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Thermodynamics - 2 - JEE MCQ


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30 Questions MCQ Test Chemistry for JEE Main & Advanced - Thermodynamics - 2

Thermodynamics - 2 for JEE 2024 is part of Chemistry for JEE Main & Advanced preparation. The Thermodynamics - 2 questions and answers have been prepared according to the JEE exam syllabus.The Thermodynamics - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Thermodynamics - 2 below.
Solutions of Thermodynamics - 2 questions in English are available as part of our Chemistry for JEE Main & Advanced for JEE & Thermodynamics - 2 solutions in Hindi for Chemistry for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Thermodynamics - 2 | 30 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Chemistry for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Thermodynamics - 2 - Question 1
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3 moles of a diatomic gas are heated from 127° C to 727° C at a constant pressure of 1 atm. Entropy change is (log 2.5 = 0 .4)

Detailed Solution for Thermodynamics - 2 - Question 1

∆S = nCplnT2/T1 + nRlnP1/P2
Since pressure is constant, so the second term will be zero.
Or ∆S = 3×7/2×8.314×2.303×log(1000/400)
= 80.42 JK-1

Thermodynamics - 2 - Question 2
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Exactly 100 J of heat was transferred reversibly to a block of gold at 25.00° C from a thermal reservoir at 25.01 °C, and then exactly 100 J of heat was absorbed reversibled from the block of gold by a thermal reservoir at 24.99° C. Thus entropy change of the system is

Detailed Solution for Thermodynamics - 2 - Question 2
  • During the first process, 100 J of heat is added at 25.01 °C, resulting in an entropy change of ΔS1 = Q/T = 100 J / 298.16 K = 0.335 JK-1.
  • In the second process, 100 J of heat is removed at 24.99 °C, leading to ΔS2 = -Q/T = -100 J / 298.15 K = -0.335 JK-1.
  • The total entropy change of the system is ΔS = ΔS1 + ΔS2 = 0.335 JK-1 - 0.335 JK-1 = 0.00 JK-1.
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Thermodynamics - 2 - Question 3
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Consider a reversible isentropic expansion of 1.0 mole of an ideal monoatomic gas from 25°C to 75°C. If the initial pressure was 1.0 bar, final pressure is 

Detailed Solution for Thermodynamics - 2 - Question 3

Isentropic process means that entropy is constant. This is true only for reversible adiabatic process.
Applying P11-γ T1γ = P21-γ T2γ (for monatomic species, γ = 5/3)
(1/P)-⅔ = (75+273/25+273)5/3
Or (1/P)-2 = (348/298)5
Or P = 1.474 bar

Thermodynamics - 2 - Question 4
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For the process, and 1 atmosphere pressure, the correct choice is

[JEE Advanced 2014]
Detailed Solution for Thermodynamics - 2 - Question 4

At 100°C and 1 atmosphere pressure H2O (l) ⇋ H2O(g) is at equilibrium. For equilibrium

*Multiple options can be correct
Thermodynamics - 2 - Question 5
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Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is (are)

[JEE Advanced 2013]
Detailed Solution for Thermodynamics - 2 - Question 5

When an ideal solution is formed process is spontaneous

According to Raoult's law, for an ideal solution

ΔH=0

ΔVmix​=0

Since there is no exchange of heat energy between system and surroundings

ΔSsurroundings​=0

ΔSsys.​=+ve

∴ From the relation

ΔG=ΔH–TΔS

ΔG=−ve

Hence, the correct options are B, C and D

Thermodynamics - 2 - Question 6
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Thermodynamics - 2 - Question 23
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Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas ? (Assume non-expansion work is zero)
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