Gibbs Free Energy

This is another thermodynamic quantity that helps in predicting the spontaneity of a process, is called Gibbs energy (G).  

Gibbs free energy is the energy needed to create a system out of nothing in an environment at both constant pressure and constant temperature.

It is defined mathematically by the equation.  

G = H - TS

Where H = heat content, S = entropy of the system, T = absolute temperature  

Solved example: Which of the following will fit into the blank?  

When two phases of the same single substance remain in equilibrium with one another at a constant P and T, their molar _________ must be equal.  

             (A) Internal energy                              (B) Enthalpy  

             (C) Entropy                                         (D) Free energy  

Solution: (D)    

Free energy change  

For isothermal process.  

      ΔG = G2 – G1 = ΔH – TΔS  

ΔG = change in Gibbs free energy of the system.  

It is that thermodynamic quantity of a system the decrease in whose value during a process is equal to the maximum possible useful work that can be obtained from the system.  

Solved example: Calculate free energy change when one mole of NaCl is dissolved in water at 25°C. Lattice energy = 700 kJ/mol.ΔS at 25°C = 26.5 Jmol–1, Hydration energy of NaCl = -696 kJ/mol.  

                        (A) -3.9 kJ                                         (B) -8kJ  

                        (C) -12kJ                                           (D) -16kJ    

Solution: (A)

Problem 1:-

Calculate the total rotational kinetic energy of all the molecules in one mole of air at 25.0ºC.

Concept:

Let us assume all the molecules in the air are diatomic. For half rotation the molecule will contribute ½ kT to its rotational kinetic energy, where k is the Boltzmann’s constant. So for one complete rotation the molecule will contribute kTkT×2= kT) to its rotational kinetic energy.

So the total rotational kinetic energy will be,

E = kT

But k= R/NA

Where R is the gas constant and NA is the Avogadro’s number.

Solution:

To find out total rotational kinetic energy E, substitute R/NA for k in the equation E = kT,

E = kT

   = (R/NA) T  

For one mole of gas (NA =1), the equation E =(R/NA)T  will be,

E = RT      

To find out the total rotational kinetic energy of all the molecules, substitute 25.0̊ C for temperature T and 8.31 J/ K for R for one mole of gas in the equation E = RT,

E = RT

   = (8.31 J/ K) (25.0̊ C)

  = (8.31 J /K)(25+273) K

  = (8.31 J/ K)(298 K)

 = 2476.38 J

Rounding off to three significant figure, the total rotational kinetic of all the molecule will be 2480 J.

Problem 2:-

Calculate the internal energy of 1 mole of an ideal gas at 250ºC.

Concept:

Internal energy (Eint) of the ideal gas is defined as the,

Eint = 3/2 nRT

Where n is the number of moles, R is the gas constant and T is the temperature.

Solution:

To find out the internal energy of the ideal gas, substitute 1 mole for n, 250̊ C for temperature T and 8.31 J/ mol. K for the gas constant R in the equation E = 3/2 nRT,

E = 3/2 nRT

   = 3/2 (1 mol) (8.31 J/ mol. K) (250̊ C)

  = (1.5)( 1 mol) (8.31 J/ mol. K) (250+273) K

  = (1.5)( 1 mol) (8.31 J/ mol. K) (523) K

 = 6519.195 J

Rounding off to three significant figures, the internal energy of the ideal gas will be 6520 J.

Problem 3:-

A cosmic-ray particle with energy 1.34 TeV is stopped in a detecting tube that contains 0.120 mpl of neon gas. Once this energy is distributed among all the atoms, by how much is the temperature of the neon increased? 

Concept:

We can assume neon is an ideal gas. The internal energy (ΔEint) of the ideal gas is defined as the,

ΔEint = 3/2 nR ΔT

Where n is the number of moles, R is the gas constant and ΔT is the increase in temperature.

