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Thermodynamics

19 JEE Main questions · Physics

Thermodynamic processes

  1. PhysicsJEE Main 21 Jan 2026, Shift 2·Q25·Numerical·Easy–moderate
    Open
    A diatomic gas (γ=1.4\gamma = 1.4) does 100 J of work when it is expanded isobarically. Then the heat given to the gas is J.
    We corrected a printing error in the original paper

    Source printed “the heat given to the gas ___ J” (“is” missing).

  2. PhysicsJEE Main 22 Jan 2026, Shift 1·Q16·MCQ·Easy–moderate
    Open
    The volume of an ideal gas increases 8 times and temperature becomes (14)th\left(\frac14\right)^{\text{th}} of initial temperature during a reversible change. If there is no exchange of heat in this process (ΔQ=0\Delta Q = 0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):
    1. (1)COX2\ce{CO2}
    2. (2)OX2\ce{O2}
    3. (3)NHX3\ce{NH3}
    4. (4)He
  3. PhysicsJEE Main 23 Jan 2026, Shift 2·Q14·MCQ·Moderate
    Open
    One mole of an ideal diatomic gas expands from volume VV to 2V2V isothermally at a temperature 27∘27^\circC and does WW joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27∘27^\circC doing the same amount of work WW, then its final temperature will be (close to) ∘^\circC.
    1. (1)−189-189
    2. (2)−56-56
    3. (3)−30-30
    4. (4)−117-117
  4. PhysicsJEE Main 24 Jan 2026, Shift 2·Q18·MCQ·Easy–moderate
    Open
    10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from P1P_1 to P2P_2 is α\alpha Joule (P1=21.7P_1 = 21.7 Pa and P2=30P_2 = 30 Pa, Cv=21C_v = 21 J/K.mol, R=8.3R = 8.3 J/mol.K). The value of α\alpha is .
    1. (1)24
    2. (2)15
    3. (3)21
    4. (4)28
  5. PhysicsJEE Main 28 Jan 2026, Shift 1·Q12·MCQ·Moderate
    Open
    In the following pp-VV diagram the equation of state along the curved path is given by (V−2)2=4ap(V-2)^2 = 4ap where aa is a constant. The total work done in the closed path is
    1. (1)−1a-\dfrac{1}{a}
    2. (2)+13a+\dfrac{1}{3a}
    3. (3)12a\dfrac{1}{2a}
    4. (4)−13a-\dfrac{1}{3a}
  6. PhysicsJEE Main 28 Jan 2026, Shift 2·Q22·Numerical·Easy–moderate
    Open
    A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle ABC is J.
  7. PhysicsJEE Main 2 Apr 2026, Shift 1·Q22·Numerical·Moderate
    Open
    A vessel contains 0.15 m30.15\ \mathrm{m^3} of a gas at pressure 8 bar and temperature 140∘140^\circC with cp=3Rc_p = 3R and cv=2Rc_v = 2R. It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is kJ. (RR is gas constant)
  8. PhysicsJEE Main 6 Apr 2026, Shift 1·Q25·Numerical·Moderate
    Open
    A certain gas is isothermally compressed to (13)rd\left(\dfrac13\right)^{\text{rd}} of its initial volume (V0=3V_0 = 3 litre) by applying required pressure. If the bulk modulus of the gas is 3×105 N/m23\times10^{5}\ \mathrm{N/m^2}, the magnitude of work done on the gas is J.
  9. PhysicsJEE Main 6 Apr 2026, Shift 2·Q5·MCQ·Easy–moderate
    Open
    A cylinder with adiabatic walls is closed at both ends and is divided into two compartments by a frictionless adiabatic piston. Ideal gas is filled in both (left and right) the compartments at same PP, VV, TT. Heating is started from left side until pressure changes to 27P8\dfrac{27P}{8}. If initial volume of each compartment was 9 litres then the final volume in right-hand side compartment is litres. (for this ideal gas CP/CV=1.5C_P/C_V = 1.5)
    1. (1)3
    2. (2)4
    3. (3)14
    4. (4)9
  10. PhysicsJEE Main 8 Apr 2026, Shift 2·Q11·MCQ·Moderate
    Open
    Initial pressure and volume of a monoatomic ideal gas are PP and VV. The change in internal energy of this gas in adiabatic expansion to volume Vfinal=27 VV_{\text{final}} = 27\,V is J.
    1. (1)−2PV(33−1)-2PV\left(3\sqrt3 - 1\right)
    2. (2)43PV\dfrac43 PV
    3. (3)−43PV-\dfrac43 PV
    4. (4)34PV\dfrac34 PV

