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Physics Exam  >  Physics Test  >  Topic wise Tests for IIT JAM   >  Thermodynamic Potential NAT Level – 1 - Physics MCQ

Thermodynamic Potential NAT Level – 1 - Free MCQ Test with solutions


MCQ Practice Test & Solutions: Thermodynamic Potential NAT Level – 1 (10 Questions)

You can prepare effectively for Physics Topic wise Tests for IIT JAM Physics with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Thermodynamic Potential NAT Level – 1". These 10 questions have been designed by the experts with the latest curriculum of Physics 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 10

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*Answer can only contain numeric values
Thermodynamic Potential NAT Level – 1 - Question 1
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A system of N localized, non-interacting spin 1/2 ions of magnetic moment in each is kept in an external  magnetic field H. If the system is in equilibrium at temperature T, the Helmholtz free energy of the system is given as   Calculate the value of α?


Detailed Solution: Question 1

Partition function 
Vibrational partition function in this 
where 



For the whole system  Z(v) = [Z(v)]N
Helmholtz free energy F = –NkB TlnZ

⇒ α = 2
The correct answer is: 2

*Answer can only contain numeric values
Thermodynamic Potential NAT Level – 1 - Question 2
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The free energy of a photon gas enclosed in a volume V is given by  where a is the constant and T is the temperature of the gas.
The chemical potential of the photon gas is?


Detailed Solution: Question 2


and  
No term depends on N
hence 
µ = 0
The correct answer is: 0

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Thermodynamic Potential NAT Level – 1 - Question 3
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Calculate the standard Gibb’s free energy change for the formation of methane from carbon and hydrogen at 298 K in the units of kJ/mol. Given that ΔH is –74.9 kJ/mol and ΔS = -80.75J{\rm{/K - mol}}.\)


Detailed Solution: Question 3

Given  ΔH = –74.9 kJ/mol
ΔS = –80.75 J/K-mol
Now use the equation
ΔS = ΔH – TΔS
⇒ ΔS = –74.9 kJ/mol +298 K (80.75/K-mol) × (1 kJ/1000J)
⇒ ΔS = – 50.9 kJ/mol
The correct answer is: -50.9

*Answer can only contain numeric values
Thermodynamic Potential NAT Level – 1 - Question 4
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Calculate the depression of melting point of ice (in K) produced by 1 atm increase of pressure, given that Lice = 80 cal gm-1 and specific volume of ice and water at 0°C are 1.091 cm3 and 1 cm3 respectively


Detailed Solution: Question 4

1 atm = 105 N/m2
 = 106 dynes/cm2



The correct answer is: 0.007429

*Answer can only contain numeric values
Thermodynamic Potential NAT Level – 1 - Question 5
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To change the melting point of ice by 1K, what should be the change in the pressure, given that L = 80 cal gm-1 and specific volume of ice and water at 0°C are 1 cm3 and 1.091 cm3 respectively. Give your answer in atm.


Detailed Solution: Question 5


dT = 1K

dp = 1.3525 × 108 dynes/cm2
⇒dp = 135.25 atm.
The correct answer is: 135.25

*Answer can only contain numeric values
Thermodynamic Potential NAT Level – 1 - Question 6
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Calculate the value of ΔH for an isothermal process?


Detailed Solution: Question 6

ΔH = Cp ΔT
ΔT = 0
⇒ ΔH = 0
The correct answer is: 0

*Answer can only contain numeric values
Thermodynamic Potential NAT Level – 1 - Question 7
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If Z be the partition function and N represent number of particles the chemical potential is given as  Find the value of α.


Detailed Solution: Question 7

dF = dU – pdV + µdN

Since F = –kT ln Z

⇒ α = –1
The correct answer is: -1

*Answer can only contain numeric values
Thermodynamic Potential NAT Level – 1 - Question 8
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A solid melts into liquid. The relation between the pressure p and the temperature T of the phase transition is P = –2T + P0 , the entropy charge associated with the phase transition is 1 Joule mol-1 K-1 . The Claussius-Clapeyron equation for the latent heat latent heat  then  is equal to?


Detailed Solution: Question 8

p = –2T + p0


ΔS = 1



The correct answer is: 0.5

*Answer can only contain numeric values
Thermodynamic Potential NAT Level – 1 - Question 9
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It the partition function of a harmonic oscillator with frequency  at temperature T in  then the free energy of 8 such independent oscillators in terms of  is


Detailed Solution: Question 9

The Helmholtz free energy N independent harmonic oscillator in terms of partition function Z is
F = –NkT ln Z
here 


N = 8

The correct answer is: 16

*Answer can only contain numeric values
Thermodynamic Potential NAT Level – 1 - Question 10
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A system of N non-interacting and distinguishable particles of spin1 is in thermodynamic equilibrium. The entropy of the system turns out to be of the form NkBln x. Find the value of x?


Detailed Solution: Question 10

Number of microstates  Ω = 3N
Entropy is given by 
∴ S = NKBln 3
⇒ x = 3
The correct answer is: 3

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About this Physics Practice Test

This test contains 10 MCQs on Thermodynamic Potential NAT Level – 1 with detailed solutions for each question. It is part of the Topic wise Tests for IIT JAM Physics course on EduRev, designed to align with the current Physics syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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