From the equation ΔEint = 3/2 nR ΔT, the increase in temperature ΔT will be,

ΔT = 2/3 ΔEint/nR

Solution:

To obtain the increase in temperature ΔT, substitute 1.34 TeV for ΔEint, 0.120 mol for number of mole n and 8.31 J/ mol. K for the gas constant R in the equation ΔT = 2/3 ΔEint/nR,

ΔT = 2/3 ΔEint/nR

     = 2/3 (1.34 TeV) (1012 eV/1 TeV) (1.6×10-19 J/eV) / (0.120 mol) (8.31 J/ mol. K)

     = 1.43×10-7 K

From the above observation we conclude that, the increased temperature of the neon will be 1.43×10-7 K.

Problem 4:-

A mixture of 1.78 kg of water and 262 g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass, is 1:1 at 0ºC. (a) Calculate the entropy change of the system during this process. (b) The system is then returned to the first equilinrium state, but in an reversible way. Calculate the entropy change of the system during this process. (c) Show that your answer is consistent with the second law of thermodynamics.

Concept:

The entropy change ΔS for a reversible isothermal process is defined as,

ΔS = Q/T

     = -mL/T

Here m is the mass, L is the latent heat and T is the temperature.

Solution:

  1. Mass of water = 1.78 kg

Mass of ice = 262 g

So the total mass of ice and water mixture will be,

Mass of ice-water mixture = (Mass of water) + (Mass of ice)

                                            = (1.78 kg) + (262 g)

                                            = (1.78 kg) + (262 g×10-3 kg/1 g)

                                            = 1.78 kg + 0.262 kg

                                            = 2.04 kg

If eventually the ice and water have the same mass, then the final state will have 1.02 kg (2.04 kg/2) of each.

Thus the mass of the water that changed into ice m will be the difference of mass of water mw  and mass of final state ms.

So, m = mw - ms

To obtain mass of water that changed into ice m, substitute 1.78 kg for mass of water mw and 1.02 kg for mass of final state ms in the equation m = mw - ms,

m = mw - ms

   = 1.78 kg – 1.02 kg

  = 0.76 kg

The change of water at 0̊ C to ice at 0̊ C  is isothermal.

To obtain the change in entropy ΔS of the system during this process, substitute 0.76 kg for mass m, 333×103 J/kg for heat of fusion of water L and 273 K for T in the equation ΔS = -mL/T,

ΔS = -mL/T

     = -(0.76 kg) (333×103 J/kg )/(273 K)

     = -927 J/K

From the above observation we conclude that, the change in entropy ΔS of the system during this process will be -927 J/K.

 
  1. Now the system is returned to the first equilibrium state, but in an irreversible way. Thus the change in entropy ΔS of the system during this process is equal to the negative of previous case.

So, ΔS = -(- 927 J/K)

           = 927 J/K

From the above observation we conclude that, the change in entropy ΔS of the system would be 927 J/K.

  1. In accordance to second law of thermodynamics, entropy change ΔS is always      zero.

The total change in entropy will be,

ΔS = (-927 J/K) + (927 J/K)

      = 0

From the above observation we conclude that, our answer is consistent with the second law of thermodynamics.

Problem 5:-

Water standing in the open at 32ºC evaporates because of the escape of some of the surface molecules. The heat vaporization is approximately equal to εn, where ε is the average energy of the escaping molecules and n is the number of molecules per kilogram. (a) Find ε. (b) What is ratio of ε to the average kinetic energy of H2O molecules, assuming that the kinetic energy is related to temperature in the same way as it is for gases?

Concept:

Number of molecules n is equal to the Avogadro’s number NA divided by molecular weight M.

So, n = NA/M

The average kinetic energy Eav of a diatomic molecule like H2O is defined as,

Eav = 3/2 kT

Here k is the Boltzmann constant and T is the temperature.

