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Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - JEE MCQ


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7 Questions MCQ Test 35 Years Chapter wise Previous Year Solved Papers for JEE - Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced

Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced for JEE 2024 is part of 35 Years Chapter wise Previous Year Solved Papers for JEE preparation. The Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced questions and answers have been prepared according to the JEE exam syllabus.The Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced below.
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*Multiple options can be correct
Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 1

Identify the intensive quantities from the following: (1993 - 1 Mark)

Detailed Solution for Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 1

Properties independent of mass are intensive properties. Hence (b) and (d) which are independent of mass are the obvious choices.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 2

The following is (are) endothermic reaction(s):                                          (1999 - 3 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 2

Consider the options:-
A) Combustion is always an exothermic process.
B) Decomposition of water requires energy. Hence it is endothermic.
C) Dehydrogenation is an endothermic reaction since energy is required to break the C−H bonds to convert ethane to ethylene.
D) Graphite is more stable than diamond and energy is needed to convert graphite into diamond. The extra energy that has to be provided ends up in the tetrahedral structure of diamond.


Hence options B,C and D are the right answer.

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*Multiple options can be correct
Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 3

Among the following the state function(s) is (are) (2009)

Detailed Solution for Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 3

Internal energy and molar enthalpy are state functions. Work (reversible or irreversible) is a path function.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 4

Among the following, the intensive property is (properties are) (2010)

Detailed Solution for Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 4

Mass independent properties (molar conductivity and electromotive force) are intensive properties.
Resistance and heat capacity are mass dependent, hence extensive properties.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 5

For an ideal gas, consider only P–V work in going from  an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure.
Which of the following choice(s) is (are) correct? [Take ΔS as change in entropy and w as work done]. (2012)

Detailed Solution for Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 5

ΔS x→z = ΔS x→y +ΔS y→ z [Entropy is a state function, hence additive]

Wx→y→z =Wx→y [Work done in y → z,  is zero because it is an isochoric process].

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 6

The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct ?

 

Detailed Solution for Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 6

T1 = T2 because process is isothermal.
Work done in adiabatic process is less than in isothermal process because area covered by isothermal curve is more than the area covered by the adiabatic curve.
In adiabatic process expansion occurs by using internal energy, hence, it decreases while in isothermal process temperature remains constant that's why no change in internal energy.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 7

An ideal gas in a thermally insulated vessel at internal pressure = P1,  volume = V1 and absolute temperature = T1 expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are P2, V2 and T2, respectively. For this expansion,    (JEE Adv. 2014)

Detailed Solution for Test: MCQs (One or More Correct Option): Thermodynamics | JEE Advanced - Question 7

Since the vessel is thermally insulated, q = 0 Further since, Pext = 0, so w = 0, hence ΔU = 0
Since ΔT = 0, T2 = T1, and P2V2 = P1V1
However, the process is adiabatic irreversible, so we can’t apply P2V2γ = P1V1γ.

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