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Test: Thermodynamics (October 10) - NEET MCQ


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10 Questions MCQ Test - Test: Thermodynamics (October 10)

Test: Thermodynamics (October 10) for NEET 2024 is part of NEET preparation. The Test: Thermodynamics (October 10) questions and answers have been prepared according to the NEET exam syllabus.The Test: Thermodynamics (October 10) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Thermodynamics (October 10) below.
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Test: Thermodynamics (October 10) - Question 1

First law of thermodynamics is consequence of conservation of [1988]

Detailed Solution for Test: Thermodynamics (October 10) - Question 1

The first law of thermodynamics is just a conservation of energy.

Test: Thermodynamics (October 10) - Question 2

A thermodynamic process is shown in the figure.
The pressures and volumes corresponding to some points in the figure are

PA = 3 × 104 Pa
VA = 2 × 10-3 m3
PB = 8 × 104 Pa
VD = 5 × 10–3 m3.
In process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be [1991]  

Detailed Solution for Test: Thermodynamics (October 10) - Question 2

Since AB is an isochoric process, so, no work is done. BC is isobaric process,
∴ W = PB × (VD – VA) = 240 J
ΔQ = 600 + 200 = 800 J
Using ΔQ = ΔU + ΔW
⇒ ΔU = ΔQ – ΔW = 800 – 240 = 560 J

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Test: Thermodynamics (October 10) - Question 3

If for a gas, , the gas is made up of molecules which are [1992]

Detailed Solution for Test: Thermodynamics (October 10) - Question 3

Since  hence gas is monoatomic.

Test: Thermodynamics (October 10) - Question 4

An ideal gas A and a real gas B have their volumes increased from V to 2V under isothermal conditions. The increase in internal energy [1993]

Detailed Solution for Test: Thermodynamics (October 10) - Question 4

Under isothermal conditions, there is no change in internal energy.

Test: Thermodynamics (October 10) - Question 5

Which of the following is not thermodynamical function ? [1993]

Detailed Solution for Test: Thermodynamics (October 10) - Question 5

Work done is not a thermodynamical function.

Test: Thermodynamics (October 10) - Question 6

An ideal carnot engine, whose efficiency is 40% receives heat at 500 K. If its efficiency is 50%, then the intake temperature for the same exhaust temperature is [1995]

Detailed Solution for Test: Thermodynamics (October 10) - Question 6

Efficiency of carnot engine (η1) = 40% = 0.4; Initial intake temperature (T1) = 500K and new efficiency (η2) = 50% = 0.5.
Efficiency 

Therefore in first case,

⇒  T2 = 0.6×500=300K

And in second case, 

Test: Thermodynamics (October 10) - Question 7

An ideal gas undergoing adiabatic change has the following pressure-temperature relationship [1996]

Detailed Solution for Test: Thermodynamics (October 10) - Question 7

We know that in adiabatic process, PVγ = constant ....(1)
From ideal gas equation, we know that PV = nRT

....(2)
Puttingt the value from equation  (2) in equation (1),

  constant

P(1 – γ) Tγ = constant

Test: Thermodynamics (October 10) - Question 8

A diatomic gas initially at 18ºC is compressed adiabatically to one eighth of its original volume.The temperature after compression will be [1996]

Detailed Solution for Test: Thermodynamics (October 10) - Question 8

Initial temperature (T1) = 18°C = 291 K
Let Initial volume (V1) = V

Final volume 

According to adiabatic process, TVγ – 1 = constant According to question,

Test: Thermodynamics (October 10) - Question 9

The efficiency of a Carn ot engine operating between the temperatures of 100ºC and –23ºC will be
[1997]

Detailed Solution for Test: Thermodynamics (October 10) - Question 9

T1 = –23°C = 250 K,   T2 = 100°C = 373K

Test: Thermodynamics (October 10) - Question 10

We consider a thermodynamic system. If ΔU represents the increase in its internal energy and W the work done by the system, which of the following statements is true? [1998]

Detailed Solution for Test: Thermodynamics (October 10) - Question 10

ΔQ = ΔU + W
For adiabatic process, ΔQ = 0
ΔU = –W

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