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Using Maxwell relation

T V

5.3.5 Second Energy Equation PdV

Using Maxwell relation

P

∂ This is popularly known as second energy

equation

Application of second energy equation

If U is function of independent variable of T and P.

(

T P

)

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Example: From relation dU =TdS−PdV Derive Maxwell relation

V

Hence U is exact differential

V S

Example: A real gas which obey van der Waal’s equation of state are kept in container which has temperature T0 and volume V0. if volume of container changes to V such that temperature of gas become T what is change in entropy?

Solution: Assume C_V is specific heat of constant volume For van der Waal’s gas

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where S is integration constant ₀ Example: For Vander wall gases, prove that

V2

Solution: From first energy equation T P

Example: Prove that (a) using Maxwell relation

V

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Use Maxwell relation

P

Example: If α is thermal expansivity at constant pressure and K_p T isothermal compressibility then prove that

Solution: From Maxwell relation

P

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Use Maxwell relation

V

Example: Prove that

(a) _{P V}

(b) For the Vander Waal’s gas prove that 1 2

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Differentiate (B) with respect to V

( )

²

Substituting the value

P

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volume expansion.

Solution: Let _{P V}

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Example: If Helmholtz free energy for radiation is given by c V

(a) What is radiation pressure

(b) If S entropy of system prove that specific heat at constant volume is given by C = 3S V

Example: The internal energy E of a system is given by

VN E bS

= 3 where b is constant and other

symbols have their used meaning.

(a) Find the temperature of system (b) Find Pressure of system

Solution: From first law of thermodynamics

VN

Example: Consider an Ideal gas where entropy is given by

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⎥⎦⎤

⎢⎣⎡ + +

= n

R V n R U

S n 5 ln 2 ln

2 σ

where n = number of moles, R = universal gas constant, U = internal energy V = volume and σ = constant

(a) Calculate specific heat at constant pressure and volume (b) Prove that internal energy is given by U PV

2

= 5

Solution: (a) From first law of thermodynamics

PdV dU

TdS = − , 1 P

dS dU dV

T T

= −

T U

S

V

= 1

⎟⎠

⎜ ⎞

⎝

⎛

∂

∂

T U

R n5 1

2 = U nRT

2

= 5

T nR C_V U

2

= 5

⎟⎠

⎜ ⎞

⎝

⎛

∂

= ∂ ⇒C_P =C_V +R 7

2 CP nR

⇒ =

(b)

V nRT V

nRT U

U 2

5 2

5 ⇒ =

=

nRT PV =

P V = nRT

⇒

V P U

2

=5 U PV

2

= 5

⇒

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Example: Using the equation of state PV = nRT and the specific heat per mole

2 C_v =3R for

monatomic ideal gas

(a) Find Entropy of given system.

(b) Find free energy of given system , 3

nRT nRT

T nRT V F

⎛ ⎞

= −⎜⎝ − ⎟⎠+ where F₀ =T⋅S₀ is again constant

Example: From electromagnetic theory Maxwell found that the pressure P from an isotropic radiation equal to

3

1 the energy density i.e. 1 3 P U

= V where V is volume of the cavity using the first energy equation prove that

Energy densityu is proportional to T . ⁴ Solution:

T T T

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MCQ (Multiple Choice Questions) Q1. Which of the following is not a Maxwell’s thermodynamic relation?

(a)

Q2. Which one of the following thermo dynamical relations is used for certain adiabatic changes, such as the sudden compression of a liquid or sudden -stretching of a rod?

(a)

Q3. Which one of the Maxwell’s thermodynamic relations given below leads to Clausius-Clapeyron equation?

(a)

Q4. Which of the following is correct if α is volume expansivity and other variables have usual meaning in thermodynamics .

(a) ^{P 2}

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volume, temperature, internal energy, entropy, and Helmholtz free energy respectively.

Then the following relation is true

(a) S

Q8. Which of the following thermodynamic relation will give the Maxwell relation

? Q9. Which of the following is not an exact differential?

(a) dQ where Q heat absorbed (b) dU where U is internal energy (c) dS where S is entropy (d) dF where S is entropy

Q10. Which among the following sets of Maxwell relations is correct? (U-internal energy, H-enthalpy, A-Helmholtz free energy and G-Gibbs free energy)?

(a)

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Q11. When a system is held at constant temperature and pressure in a state of equilibrium, then it attains a minimum value of:

(a) internal energy (b) enthalpy (c) Helmholtz energy (d) Gibb’s free energy Q12. Given that

=

H the enthalpy of a system

=

T absolute temperature and S =entropy

TS H

G= − is the Gibbs function for the system In the case of a reversible, isotherm and isobaric process:

(a) G= constant

(b) G>0 and changes withT (c) G<0 and changes with S (d) G changes with both T and S

Q13. The Gibb’s function G in thermodynamics is defined is TS

H G= −

where H is the enthalpy, T is the temperature and S is the entropy. In an isothermal, isobaric, reversible process, G :

(a) remains constant, but not zero (b) varies linearly (c) varies non-linearly (d) is zero Q14. The internal energy E of a system is given by

VN E bS

= 3 , where b is a constant and other

symbols have their usual meaning. The temperature of this system is equal to (a) VN

bS²

(b) VN

bS²

3 (c)

N V

bS

2 3

(d)

