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JEE • PHYSICS • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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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 91-120 of 384 questions

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91 Medium
91 of 384
Application +4 | -1
What is the temperature of source in Carnot cycle of 10% efficiency when heat exhausts at 270 K?
View answer and explanation
Correct option: C
300 K
The correct answer is C (300 K) because it is the temperature of the source calculated using the Carnot cycle efficiency formula.
92 Medium
92 of 384
Understanding +4 | -1
Two cylinders of equal capacity are connected to each other via stop cock. AA contains an ideal gas at a standard temperature and pressure, BB is completely evacuated. The entire system is thermally insulated. The stop cock suddenly opens. Which one is correct option?
View answer and explanation
Correct option: D
Change in internal energy is zero
Option D (Change in internal energy is zero) is correct because in free expansion against vacuum, no work is done, and for an ideal gas, internal energy depends only on temperature.
93 Medium
93 of 384
Analyzing +4 | -1
Assertion(A): In isothermal process, whole of the heat energy supplied to the body is converted into internal energy.

Reason(R): According to the first law of thermodynamics,

ΔQ=ΔU+ΔW\Delta Q=\Delta U+\Delta W
View answer and explanation
Correct option: D
Assertion is incorrect but Reason is correct.
The correct answer is D because the assertion is incorrect (heat energy is converted into work, not internal energy in an isothermal process), but the reason (first law of thermodynamics) is correct.
94 Medium
94 of 384
Application +4 | -1
The water of volume 4 m34 \mathrm{~m}^3 at the height 20 m20 \mathrm{~m} is pressed by 2×105 N2 \times 10^5 \mathrm{~N} pressure. The work done by motor is
View answer and explanation
Correct option: B
16×105 J16 \times 10^5 \mathrm{~J}
Option B (16×105 J16 \times 10^5 \mathrm{~J}) is correct because the work done is the sum of the work done against gravity and the work done against pressure.
95 Medium
95 of 384
Application +4 | -1
10 moles of an ideal monoatomic gas at 10∘C{10^{^ \circ }}C is mixed with 20 moles of another monoatomic gas at 20∘C.{20^ \circ }C. Then the temperature of the mixture is
View answer and explanation
Correct option: D
16.6∘C{16.6^{^ \circ }}C
Option D (16.6∘C{16.6^{^\circ }}C) is correct because it is the final temperature calculated using the principle of heat exchange between the two gases.
96 Medium
96 of 384
Understanding +4 | -1
(A): Thermodynamic process in nature are irreversible.
(R): Dissipative effects can not be eliminated.
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 thermodynamic processes in nature are irreversible due to the presence of dissipative effects that cannot be eliminated.
97 Medium
97 of 384
Understanding +4 | -1
One mole of helium is adiabatically expanded from its initial state (Pi,Vi,Ti)\left(P_i, V_i, T_i\right) to its final state (Pf,Vf,Tf)\left(P_f, V_f, T_f\right). The decrease in the internal energy associated with this expansion is equal to
View answer and explanation
Correct option: A
CV(Ti−Tf){C}_{{V}}\left({T}_{{i}}-{T}_{{f}}\right)
Option A (CV(Ti−Tf){C}_{{V}}\left({T}_{{i}}-{T}_{{f}}\right)) is correct because the change in internal energy for an adiabatic process is given by ΔU=nCVΔT\Delta U = nC_V\Delta T, and in this case, the decrease is CV(Ti−Tf)C_V(T_i - T_f).
98 Medium
98 of 384
Understanding +4 | -1
The possibility of increase in the temperature of a gas without adding heat to it happens in:
View answer and explanation
Correct option: B
Adiabatic compression
Option B is correct because in adiabatic compression, work is done on the gas, increasing its internal energy and temperature without adding heat.
99 Medium
99 of 384
Application +4 | -1
The relation between energy UU, pressure PP and volume VV for an ideal gas in an adiabatic process is given by, relation U=a+bPVU=a+b P V. Find the value of adiabatic exponent (γ)(\gamma) of this gas.
View answer and explanation
Correct option: A
b+1b\frac{b+1}{b}
Option A (b+1b\frac{b+1}{b}) is correct because it is derived by comparing the given relation with the adiabatic process equation and solving for the adiabatic exponent.
