Daily Olympiad: Physics - Thermodynamics [20260512]

Challenge yourself with today's JEE Main practice! This test covers 'Thermodynamics' for Physics (JEE Main - 12). Level: Hard | Duration: 45 mins.

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1. Two moles of an ideal gas at 300 K are allowed to expand isothermally from 1 L to 5 L. Simultaneously, two moles of another ideal gas at 300 K expand isothermally from 5 L to 10 L. What is the total entropy change of the universe for this process?

Solution
Correct: C
The entropy change for isothermal expansion of an ideal gas is ΔS = nR ln(V2/V1). For the first gas: ΔS1 = 2R ln(5). For the second gas: ΔS2 = 2R ln(2). Total ΔS = 2R ln(5) + 2R ln(2) = 2R ln(10) ≈ 2R * 2.3 ≈ 4.6R. Since the processes are reversible, the entropy of the universe increases by the same amount as calculated. Correct answer: 4.6R ≈ 18.3R ln(5)/2.3 ≈ option C.

2. A Carnot engine operates between temperatures T1 and T2 (T1 > T2). If the high-temperature reservoir is cooled from T1 to T1/2 while keeping the low-temperature reservoir at T2, what is the new efficiency?

Solution
Correct: D
Efficiency of a Carnot engine is η = 1 - T2/T1. After cooling the high-temperature reservoir to T1/2, the new efficiency becomes η' = 1 - T2/(T1/2) = 1 - 2T2/T1. Thus, the correct answer is D.

3. A gas undergoes three processes: (1) Adiabatic expansion, (2) Isothermal compression, (3) Isobaric compression, returning to initial state. Which statement about the net work done by the gas is true?

Solution
Correct: B
In the adiabatic expansion (1), the gas does positive work. During isothermal compression (2), work on the gas is greater in magnitude than expansion, making net work negative. Isobaric compression (3) adds to negative work. Hence, the net work done by the gas is always negative (B).