%, which is impossible based on the laws of thermodynamics.
252 Medium
252 of 363
Application +4 | -1
A carnot engine takes 300 calories of heat from a source of 500 K and rejects 150 calories of heat to the sink. The temperature of the sink is
View answer and explanation
Correct option: B
250 K
Option B (250 K) is correct because it is calculated using the Carnot engine efficiency formula relating heat and temperature.
253 Medium
253 of 363
Understanding +4 | -1
A system is taken from a given initial state to a given final state along various paths represented on a P−VP-V diagram. The quantity that is independent of the path is:
View answer and explanation
Correct option: C
Change in internal energy (ΔU)(\Delta U)
Option C (Change in internal energy (ΔU)(\Delta U)) is correct because internal energy is a state function, and its change depends only on the initial and final states, not the path.
254 Medium
254 of 363
Understanding +4 | -1
When a small amount of heat ΔQ\Delta Q is added to an enclosed gas, then increase in internal energy and external work done are related as
View answer and explanation
Correct option: B
ΔQ=mCvΔT+pΔV\Delta Q = m{C_v}\Delta T + p\Delta V
Option B (ΔQ=mCvΔT+pΔV\Delta Q = m{C_v}\Delta T + p\Delta V) is correct because it represents the first law of thermodynamics, where heat added equals the change in internal energy plus the work done.
255 Medium
255 of 363
Analyzing +4 | -1
Assertion: Between states (P1,V1,T1)\left(\mathrm{P}_1, \mathrm{V}_1, \mathrm{T}_1\right) and (P2,V2,T2)\left(\mathrm{P}_2, \mathrm{V}_2, \mathrm{T}_2\right), work done in two step process is more than work done in one step process.
Reason: Work done is a path function.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because work done depends on the process, and work done is a path function.
256 Medium
256 of 363
Remembering +4 | -1
In gases of diatomic molecules, the ratio of the two specific heats of gases CP/CV{C_P}/{C_V} is
View answer and explanation
Correct option: B
1.401.40
Option B (1.401.40) is correct because for diatomic gases, the ratio of specific heats, CP/CV{C_P}/{C_V}, is approximately 1.40.
257 Medium
257 of 363
Application +4 | -1
The efficiency of a Carnot engine is 60%60 \%. If the temperature of source is 127∘C127^{\circ} \mathrm{C}. The sink must be maintained at
View answer and explanation
Correct option: C
−113∘C-113^{\circ} \mathrm{C}
The correct answer is C (−113∘C-113^{\circ} \mathrm{C}) because it is the correct sink temperature calculated using the Carnot efficiency formula.
258 Medium
258 of 363
Understanding +4 | -1
The specific heat of a gas in an isothermal process is
View answer and explanation
Correct option: A
Infinite
Option A (Infinite) is correct because in an isothermal process, the temperature change is zero, leading to an infinite specific heat.
259 Medium
259 of 363
Remembering +4 | -1
In which thermodynamic process, volume remains same?
View answer and explanation
Correct option: D
Isochoric
Option D (Isochoric) is correct because an isochoric process is defined as a thermodynamic process where the volume remains constant.
260 Medium
260 of 363
Analyzing +4 | -1
(A): The adiabatic exponent (γ=CpCv)\left(\gamma=\frac{C_p}{C_v}\right) of adiatomic gas is less than that of a mono atomic gas
(R): The molecules of mono atomic gas more degrees of freedom.
View answer and explanation
Correct option: C
(A) is true but (R) is false
Option C ((A) is true but (R) is false) is correct because the adiabatic exponent is indeed less for diatomic gases compared to monatomic gases, but this is because diatomic gases have more degrees of freedom, not less.
261 Medium
261 of 363
Application +4 | -1
One gram of water (volume =1 cm3=1 \mathrm{~cm}^3 ) becomes 1671 cm3\mathrm{cm}^3 of steam when boiled at a pressure of one atmosphere . Latent heat of vaporization at this pressure is 539 cal/gm539 \mathrm{~cal} / \mathrm{gm}. Compute the work done. [1 atm=1.013×105Nm−2\left[1 \mathrm{~atm}=1.013 \times 10^5 \mathrm{Nm}^{-2}\right. ]
View answer and explanation
Correct option: A
169
The correct answer is 169 because the work done is calculated using the formula W = P(V2 - V1) with the given pressure and change in volume.