2

⎟⎠

⎜ ⎞

⎝

⎛ N S

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Q15. The free energy of gas of N particles in a volume V and at a temperature T is

[

^aV

( )

^{k T N}

]

T Nk

F = _B ln _{0 B} ^5/2/ , where a is a constant and ₀ k_B denotes the Boltzmann constant. The internal energy of the gas is

(a) Nk_BT 2

3 (b) Nk_BT

2 5

(c) ^NkB^T

[

^aV

(

^kB^T

)

^N

]

^NkB^T

2 / 3

ln ₀ ^5/2 − (d) Nk_BTln

[

a0V /

( )

k_BT ^5/2

]

Q16. The entropy S of a thermodynamic system as a function of energy E is given by the following graph

The temperatures of the phases A,B and C , denoted by T ,_A T_B andT , respectively, _C satisfy the following inequalities:

(a) T_C >T_B >T_A (b) T_A >T_C >T_B (c) T_B >T_C >T_A (d) T_B >T_A >T_C

Q17. The entropy of an ideal paramagnet in a magnetic field is given approximately by

2

S =S0−cU where U is energy of the spin system and c is constant then which one is correct plot between internal energy and temperature T where −∞ < < ∞ T

(a) (b)

(c) (d)

U

T

U

T

U

T U

T

S

A

B C

→ E

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Q18. A certain system is found to have Gibbs free energy given by

5/2

G(p,T) RT ln aP (RT)

⎛ ⎞

= ⎜ ⎟

⎝ ⎠

Where a and R are constants then specific heat at constant pressure (c_p) is given by (a) ³R

2 (b)⁵R

2 (c)⁷R

2 (d) ⁹R

2

Q19. Helmholtz free energy is given by F =−CT⁴ where C is constant and T is temperature in Kelvin then which one is correct relation between specific heat at constant volume Cv and entropy S is given by

(a) 2C_V = S (b) C_V =4S (c) C_V =3S (d) Cv S 2

= 3

Q20. If C is the specific heat of the ideal gas then which of the following is correct of _V Vander wall gases for same degree of freedom .

(a) dU =C dT_V (b) a₂

dU dV

= −V

(c) _V a₂

dU C dT dV

= −V (d) _V a₂

dU C dT dV

= +V

Q21. For a Van der Waals gas the equation of the adiabatic curve in the variables T,V; (a) T

(

V −b

)

^R/CP =constant (b) T

(

V −b

)

^R/CV =constant (c) T

(

V −b

)

^−R/CP =constant (d) T

(

V −b

)

^−R/CV =constant

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MSQ (Multiple Select Questions)

Q22. If His enthalpy and G is Gibbs free energy of the thermodynamic system then which of the following is correct

(a) H = −G TS (b) H = +G TS Q23. Which of the following statements are correct .

(a) The first energy equation is be given by P

(c) van der Waal’s gas behave like a ideal as high temperature (d) for Ideal gas c_p−c_V =R

Q25. Which of the following is correct?

(a) _{P V}

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Q26. If α is volume expansivity of substance and β_T is isothermal compressibility β_S is adiabatic compressibility and it is given that

2 one of the following is correct

(a)

Q27. Which of the following is correct if all variable have usual meaning in thermodynamics (a)

(a) the increment of the internal energy of the gas is zero (b) the increment of the internal energy of the gas _⎟⎟

⎠

(c) heat exchange during the process is _⎥

⎦

(d) heat exchange during the process is

b

Q29. The free energy for a photon gas is given by ⁴ 3 VT entropy S and the pressure P of the photon gas are

(a) ³

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Solutions

MCQ (Multiple Choice Questions) Solution Ans. 1: (d)

Ans. 2: (a)

Solution: The sudden stretching of wire of compression of liquid is given by

P

Solution: The rate of change of temperature pressure is given by Clausius-Clapeyron equation which is given as T

(

V V

)

^L

This can be derived by Maxwell’s first thermodynamical relation given as

V

Solution: from Maxwell relation

T V Solution:

V T

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Ans. 6: (b) Solution:

p T

Solution: Heat exchange is path dependent, so it is not perfect differential.

Ans. 10: (b)

Solution: The change in Gibb’s free energy is given as SdT

From laws of thermodynamics is given as PdV

dU TdS= + so, G=U +PV−TS

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Solution: Gibb’s function is given as TS

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Solution: Now temperature of phase TA,TB,TC

Now dE T Solution:

V

Solution: 4CT³

T

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MSQ (Multiple Select Questions) Ans. 22: (b) and (c)

Solution: The first law of thermodynamics is given as dU+PdV =TdS

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Solution:

2

Solution: from Maxwell relation

T V

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Similarly another Maxwell relation is given by

T V

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Chapter - 6

Phase Transition and Low Temperature Physics 6.1 Third Law of Thermodynamics and Attainable of Low Temperature

The third law of thermodynamics is some time stated as follows:

It is impossible for any process, no matter how idealized, to reduce the entropy of a system to its zero point value in a finite number of

operations.

Properties of material at low temperature At T →0

CP = CV = 0 At T → 0

S → 0

At T → 0

Thermal expansion coefficient 1 0

⎟ =

⎠

⎜ ⎞

⎝

⎛

∂

= ∂

T P

V α V

=0 α

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