Thermodynamics

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Thermodynamics — solved practice questions

8 JEE Physics questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1easy

   A Carnot engine operates between a source at 600 K and a sink at 300 K. Its efficiency is:

   • A25%
   • B33%
   • C50%
   • D60%
   Show answer & solution

   Correct answer: (C) 50%

   Carnot efficiency = 1 - T_sink/T_source = 1 - 300/600 = 0.5, i.e. 50%.

2. Q2easy

   In a process a gas absorbs 100 J of heat and does 40 J of work on its surroundings. The change in its internal energy is:

   • A140 J
   • B60 J
   • C40 J
   • D100 J
   Show answer & solution

   Correct answer: (B) 60 J

   First law: dU = Q - W = 100 - 40 = 60 J.

3. Q3easy

   For an ideal monatomic gas, the ratio of molar specific heats gamma = Cp/Cv is:

   • A1.4
   • B1.33
   • C1.5
   • D5/3
   Show answer & solution

   Correct answer: (D) 5/3

   For a monatomic gas Cv = (3/2)R and Cp = (5/2)R, so gamma = 5/3 = 1.67.

4. Q4medium

   One mole of an ideal gas expands isothermally and reversibly at 300 K from volume V to 2V (R = 8.314 J/mol/K). The work done by the gas is closest to:

   • A1730 J
   • B2500 J
   • C830 J
   • D3600 J
   Show answer & solution

   Correct answer: (A) 1730 J

   W = nRT ln(V2/V1) = 1 * 8.314 * 300 * ln(2) = 1729 J, about 1730 J.

5. Q5medium

   During an isothermal process on an ideal gas, which statement is correct?

   • AQ = 0
   • BQ = W
   • CW = 0
   • DdU is nonzero
   Show answer & solution

   Correct answer: (B) Q = W

   For an ideal gas internal energy depends only on temperature, so at constant T, dU = 0 and by the first law Q = W.

6. Q6easy

   A heat engine takes in 800 J from a hot reservoir and rejects 600 J to a cold reservoir per cycle. Its efficiency is:

   • A75%
   • B60%
   • C25%
   • D50%
   Show answer & solution

   Correct answer: (C) 25%

   Work done = 800 - 600 = 200 J; efficiency = W/Q_in = 200/800 = 0.25, i.e. 25%.

7. Q7hard

   A monatomic ideal gas at 300 K expands adiabatically to twice its original volume. Its final temperature is closest to:

   • A150 K
   • B300 K
   • C260 K
   • D189 K
   Show answer & solution

   Correct answer: (D) 189 K

   For an adiabatic process T V^(gamma-1) = constant with gamma = 5/3, so T2 = 300*(1/2)^(2/3) = about 189 K.

8. Q8medium

   For a refrigerator working as a reverse Carnot engine between 250 K (cold) and 300 K (hot), the coefficient of performance is:

   • A5
   • B6
   • C1.2
   • D0.2
   Show answer & solution

   Correct answer: (A) 5

   For a Carnot refrigerator, COP = T_cold/(T_hot - T_cold) = 250/(300 - 250) = 5.

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