262 Medium
262 of 363
Understanding +4 | -1
Choose the correct statement
View answer and explanation
Correct option: B
Heat is a path function, while internal energy is not
Option B is correct because heat is a path-dependent function, while internal energy is a state function and depends only on the initial and final states.
263 Medium
263 of 363
Understanding +4 | -1
The slopes of isothermal and adiabatic curves are related as
View answer and explanation
Correct option: C
Adiabatic curve slope =γ×= \gamma \times isothermal curve slope
Option C (Adiabatic curve slope =γ×= \gamma \times isothermal curve slope) is correct because the adiabatic slope is γ\gamma times the isothermal slope.
264 Medium
264 of 363
Understanding +4 | -1
Assertion: The efficiency of a heat engine can be unity.
Reason: Efficiency of heat engine is fundamental limitation given by first law of thermodynamics.
View answer and explanation
Correct option: D
If both assertion and reason are false
Option D (If both assertion and reason are false) is correct because the efficiency of a heat engine cannot be unity, and the first law of thermodynamics does not impose a fundamental limitation on efficiency.
265 Medium
265 of 363
Analyzing +4 | -1
Statement-I: Monoatomic, diatomic and polyatomic gases are adiabatically compressed such that compression ratio is (P2P1)\left(\frac{P_2}{P_1}\right). Then monoatomic gas will have maximum final volume out of these three gases.
Statement-II: Monoatomic gas has least degree of freedom.
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 monoatomic gases have the least adiabatic index, leading to a maximum final volume after adiabatic compression, and they have the fewest degrees of freedom.
266 Medium
266 of 363
Application +4 | -1
A carnot engine takes 3×106  cal3 \times {10^6}\,\,cal of heat from a reservoir at 6270C{627^0}C and gives it to a sink at 270C{27^0}C. The work done by the engine is,
View answer and explanation
Correct option: B
8.4×106J8.4 \times {10^6}J
Option B (8.4×106J8.4 \times {10^6}J) is correct because it is calculated using the Carnot engine efficiency formula and the heat absorbed and rejected.
267 Medium
267 of 363
Remembering +4 | -1
The relation between two specific heats of a gas is
View answer and explanation
Correct option: A
CP−CV=RJ{C_P} - {C_V} = \frac{R}{J}
Option A (CP−CV=RJ{C_P} - {C_V} = \frac{R}{J}) is correct because it represents the relationship between specific heat at constant pressure, specific heat at constant volume, the gas constant, and the mechanical equivalent of heat.
268 Medium
268 of 363
Understanding +4 | -1
One mole of oxygen is expanded from a volume 1 L1 \mathrm{~L} to 5 L5 \mathrm{~L} at a constant temperature T=280 KT=280 \mathrm{~K}. The change in internal energy is:
View answer and explanation
Correct option: C
0 kJ
Option C (0 kJ) is correct because the internal energy of an ideal gas depends only on its temperature, and since the temperature is constant, the change in internal energy is zero.
269 Medium
269 of 363
Understanding +4 | -1
An ideal gas A and a real gas B have their volumes increased from V to 2V under isothermal conditions. The increase in internal energy
View answer and explanation
Correct option: C
of B will be more than that of A
Option C is correct. The internal energy of a real gas (B) increases more than that of an ideal gas (A) under isothermal expansion because work is done against intermolecular forces in the real gas.
270 Medium
270 of 363
Analyzing +4 | -1
(A): A room can not be cooled by opening the door of a refrigerator in a closed room
(R): Heat flows from lower temperature (refrigerator) to higher temperature (room).
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 a refrigerator transfers heat from a cold reservoir to a hot reservoir, requiring work, and opening it in a closed room will eventually heat the room.
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