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IIT JAM Exam  >  IIT JAM Tests  >  First Law Of Thermodynamics NAT Level – 2 - IIT JAM MCQ

First Law Of Thermodynamics NAT Level – 2 - IIT JAM MCQ


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10 Questions MCQ Test - First Law Of Thermodynamics NAT Level – 2

First Law Of Thermodynamics NAT Level – 2 for IIT JAM 2024 is part of IIT JAM preparation. The First Law Of Thermodynamics NAT Level – 2 questions and answers have been prepared according to the IIT JAM exam syllabus.The First Law Of Thermodynamics NAT Level – 2 MCQs are made for IIT JAM 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for First Law Of Thermodynamics NAT Level – 2 below.
Solutions of First Law Of Thermodynamics NAT Level – 2 questions in English are available as part of our course for IIT JAM & First Law Of Thermodynamics NAT Level – 2 solutions in Hindi for IIT JAM course. Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free. Attempt First Law Of Thermodynamics NAT Level – 2 | 10 questions in 45 minutes | Mock test for IIT JAM preparation | Free important questions MCQ to study for IIT JAM Exam | Download free PDF with solutions
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First Law Of Thermodynamics NAT Level – 2 - Question 1
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5000J of heat are added to two moles an ideal monoatomic gas, initially at a temperature of 500K, while the gas performs 7500J of work. What is the final temperature of the gas ? (in Kelvin)


Detailed Solution for First Law Of Thermodynamics NAT Level – 2 - Question 1

ΔU = ΔQ – ΔW = 5000 - 7500
ΔU = –2500 J

⇒ ΔT = –100 K
⇒ T = 500 - 100 = 400 K
The correct answer is: 400

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First Law Of Thermodynamics NAT Level – 2 - Question 2
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For a diatomic ideal gas near room temperature, what fraction of heat supplied is available for external work if gas is expanded at constant pressure


Detailed Solution for First Law Of Thermodynamics NAT Level – 2 - Question 2

In the process of expansion at constant pressure p, assuming that the volume increases from V1 to V2 and temperature change from T1 to T2 , we have
pV1 = nRT1
pV2 = nRT2
In this process, the work done by the system on the outside world is W = p(V2 -V1 )= nRΔT and increase in internal energy of the system is
ΔU = CvΔT
∴ 
The correct answer is: 0.286

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First Law Of Thermodynamics NAT Level – 2 - Question 3
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A gas has pressure p and volume V. It is now compressed adiabatically to 1/32 times the original volume. If (32)1.4 = 128, the final pressure in terms of p is 


Detailed Solution for First Law Of Thermodynamics NAT Level – 2 - Question 3

For an adiabatic process,

⇒ 
⇒ p = (32)1.4 p
⇒ p1 = 128p.
The correct answer is: 128

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First Law Of Thermodynamics NAT Level – 2 - Question 4
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1cm3 of water at its boiling point absorbs 540 calories of heat to become steam with a volume of 1671 cm3. If the atmospheric pressure is 1.013 × 105 N/m2 and mechanical equivalent of heat = 4.19 J/calorie, the energy spent in this process in overcoming intermolecular forces is (in calories)

Detailed Solution for First Law Of Thermodynamics NAT Level – 2 - Question 4

Energy spent in overcoming intermolecular forces
ΔU = ΔQ – ΔW
= ΔQ – p(V2 – V1)

= 500cal.
The correct answer is: 500

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First Law Of Thermodynamics NAT Level – 2 - Question 5
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An ideal gas at 27ºC is compressed adiabatically to 8/27 of its original volume. If  γ = 5/3 then the rise in temperature (in Kelvin) of the gas __.

Detailed Solution for First Law Of Thermodynamics NAT Level – 2 - Question 5

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First Law Of Thermodynamics NAT Level – 2 - Question 6
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Two systems with heat capacities 100cal/gm-K and 200cal/gm-K interact thermally and come to a common temperature 400K. If the initial temperature of system 1 is 500K, what was the initial temperature of systems (in Kelvin) scale 

Detailed Solution for First Law Of Thermodynamics NAT Level – 2 - Question 6

Let initial temperature of system 2 is T2. According to conservation of energy, we know that heat released from one system is equal to the heat absorbed by the other system, i.e.
C1 (Tf – T1 ) = C2 (T2 –Tf )

= –50+400
T2 = 350K
The correct answer is: 350

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First Law Of Thermodynamics NAT Level – 2 - Question 7
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During an adiabatic processes, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio Cp /CV for the gas is

Detailed Solution for First Law Of Thermodynamics NAT Level – 2 - Question 7


But for an adiabatic processes,


⇒
⇒ 
The correct answer is: 1.5

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First Law Of Thermodynamics NAT Level – 2 - Question 8
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When an ideal gas  is heated under constant pressure, then what percentage of given heat energy will be utilized in doing external work?

Detailed Solution for First Law Of Thermodynamics NAT Level – 2 - Question 8


∴ Percentage energy utilized in doing external work =  
The correct answer is: 40

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First Law Of Thermodynamics NAT Level – 2 - Question 9
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5 mole of hydrogen gas is heated from 30ºC to 60ºC at constant pressure. Heat given to the gas is in calories (given R = 2cal/mole-ºC)

Detailed Solution for First Law Of Thermodynamics NAT Level – 2 - Question 9





= 1050 calories
The correct answer is: 1050

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First Law Of Thermodynamics NAT Level – 2 - Question 10
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Three liquids with masses m1, m2, m3 are thoroughly mixed. If their specific heats are C1, C2 and C3 and their temperature T1, T2, T3, then the temperature of the mixture is given that

Detailed Solution for First Law Of Thermodynamics NAT Level – 2 - Question 10

Let the final temperature be T K. Total heat supplied by 3 liquids in coming down to

Total heat used by three liquids in raising temperature from 0º to TºC

from (1) & (2),

⇒ 


T = 558.8K
The correct answer is: 558.8

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