100 Medium
100 of 384
Understanding +4 | -1
Even Carnot engine cannot give 100% efficiency because we cannot
View answer and explanation
Correct option: C
Reach absolute zero temperature
Option C (Reach absolute zero temperature) is correct because achieving 100% efficiency in a Carnot engine requires the cold reservoir to be at absolute zero, which is impossible.
101 Medium
101 of 384
Remembering +4 | -1
The mechanical equivalent of heat JJ is
View answer and explanation
Correct option: C
A conversion factor
Option C (A conversion factor) is correct because the mechanical equivalent of heat, J, serves as a conversion factor between heat and mechanical energy.
102 Medium
102 of 384
Understanding +4 | -1
A heat engine is a device
View answer and explanation
Correct option: B
Which converts heat energy into mechanical energy
Option B (Which converts heat energy into mechanical energy) is correct because a heat engine's primary function is to transform thermal energy into mechanical work.
103 Medium
103 of 384
Application +4 | -1
An ideal monoatomic gas at 27∘C{27^ \circ }C is compressed adiabatically to 8/27 times of its present volume. The increase in temperature of the gas is
View answer and explanation
Correct option: A
375∘C{375^ \circ }C
Option A (375∘C{375^ \circ }C) is correct because the temperature increase is calculated using the adiabatic process equation and the given volume compression ratio.
104 Medium
104 of 384
Understanding +4 | -1
A container of volume 1  m31\,\,{{\text{m}}^3} is divided into two equal compartments, one of which contains an ideal gas at 400 K. The other compartment is vacuum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would be
View answer and explanation
Correct option: B
400 K
Option B (400 K) is correct because in free expansion, the temperature remains constant as no work is done and the system is thermally isolated.
105 Medium
105 of 384
Application +4 | -1
When heat energy of 1500joules1500joules, is supplied to a gas at constant pressure2.1×105N/m22.1 \times {10^5}N/{m^2}, there was an increase in its volume equal to 2.5×10−3N/m22.5 \times {10^{ - 3}}N/{m^2}. The increase in internal energy of the gas in joules is
View answer and explanation
Correct option: C
975
Option C (975) is correct because it is the result of subtracting the work done by the gas from the total heat supplied, according to the first law of thermodynamics.
106 Medium
106 of 384
Understanding +4 | -1
(A): During an adiabatic compression temperature of an ideal gas increases.
(R): Internal energy of a gas can be changed even without addition or withdrawn of heat but by doing work.
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 during adiabatic compression, work is done on the gas, increasing its internal energy and thus its temperature. The reason correctly explains this.
107 Medium
107 of 384
Application +4 | -1
When a mole of an ideal gas expands isothermally at 27∘C27^\circ C to double its original volume, the work done by the gas is [R=8.314   deg⁡−1 mol−1][log⁡102=0.3010]\left[ {R = 8.314\,\,\,{{\deg }^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}} \right]\left[ {\log _{10}^2 = 0.3010} \right]
View answer and explanation
Correct option: C
1726 J
Option C (1726 J) is correct because it is the work done calculated using the isothermal expansion formula with the given values.
108 Medium
108 of 384
Analyzing +4 | -1
Assertion: first law of thermodynamics allows many processes which actually don't happen.
Reason: First law of thermodynamics must not be violated for any process to happen.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because the first law allows many processes that don't happen, and the first law must hold for any process.
109 Medium
109 of 384
Remembering +4 | -1
In which of the following processes, heat is neither absorbed nor released by a system?
View answer and explanation
Correct option: B
Adiabatic
The correct answer is B (Adiabatic) because an adiabatic process is defined as one in which there is no heat exchange between the system and its surroundings.
110 Medium
110 of 384
Understanding +4 | -1
When gas in a vessel expands its internal energy decreases. The process involved is
View answer and explanation
Correct option: C
Adiabatic
Option C (Adiabatic) is correct because in an adiabatic process, there is no heat exchange with the surroundings, and the internal energy decreases as the gas expands.
111 Medium
111 of 384
Understanding +4 | -1
Statement-I: According to Joule, heat and work are related
Statement-II: For every 1 Cal1 \mathrm{~Cal} of heat we can get 4.186 J4.186 \mathrm{~J} of mechanical work.
