JEE Exam  >  JEE Tests  >  Thermodynamics - 2 - JEE MCQ

Thermodynamics - 2 - JEE MCQ


Test Description

30 Questions MCQ Test - Thermodynamics - 2

Thermodynamics - 2 for JEE 2024 is part of JEE 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 course for JEE & Thermodynamics - 2 solutions in Hindi for JEE 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 for JEE Exam | Download free PDF with solutions
Thermodynamics - 2 - Question 1

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

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

1 Crore+ students have signed up on EduRev. Have you? Download the App
Thermodynamics - 2 - Question 3

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

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

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


Detailed Solution for Thermodynamics - 2 - Question 6


Thermodynamics - 2 - Question 7


Detailed Solution for Thermodynamics - 2 - Question 7


Thermodynamics - 2 - Question 8


Detailed Solution for Thermodynamics - 2 - Question 8

Thermodynamics - 2 - Question 9


Detailed Solution for Thermodynamics - 2 - Question 9


Thermodynamics - 2 - Question 10


Detailed Solution for Thermodynamics - 2 - Question 10

Thermodynamics - 2 - Question 11


Detailed Solution for Thermodynamics - 2 - Question 11


Thermodynamics - 2 - Question 12


Detailed Solution for Thermodynamics - 2 - Question 12


Thermodynamics - 2 - Question 13


Detailed Solution for Thermodynamics - 2 - Question 13


Thermodynamics - 2 - Question 14


Detailed Solution for Thermodynamics - 2 - Question 14


Thermodynamics - 2 - Question 15


Detailed Solution for Thermodynamics - 2 - Question 15


Thermodynamics - 2 - Question 16


Detailed Solution for Thermodynamics - 2 - Question 16

Thermodynamics - 2 - Question 17


Detailed Solution for Thermodynamics - 2 - Question 17


Thermodynamics - 2 - Question 18


Detailed Solution for Thermodynamics - 2 - Question 18


Thermodynamics - 2 - Question 19


Detailed Solution for Thermodynamics - 2 - Question 19


Thermodynamics - 2 - Question 20


Detailed Solution for Thermodynamics - 2 - Question 20


Thermodynamics - 2 - Question 21


Detailed Solution for Thermodynamics - 2 - Question 21


Thermodynamics - 2 - Question 22


Detailed Solution for Thermodynamics - 2 - Question 22


Thermodynamics - 2 - Question 23

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)

Detailed Solution for Thermodynamics - 2 - Question 23


Thermodynamics - 2 - Question 24


Detailed Solution for Thermodynamics - 2 - Question 24


Thermodynamics - 2 - Question 25


Detailed Solution for Thermodynamics - 2 - Question 25


Thermodynamics - 2 - Question 26

Detailed Solution for Thermodynamics - 2 - Question 26


Thermodynamics - 2 - Question 27


Detailed Solution for Thermodynamics - 2 - Question 27


Thermodynamics - 2 - Question 28


Detailed Solution for Thermodynamics - 2 - Question 28


Thermodynamics - 2 - Question 29


Detailed Solution for Thermodynamics - 2 - Question 29

Thermodynamics - 2 - Question 30


Detailed Solution for Thermodynamics - 2 - Question 30


Information about Thermodynamics - 2 Page
In this test you can find the Exam questions for Thermodynamics - 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Thermodynamics - 2, EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE