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Thermodynamics - 1 - JEE MCQ


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30 Questions MCQ Test Chemistry for JEE Main & Advanced - Thermodynamics - 1

Thermodynamics - 1 for JEE 2024 is part of Chemistry for JEE Main & Advanced preparation. The Thermodynamics - 1 questions and answers have been prepared according to the JEE exam syllabus.The Thermodynamics - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Thermodynamics - 1 below.
Solutions of Thermodynamics - 1 questions in English are available as part of our Chemistry for JEE Main & Advanced for JEE & Thermodynamics - 1 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 - 1 | 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 - 1 - Question 1
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The pressure-volume work for an ideal gas can be calculated using the expression

This type of work can also be calculated using the area under the curve within the specified limits. When an ideal gas is compressed, (I) reversibly or (II) irreversibly, then

Detailed Solution for Thermodynamics - 1 - Question 1

Work done is the area under the P−V curve. It can be seen in the curve above that the area under the P−V curve for irreversible compression of the gas is more than the area under the curve for reversible compression.
Thus, work done for irreversible compression is more than that for reversible compression.

Thermodynamics - 1 - Question 2
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In the following case

 expanded in reversible adiabatic condition to make volume 8 times ( γ = 1.33). Find the  Work Done and Temperature.

Detailed Solution for Thermodynamics - 1 - Question 2

As q = 0, we have adiabatic process.
V1 = 100L and V2 = 800L
T1 = 300K and T2 = not given
For NH3, γ = 4/3
Applying TVγ = constant
(300)(100)4/3-1 = (T)(800)4/3-1
T = 300/2 = 150K
W= nR(T2-T1)/γ-1
= 1×8.314×150/(4/3-1)
= 3714 J
= 900 cal

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Thermodynamics - 1 - Question 3
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1.0 mole of a monoatomic ideal gas is expanded from state I to state II at 300 K.

Thus, work done is

Detailed Solution for Thermodynamics - 1 - Question 3

Work done in isothermal process
W= -2.303 nRT log P1/P2
here n=1 and R =8.314  and T= 300 P1=4and P2=2.
W= -1724 J 
So option A is correct.

Thermodynamics - 1 - Question 4
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1 mole of a diatomic gas is contained in a piston. It gains 50.0 J of heat and work is done on the surrounding by the system is -100 J. Thus,
*Multiple options can be correct
Thermodynamics - 1 - Question 5
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A sample containing 1.0 mole of an ideal gas is expanded isothermally and reversibly to ten time of its original volume, in two separate experiments. The expansion is carried out 300 K and at 600 K, respectively. Choose the correct option.
Detailed Solution for Thermodynamics - 1 - Question 5

Work done in isothermal process, w = nRTln(V2/V1)
w600/w300 = 1×R×600×ln(10)/ 1×R×300ln(10) = 2
∆U = 0 for isothermal processes.

Thermodynamics - 1 - Question 6
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Thermodynamics - 1 - Question 7
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Thermodynamics - 1 - Question 8
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During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. The change in internal energy, ΔU (in kJ) is :
Detailed Solution for Thermodynamics - 1 - Question 8

Work done on system = +10 kJ
Heat escaped = –2kI
ΔU = q + w
= 10 – 2 = 8 KJ.

Thermodynamics - 1 - Question 9
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Thermodynamics - 1 - Question 11
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Thermodynamics - 1 - Question 23
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For silver, Cp(JK–1 mol–1) = 23 + 0.01 T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of ΔH will be close to -
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