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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 241-270 of 384 questions

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241 Medium
241 of 384
Analyzing +4 | -1
Assertion: When a bullet is fired from a gun, the bullet pierces a wooden block and stops, decreasing the temperature of the bullet and the surrounding layers of wood.
Reason: Temperature is related to the energy of motion of the bullet as a whole.
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 assertion is false (temperature increases) and the reason is false (temperature relates to internal motion).
242 Medium
242 of 384
Remembering +4 | -1
In an adiabatic process, the quantity which remains constant is
View answer and explanation
Correct option: D
Total heat of the system
Option D (Total heat of the system) is correct because, by definition, an adiabatic process involves no heat exchange with the surroundings, thus the total heat remains constant.
243 Medium
243 of 384
Understanding +4 | -1
Assertion: In a thermo flask, inner surface is shiny.
Reason: Shiny surface is poor radiator of heat.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because a shiny surface is a poor radiator of heat, which helps to minimize heat loss in a thermos flask.
244 Medium
244 of 384
Remembering +4 | -1
Which property of the system does not change in an adiabatic process
View answer and explanation
Correct option: D
Total amount of heat
Option D (Total amount of heat) is correct because, by definition, an adiabatic process is one in which there is no heat exchange between the system and its surroundings.
245 Medium
245 of 384
Understanding +4 | -1
(A): If an ideal gas expands in vaccum in an insulated chamber, ΔQ,ΔU,ΔW\Delta Q, \Delta U, \Delta W all are zero.
(R): Temperature of gas remains constant.
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 both the assertion and the reason are true, but the reason (temperature remains constant) is not the correct explanation for the assertion (ΔQ,ΔU,ΔW\Delta Q, \Delta U, \Delta W are zero).
246 Medium
246 of 384
Understanding +4 | -1
In the equation for I law of thermodynamics dQ=du+dwd Q=d u+d w

a) dQdQ depends on path

b) dwdw depends on path

c) dudu depends on path
View answer and explanation
Correct option: B
Both a, b are correct
Option B is correct because heat and work are path-dependent, while internal energy is a state function and depends only on the initial and final states.
247 Medium
247 of 384
Application +4 | -1
Two Carnot engines AA and BB are operated in succession. The first one,AA receives heat from a source at T1=800K{T_1} = 800K and rejects to sink at T2K{T_2}K . The second engine BB receives heat rejected by the first engine and rejects to another sink at T3=300K{T_3} = 300K. If the work outputs of two engines are equal, then the value of T2{T_2} is
View answer and explanation
Correct option: C
550K550K
Option C (550K550K) is correct because the intermediate temperature is calculated by equating the work outputs of the two Carnot engines.
248 Medium
248 of 384
Understanding +4 | -1
For an engine operating between t1∘C{t_1}^ \circ C and t2∘C{t_2}^ \circ C, the efficiency will be
View answer and explanation
Correct option: D
t1−t2t1+273\frac{{{t_1} - {t_2}}}{{{t_1} + 273}}
Option D (t1−t2t1+273\frac{{{t_1} - {t_2}}}{{{t_1} + 273}}) is correct because the efficiency of an engine is calculated using absolute temperatures (Kelvin).
249 Medium
249 of 384
Analyzing +4 | -1
100 mol100 \mathrm{~mol} of an ideal gas is heated from 10∘10^{\circ} to 20∘C20^{\circ} \mathrm{C} keeping its (i) volume constant (ii) pressure constant. Let ΔUa\Delta U_a and ΔUb\Delta U_b denote the change in the internal energy of the gas due to process (i) and (ii), respectively. Then, which of the following shall be true?
View answer and explanation
Correct option: B
ΔUa>ΔUb\Delta U_a > \Delta U_b
Option A (ΔUa=ΔUb\Delta U_a=\Delta U_b) is incorrect. For a constant volume process, all heat added goes into internal energy. For a constant pressure process, some heat is used for work, so the internal energy change is less.
250 Medium
250 of 384
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.
251 Medium
251 of 384
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.
252 Medium
252 of 384
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.
253 Medium
253 of 384
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.
254 Medium
254 of 384
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.
255 Medium
255 of 384
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.
256 Medium
256 of 384
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.
257 Medium
257 of 384
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.
258 Medium
258 of 384
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}.
259 Medium
259 of 384
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.
260 Medium
260 of 384
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.
261 Medium
261 of 384
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.
262 Medium
262 of 384
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.
263 Medium
263 of 384
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.
264 Medium
264 of 384
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.
265 Easy
265 of 384
Understanding +4 | -1
The internal energy of a system remains constant
when it undergoes

(A) a cycle process

(B) an isothermal process

(C) an adiabatic process

(D) any process in which the heat given out by the
system is equal to the work done on the system

Choose the correct statement
View answer and explanation
Correct option: D
A, B, D
Option A, B, D is correct because internal energy remains constant in a cyclic process, an isothermal process, and any process where heat given out equals work done.
266 Medium
266 of 384
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.
267 Medium
267 of 384
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.
268 Medium
268 of 384
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.
269 Medium
269 of 384
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.
270 Medium
270 of 384
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.
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