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First Law Of Thermodynamic NAT Level - 1 - Physics MCQ


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10 Questions MCQ Test Topic wise Tests for IIT JAM Physics - First Law Of Thermodynamic NAT Level - 1

First Law Of Thermodynamic NAT Level - 1 for Physics 2024 is part of Topic wise Tests for IIT JAM Physics preparation. The First Law Of Thermodynamic NAT Level - 1 questions and answers have been prepared according to the Physics exam syllabus.The First Law Of Thermodynamic NAT Level - 1 MCQs are made for Physics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for First Law Of Thermodynamic NAT Level - 1 below.
Solutions of First Law Of Thermodynamic NAT Level - 1 questions in English are available as part of our Topic wise Tests for IIT JAM Physics for Physics & First Law Of Thermodynamic NAT Level - 1 solutions in Hindi for Topic wise Tests for IIT JAM Physics course. Download more important topics, notes, lectures and mock test series for Physics Exam by signing up for free. Attempt First Law Of Thermodynamic NAT Level - 1 | 10 questions in 30 minutes | Mock test for Physics preparation | Free important questions MCQ to study Topic wise Tests for IIT JAM Physics for Physics Exam | Download free PDF with solutions
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First Law Of Thermodynamic NAT Level - 1 - Question 1
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If R = universal gas constant, the amount of heat needed to raise the temperature of 2 mole of an ideal monoatomic gas from 273K to 373K when no work is done (in units of R)


Detailed Solution for First Law Of Thermodynamic NAT Level - 1 - Question 1

ΔQ = ΔQ = μCVΔT
⇒ 
The correct answer is: 300

*Answer can only contain numeric values
First Law Of Thermodynamic NAT Level - 1 - Question 2
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The weight of a person is 60kg. If he get 105 calories heat through food and the efficiency of his body is 28%, then up to how much height, he can climb (in meters)?


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

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First Law Of Thermodynamic NAT Level - 1 - Question 3
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A thermodynamic process is shown in the figure. The pressure and volumes corresponding to some points in the figure are : 

 

In process AB, 600 Joules of heat is added to the system and in process BC, 200J of heat is added to the system. The change in the internal energy of the system in process AC would be? (in Joules)


Detailed Solution for First Law Of Thermodynamic NAT Level - 1 - Question 3

From the graph, WAB = 0
and 

Now 
From the first law of thermodynamics,

The correct answer is: 560

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First Law Of Thermodynamic NAT Level - 1 - Question 4
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How much energy (in 107 Joules) is absorbed by 10kg molecule of an ideal gas if it expands from its initial pressure of 8 atm to 4 atm at a constant temperature of 27ºC.

Detailed Solution for First Law Of Thermodynamic NAT Level - 1 - Question 4



W = 1.728×107 J
Amount of energy absorbed = Work done
= 1.728×107J
The correct answer is: 1.728

*Answer can only contain numeric values
First Law Of Thermodynamic NAT Level - 1 - Question 5
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A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is (in Joule)

Detailed Solution for First Law Of Thermodynamic NAT Level - 1 - Question 5

Work done by the system
 = Area of the shaded portion on p-V diagram.
= (300–100)×10-6 × (200–100)×103
= 20J
Since, process is anticlockwise, work done = –20J.
The correct answer is: -20

*Answer can only contain numeric values
First Law Of Thermodynamic NAT Level - 1 - Question 6
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One mole of O2 gas having a volume equal to 22.4 liters at 0ºC and 1atm pressure is compressed isothermally so that its volume reduces to 11.2 liters. The work done in this process is (in Joules)?

Detailed Solution for First Law Of Thermodynamic NAT Level - 1 - Question 6

Work done  

= –1572.5 J
The correct answer is: -1572.5

*Answer can only contain numeric values
First Law Of Thermodynamic NAT Level - 1 - Question 7
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A gas is compressed at a constant pressure of 50N/m2 from a volume of 10m3 to a volume of 4m3. Energy of 100J is then added to the gas by heating. Then increase in its internal energy is?

Detailed Solution for First Law Of Thermodynamic NAT Level - 1 - Question 7

ΔQ = ΔU + ΔW
⇒ 100 = ΔU + 50(4 - 10)
⇒ ΔU = 100 + 300 = 400J
The correct answer is: 400

*Answer can only contain numeric values
First Law Of Thermodynamic NAT Level - 1 - Question 8
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A block of mass 100gm slides on a rough horizontal surface. If the speed of the block decreases form 10m/s to 5m/s, then thermal energy developed in the process in Joules is?

Detailed Solution for First Law Of Thermodynamic NAT Level - 1 - Question 8

Thermal energy developed = loss in Kinetic energy

= 3.75J
The correct answer is: 3.75

*Answer can only contain numeric values
First Law Of Thermodynamic NAT Level - 1 - Question 9
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When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases. the internal energy of the gas is?

Detailed Solution for First Law Of Thermodynamic NAT Level - 1 - Question 9


The correct answer is: 0.7143

*Answer can only contain numeric values
First Law Of Thermodynamic NAT Level - 1 - Question 10
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If CV = 4.96 cal/mole-K, then increase in internal energy when temperature of 2moles of this gas is increased from 340K to 342K (in cal)?

Detailed Solution for First Law Of Thermodynamic NAT Level - 1 - Question 10

Increase in internal energy
ΔU = μCVΔT
= 2 × 4.96 × (342 – 340)
= 19.84 cal
The correct answer is: 19.84

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