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Statement-I is true and statement-II is true. Joule's law relates heat and work, and 1 calorie is approximately 4.186 Joules.
112 Medium
112 of 384
Application +4 | -1
A cylinder of mass 1kg1kg is given heat of 20000J20000J at atmospheric pressure. If initially temperature of cylinder is 20∘C{20^ \circ }C, then work done by the cylinder will be (Given that Specific heat of cylinder =400Jkg−1= 400J k{g^{ - 1}}), Coefficient of volume expansion =9×10−5∘C−1= 9 \times {10^{ - 5}}^ \circ {C^{ - 1}}, Atmospheric pressure=105N/m2= {10^5}N/{m^2} and density of cylinder 9000kg/m39000kg/{m^3})
View answer and explanation
Correct option: B
0.05J0.05J
Option B (0.05J) is correct because it is calculated using the formula for work done by a gas at constant pressure.
113 Medium
113 of 384
Application +4 | -1
A Carnot engine whose low-temperature reservoir is at 27∘C{27^ \circ }C has efficiency 37.5%37.5\%. The high-temperature reservoir is at
View answer and explanation
Correct option: D
207∘C{207^ \circ }C
Option D (207∘C{207^ \circ }C) is correct because it is the high-temperature reservoir calculated using the Carnot engine efficiency formula.
114 Medium
114 of 384
Understanding +4 | -1
Assertion: Specific heat of any substance remains constant at all temperatures.

Reason : It is given by s=1m⋅dQdTs=\frac{1}{m} \cdot \frac{d Q}{d T}
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D (If Assertion is false but the Reason is true.) is correct because specific heat is temperature-dependent, and the given formula for specific heat is correct.
115 Medium
115 of 384
Understanding +4 | -1
Following statements are given:
(A) The average kinetic energy of a gas molecule
decreases when the temperature is reduced.
(B) The average kinetic energy of a gas molecule
increases with increase in pressure at constant
temperature.
(C) The average kinetic energy of a gas molecule
decreases with increase in volume.
(D) Pressure of a gas increases with increase in
temperature at constant pressure.
(E) The volume of gas decreases with increase in
temperature.
Choose the correct answer from the options given
below:
View answer and explanation
Correct option: A
(A) only
Option A ((A) only) is correct because the average kinetic energy of a gas molecule is directly proportional to temperature.
116 Medium
116 of 384
Analyzing +4 | -1
An ideal monoatomic gas undergoes process PV1.25=P V^{1.25}= Constant . Then
View answer and explanation
Correct option: D
Heat is absorbed by gas if volume increases
Heat is absorbed by gas if volume increases (Option D) because the process is polytropic with an index less than the adiabatic index, implying that as volume increases, temperature decreases, leading to heat absorption.
117 Medium
117 of 384
Understanding +4 | -1
An ideal gas undergoes a reversible isothermal expansion from a state i to f. The change in entropy Δs(i→f)\Delta s\left( {i \to f} \right) is
View answer and explanation
Correct option: B
> 0
Option B (> 0) is correct because in a reversible isothermal expansion, the volume increases, leading to an increase in entropy.
118 Medium
118 of 384
Understanding +4 | -1
If cpc_p and cvc_v denote the specific heats (per unit mass of an ideal gas of molecular weight MM), then
View answer and explanation
Correct option: C
cp−cv=R/Mc_p-c_v=R/M
Option C is correct because the relationship between specific heats and the gas constant is cp - cv = R/M.
119 Medium
119 of 384
Understanding +4 | -1
The specific heat at constant pressure is greater than that of the same gas at constant volume because
View answer and explanation
Correct option: B
At constant pressure, work is done in expanding the gas
Option B (At constant pressure, work is done in expanding the gas) is correct because at constant pressure, some of the heat supplied is used to do work against the external pressure, in addition to increasing the internal energy.
120 Medium
120 of 384
Remembering +4 | -1
The thermodynamic process, in which internal energy of the system remains constant is
View answer and explanation
Correct option: C
Isothermal
Option C (Isothermal) is correct because an isothermal process is defined as a thermodynamic process in which the temperature, and therefore the internal energy, of the system remains constant.
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