Laws of thermodynamics

  1. PhysicsJEE Main 21 Jan 2026, Shift 1·Q24·Numerical·Easy–moderate
    Open
    10 mole of oxygen is heated at constant volume from 30∘30^\circC to 40∘40^\circC. The change in the internal energy of the gas is cal. (The molecular specific heat of oxygen at constant pressure, Cp=7C_p = 7 cal./mol ∘^\circC and R=2R = 2 cal./mol ∘^\circC.)
  2. PhysicsJEE Main 23 Jan 2026, Shift 2·Q1·MCQ·Moderate
    Open
    The internal energy of a gas is 3nRT. One mole of this gas is kept in a cylinder having internal cross section area of 17 cm217\ \mathrm{cm^2} and fitted with a light movable frictionless piston. The gas is heated slowly by supplying 126 J heat. If the temperature rises by 4 ∘4\ ^\circC, then the piston will move cm.
    (atmospheric pressure =105= 10^{5} Pa)
    1. (1)14.5
    2. (2)1.55
    3. (3)15.5
    4. (4)1.45
    We corrected a printing error in the original paper

    Source printed “The internal energy of a monoatomic gas is 3nRT. One mole of helium is kept”; corrected to the intended “The internal energy of a gas is 3nRT. One mole of this gas is kept” (a monoatomic gas has U=32nRTU = \tfrac32 nRT, which gives 44.8 cm, no option; the key 15.5 uses U=3nRTU = 3nRT); “suppling”.

  3. PhysicsJEE Main 24 Jan 2026, Shift 1·Q4·MCQ·Moderate
    Open
    Density of water at 4∘4^\circC and 20∘20^\circC are 1000 kg/m31000\ \mathrm{kg/m^3} and 998 kg/m3998\ \mathrm{kg/m^3} respectively. The increase in internal energy of 4 kg water when it is heated from 4∘4^\circC to 20∘20^\circC is J.
    (Specific heat capacity of water =4.2 kJ kg−1 ∘C−1= 4.2\ \mathrm{kJ\,kg^{-1}\,{}^\circ C^{-1}} and 1 atmospheric pressure =105= 10^{5} Pa)
    1. (1)315826.2
    2. (2)234699.2
    3. (3)258700.8
    4. (4)268799.2
    We corrected a printing error in the original paper

    Source printed “1000 kg/m3^3 and respectively” and “4.2 J/kg.”; corrected to the intended “1000 kg/m3^3 and 998 kg/m3^3 respectively” and “4.2 kJ kg−1^{-1} ∘^\circC−1^{-1}” (the values the key uses).

  4. PhysicsJEE Main 24 Jan 2026, Shift 2·Q21·Numerical·Easy–moderate
    Open
    When 300 J of heat given to an ideal gas with Cp=72RC_p = \dfrac72 R its temperature raises from 20 ∘20\ ^\circC to 50 ∘50\ ^\circC keeping its volume constant. The mass of the gas is (approximately) g. (R=8.314R = 8.314 J/mol.K).

    Flawed question. Asks for the mass of the gas but no molar mass is given (only n≈0.481n \approx 0.481 mol follows); the printed key 481 matches no stated quantity, so there is no evident intended version. It is shown for completeness and is left out of practice tests.

