Test: Thermodynamics (October 10) - NEET MCQ

Test: Thermodynamics (October 10) - NEET MCQ

Test Description

10 Questions MCQ Test Daily Test for NEET Preparation - Test: Thermodynamics (October 10)

Test: Thermodynamics (October 10) for NEET 2024 is part of Daily Test for NEET Preparation 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.
Solutions of Test: Thermodynamics (October 10) questions in English are available as part of our Daily Test for NEET Preparation for NEET & Test: Thermodynamics (October 10) solutions in Hindi for Daily Test for NEET Preparation course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Thermodynamics (October 10) | 10 questions in 20 minutes | Mock test for NEET preparation | Free important questions MCQ to study Daily Test for NEET Preparation for NEET Exam | Download free PDF with solutions
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

 1 Crore+ students have signed up on EduRev. Have you?
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

Daily Test for NEET Preparation

12 docs|366 tests
Information about Test: Thermodynamics (October 10) Page
In this test you can find the Exam questions for Test: Thermodynamics (October 10) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Thermodynamics (October 10), EduRev gives you an ample number of Online tests for practice

Daily Test for NEET Preparation

12 docs|366 tests