Solution:

(a)

As heat of vaporization Lv is approximately equal to εn, where ε is the average energy of the escaping molecules and n is the number of molecules per kilogram, so,

Lv = εn

Or,

ε = Lv/n

   = Lv/( NA/M)

   = LvM / NA

To obtain the average energy ε, substitute 2256×103 J/kg for latent heat of vaporization of water molecule (H2O) Lv, 0.018 kg/mol for M and 6.03×1023/mol for NA in the equation ε = LvM / NA, we get,

ε = LvM / NA

 = (2256×103 J/kg)( 0.018 kg/mol)/( 6.03×1023)

= 6.75×10-20 J

From the above observation we conclude that, the value of average energy ε would be 6.75×10-20 J.

(b)

As, Eav = 3/2 kT, therefore the ratio of ε to the average kinetic energy Eav of H2O molecule will be,

ε / Eav = ε / (3/2 kT)

           = 2 ε / 3 kT

To obtain the ratio of ε to the average kinetic energy Eav of H2O molecule, substitute 6.75×10-20 J for ε , 1.38×10-23 J/K for k and 32̊ C for T in the equation ε / Eav= 2 ε / 3 kT, we get,

ε / Eav= 2 ε / 3 kT

           = 2(6.75×10-20 J)/3(1.38×10-23 J/K)( 32̊ C)

           = 2(6.75×10-20 J)/3(1.38×10-23 J/K)( 32+273) K)

          =2(6.75×10-20 J)/3(1.38×10-23 J/K)( 305K)

          = 10.7

From the above observation we conclude that, the ratio of ε to the average kinetic energy Eav of H2O molecule would be 10.7.

To know more about the study material of engineering and medical exams, please fill up the form given below:

 

 

 

 

 

 

 

 

 

 

 

      

 

Name
Email Id
Mobile

Exam
Target Year

Related Resources
Third Law of Thermodynamics

THIRD LAW OF THERMODYNAMICS:- In all heat engines,...

Level 1 Objective Problems Of Thermodynamics

Solved Examples on Thermodynamics:- Problem 1 :- A...

Enthalpy of Reaction

Enthalpy of Reaction It is the enthalpy change...

Enthalpy of System

Enthalpy of a System The quantity U + PV is known...

Heat Capacity and Specific Heat

Specific Heat Capacity or Specific Heat [c]:- It...

Thermodynamic Process and their Types

Thermodynamic change or Thermodynamic Process:-...

HESS Law

HESS’S LAW This law states that the amount...

Level 2 Objective Problems Of Thermodynamics

Level 2 Objective Problems of Thermodynamics Level...

Application of Bond Energies

Application of bond energies (i) Determination of...

Work done during isothermal expansion

Work Done During Isothermal Expansion:- Consider...

Introduction to Thermodynamics

Introduction to Thermodynamics:- Thermodynamics:-...

Application of Hess Law

Application of Hess's Law 1. Calculation of...

Relationship-Free Energy and Equilibrium Constant

Relationship between free Energy and Equilibrium...

First Law of Thermodynamics

The First Law of Thermodynamics:- The first law of...

Objective Questions of Thermodynamics

Objective Questions of Thermodynamics and Answers...

Work done during adiabatic expansion

Work Done During Adiabatic Expansion:- Consider...

Macroscopic Extensive Intensive Properties

Macroscopic Properties He properties associated...

Solved Problems Part 1

Solved Problems on Specific Heat, Latent Heat and...

Solved Sample Problems Based on Thermodynamics

Solved Problems on Thermodynamics:- Problem 1:- A...

State of System

Thermodynamic State of a System and Macroscopic...

Second Law of Thermodynamics

Second Law of Thermodynamics:- Entropy:- The...

Specific Heat Capacity and Its Relation with Energy

Specific Heat Capacity and Its Relation with...

Reversible Irreversible Process

Reversible and Irreversible Process:- Reversible...

Miscellaneous Exercises Part I

Miscellaneous Exercises Thermal Physics:- Problem...

Bomb Calorimeter

BOMB CALORIMETER The bomb calorimeter used for...