  5. PhysicsJEE Main 2 Apr 2026, Shift 1·Q8·MCQ·Easy–moderate
    Open
    Heat is supplied to a diatomic gas at constant pressure. Then the ratio of ΔQ:ΔU:ΔW\Delta Q : \Delta U : \Delta W is .
    1. (1)2 : 3 : 5
    2. (2)5 : 3 : 2
    3. (3)2 : 5 : 7
    4. (4)7 : 5 : 2
  6. PhysicsJEE Main 2 Apr 2026, Shift 2·Q23·Numerical·Easy–moderate
    Open
    5 moles of unknown gas is heated at constant volume from 10∘10^\circC to 20∘20^\circC. The molar specific heat of this gas at constant pressure cp=8c_p = 8 cal/mol. ∘^\circC and R=8.36R = 8.36 J/mol. ∘^\circC. The change in the internal energy of the gas is calorie.
  7. PhysicsJEE Main 5 Apr 2026, Shift 1·Q7·MCQ·Easy–moderate
    Open

    Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

    Assertion A : Change in internal energy of a system containing nn mole of ideal gas can be written as ΔU=nCv(Tf−Ti)=nRγ−1(Tf−Ti)\Delta U = nC_v(T_f - T_i) = \dfrac{nR}{\gamma - 1}(T_f - T_i), where γ=CPCv\gamma = \dfrac{C_P}{C_v}, TiT_i = initial temperature, TfT_f = final temperature.

    Reason R : Relation between degree of freedom ff and γ (=Cp/Cv)\gamma\ (= C_p/C_v) is (γ=1+2f)\left(\gamma = 1 + \dfrac{2}{f}\right)

    Choose the correct answer from the options given below

    1. (1)Both A and R are true and R is the correct explanation of A
    2. (2)Both A and R are true but R is NOT the correct explanation of A
    3. (3)A is true but R is false
    4. (4)A is false but R is true
    We corrected a printing error in the original paper

    Source labelled the two statements “Statement I” and “Statement II” (the lead-in and options use Assertion A / Reason R); source printed “ttue” in option (4).

  8. PhysicsJEE Main 5 Apr 2026, Shift 1·Q8·MCQ·Easy–moderate
    Open
    Consider the following statements:
    • (A)Zeroth law of thermodynamics gives concept of temperature
    • (B)First law of thermodynamics gives concept of internal energy
    • (C)In isothermal expansion of ideal gas, ΔQ≠ΔW\Delta Q \neq \Delta W
    • (D)Product of intensive and extensive variables is extensive
    • (E)The ratio of any extensive variable to mass will be an extensive variable
    Choose the correct combination of statements from the options given below:
    1. (1)C, D and E Only
    2. (2)A, B and C Only
    3. (3)A, B and D Only
    4. (4)B, C and D Only
  9. PhysicsJEE Main 8 Apr 2026, Shift 2·Q10·MCQ·Moderate
    Open
    One mole of diatomic gas having rotational modes only is kept is a cylinder with a piston system. The cross-section area of the cylinder is 4 cm24\ \mathrm{cm^2}. The gas is heated slowly to raise the temperature by 1.2 ∘1.2\ ^\circC during which the piston moves by 25 mm. The amount of heat supplied to the gas is J. (Atmospheric pressure =100= 100 kPa, R=8.3R = 8.3 J/mol.K) (Neglect mass of the piston)
    1. (1)24.8
    2. (2)25
    3. (3)15.04
    4. (4)29.98

    Flawed question. The data are inconsistent: with a free piston at 100 kPa, PΔVP\Delta V must equal nRΔT=9.96nR\Delta T = 9.96 J, but a 25 mm travel gives 1.0 J; no option equals the consistent Q=34.9Q = 34.9 J, and the key 25 is only the option nearest to 24.9+1.0=25.924.9 + 1.0 = 25.9 J. It is shown for completeness and is left out of practice tests.