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First Law Of Thermodynamics MSQ - Physics MCQ


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First Law Of Thermodynamics MSQ for Physics 2024 is part of Topic wise Tests for IIT JAM Physics preparation. The First Law Of Thermodynamics MSQ questions and answers have been prepared according to the Physics exam syllabus.The First Law Of Thermodynamics MSQ MCQs are made for Physics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for First Law Of Thermodynamics MSQ below.
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*Multiple options can be correct
First Law Of Thermodynamics MSQ - Question 1

In case of isothermal expansion, which of the following are not forbidden?
Select one or more:

Detailed Solution for First Law Of Thermodynamics MSQ - Question 1

Let psurrdV  work done by an ideal gas in the three types of expansion.
psurr is a pressure surrounding
Since 
The maximum work is done when the gas pressure

For reversible expansion 

Path 2 corresponds to a permitted or irreversible expansion, in this case, work done

In a forbidden expansion i.e. one against an external pressure greater than the gas pressure, the work done by the gas would have to be grater than the maximum work, such an expansion is never observed.
∴ In case of isothermal expansion
Permitted process : 
Reversible process : 
Forbidden process : 
The correct answers are:

*Multiple options can be correct
First Law Of Thermodynamics MSQ - Question 2

A thermos bottle containing coffee is vigorously shaken and thereby the temperature of coffee rises. Regard the coffee as the system.
Select one or more:

Detailed Solution for First Law Of Thermodynamics MSQ - Question 2

The heat has not been transferred to coffee which is thermally insulated by shaking, work has been done on coffee (system) against the viscous forces in it.
According to first law,
ΔU = Q – W
Here Q = 0
And W is negative as work is done on the system so that
ΔU = 0 – (–W)
i.e. ΔU is positive. The internal energy of system (coffee) increases.
Q = 0, W = -ve, ΔU = +ve
The correct answers are: Q = 0, W = –ve, ΔU = +ve

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*Multiple options can be correct
First Law Of Thermodynamics MSQ - Question 3

Which type of ideal gas will have largest value of CP – CV?
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Detailed Solution for First Law Of Thermodynamics MSQ - Question 3

For an ideal gas,

Whatever be the number of atoms each gas molecule has
The correct answers are: Monatomic, Diatomic, Polyatomic

*Multiple options can be correct
First Law Of Thermodynamics MSQ - Question 4

Which among the following statements are correct?
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Detailed Solution for First Law Of Thermodynamics MSQ - Question 4

Energy changes with mass of the body, therefore, it is an extensive property.
Specific energy is the energy of the system per unit mass of the system, therefore, it will become intensive property. Heat capacity is product of specific heat and mass of the body. It depends on mass of the system,
∴ heat capacity is an extensive property. Internal energy of the system is independent of the path followed by system. It has fixed value along the path
∴ energy is the point function
The correct answers are: Energy is an extensive property, Energy is a point function, Heat capacity is an extensive property

*Multiple options can be correct
First Law Of Thermodynamics MSQ - Question 5

Which of the following processes must violate the first law of thermodynamics?
Select one or more:

Detailed Solution for First Law Of Thermodynamics MSQ - Question 5


ΔU = ΔE

*Multiple options can be correct
First Law Of Thermodynamics MSQ - Question 6

1 mole of ideal monoatomic gas at 27ºC expands under adiabatic conditions at a pressure of 1.5 atm from a volume of 4dm3 to 16dm3.
Select one or more:

Detailed Solution for First Law Of Thermodynamics MSQ - Question 6

Since processes is adiabatic q = 0.
As the gas expands against constant external processes

= –1.5(16–4) = –18atm-dm3
ΔU = Q + W = 0 + (–18) = –18atm-dm3
The correct answers are:  W = –18atm – dm3 , ΔU = –18atm – dm3

*Multiple options can be correct
First Law Of Thermodynamics MSQ - Question 7

In an isothermal expansion of an ideal gas,
Select one or more:

Detailed Solution for First Law Of Thermodynamics MSQ - Question 7

For one mole of an ideal gas

⇒ dU = CVdT
For isothermal process, T is constant

Also ΔH = ΔU + Δ(pV)
pV = RT

= ΔU + RΔT
ΔH = 0.
The correct answers are: ΔU = 0, ΔH = 0

*Multiple options can be correct
First Law Of Thermodynamics MSQ - Question 8

Which of the following are path function?
Select one or more:

Detailed Solution for First Law Of Thermodynamics MSQ - Question 8

Heat transfer as well as work transfer between the system and surrounding depends upon the path by which the process is occurred. Therefore heat energy and work energy are path function. The change in internal energy ΔE remains constant, no matter which path is followed by a system to undergo a change of a certain state. Thus, internal energy is a point function or state function.
The correct answers are: Heat Energy, Work

*Multiple options can be correct
First Law Of Thermodynamics MSQ - Question 9

A resistor immersed in running water carries an electric current. Consider the resistor as the system.
Select one or more:

Detailed Solution for First Law Of Thermodynamics MSQ - Question 9

There is no flow of heat into the resistor and work is done on the resistor (system).
According to first law as applied to the system, we write,
ΔU = Q – W
For resistor Q = 0
W = –ve. Thus
ΔU = 0 – (–W) = W
∴ Q = 0; W = –ve, ΔU = +ve
The correct answers are: Q = 0, W = –ve, ΔU = +ve

*Multiple options can be correct
First Law Of Thermodynamics MSQ - Question 10

Consider that 214 J of work done on a system, and 293 J of heat are extracted from the system. Then
Select one or more:

Detailed Solution for First Law Of Thermodynamics MSQ - Question 10

Since work is done on the system, therefore, algebraic sign of work done will be positive and magnitude of work done is 214J.
Since heat is extracted from the system, 
∴ Q = –293 J
Now ΔU = Q + W
= –293 J + 214 J
ΔU = –79 J
The correct answers are: Q = –293 J, ΔU = –79 J

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