Maxwell's Relations MCQ Level – 2 - IIT JAM MCQ

Adiabatic volume coefficient of expansion,

Isobaric volume coefficient of expansion,



But, from Maxwell's relation





But 
and 

or 
or 
The correct answer is: 

Joule - Thomson coefficient µ,

For 1 mole of perfect gas, the equation of state is pV = RT
Differentiating it with respect to T, taking P constant,
We have,  
or 
or  
Substituting this result in equation (1) we get,
Joule-Thomson coefficient, μ = 0
The correct answer is: 0

Here, ∂H = 540 cal.
= 540 × 4.2 × 10⁷ erg.
∂V = 1676cc
Boiling Temperature T = 373 K
Temperature when water is to be boil = 150°C
= 150 + 273 = 423 K.
∂T = 423 – 373 = 50 K
∂p = ?
Applying these value in the maxwell's thermodynamical relation



= 1.814 × 10⁶ dynes/cm²
= 1.814 atmosphere
Therefore, the pressure at which water would boil at 150°C
= 1.814 + 1,000 = 2.814 atmospheric pressure.
The correct answer is: 2.814

Here, change in pressure, dP = 80 – 76
= 4 cm of Hg
= 4 × 13.6 × 980 dynes/cm²
Normal Boiling point, T = 80°C]
= 80 + 273 = 353 K
L = 980 joule/g
= 380 × 10⁷ erg/g.



= 1.233 K
=1.233°C
The boiling point of benzene at a pressure of 80 cm of Hg
= 80 + 1.233
= 81.233°C
The correct answer is: 81.233°C

From the maxwell's relation


But dH = dU + pdV

or 
But from perfect gas equation
pV = RT

Putting in (1)


The correct answer is: 0

Isothermal elasticity, 
Adiabatic elasticity 




But from the first four Maxwell's equation


∴ Substituting these values in equation (1)


The correct answer is: 

For a change of state from liquid to vapour
 ...(i)
Here S₁ and S₂ are the entropies in the liquid and vapour states respectively Differentiating equation (1) with respect to T.


This equation is known as clausius latent heat equation.
The correct answer is:  

Condition for a change to occur in any processes
dU – TdS ≤ pdV
d(U + pV – TS) ≤ 0
d(H – TS) ≤ 0
as [H = U + pV]
G = H – TS
or dG ≤ 0
It means is isothermal, Isobaric and reversible process, dG = 0 or G = constant
The correct answer is: remains constant but not zero

Let the entropy S of a thermodynamic system be a function of temperature T and volume v, i.e. S = f(T, V)
Since, dS is perfect differential, we can write

Multiplying by T, we get

For any substance, the specific heat at constant volume is given by

Using dQ = TdS
Also from Maxwell's equation, we have


The correct answer is: 

Adiabatic pressure coefficient of expansion,

isochoric pressure coefficient of expansion.


But, From Maxwell's relation


The correct answer is: