Gibbs Free EnergyThis 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 = G

_{2}– G_{1}= ΔH – TΔSΔG = 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

kTto its rotational kinetic energy, wherekis the Boltzmann’s constant. So for one complete rotation the molecule will contributekT(½kT×2=kT) to its rotational kinetic energy.So the total rotational kinetic energy will be,

E=kTBut

k=R/N_{A}Where R is the gas constant and

N_{A}is the Avogadro’s number.

Solution:To find out total rotational kinetic energy

E, substituteR/N_{A}forkin the equationE=kT,

E=kT= (

R/N_{A})TFor one mole of gas (

N_{A}=1), the equationE=(R/N_{A})Twill be,

E=RTTo find out the total rotational kinetic energy of all the molecules, substitute 25.0

^{°}C for temperatureTand 8.31 J/ K forRfor one mole of gas in the equationE=RT,

E=RT= (8.31 J/ K) (25.0

^{°}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 (

E_{int}) of the ideal gas is defined as the,

E_{int}= 3/2nRTWhere

nis the number of moles,Ris the gas constant andTis the temperature.

Solution:To find out the internal energy of the ideal gas, substitute 1 mole for

n, 250^{°}C for temperatureTand 8.31 J/ mol. K for the gas constantRin the equationE= 3/2nRT,

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

^{°}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 (Δ

E_{int}) of the ideal gas is defined as the,Δ

E_{int}= 3/2nR ΔTWhere

nis the number of moles,Ris the gas constant and ΔTis the increase in temperature.From the equation Δ

E_{int}= 3/2nR ΔT, the increase in temperatureΔTwill be,

ΔT= 2/3 ΔE_{int}/nR

Solution:To obtain the increase in temperature

ΔT, substitute 1.34 TeV for ΔE_{int}, 0.120 mol for number of molenand 8.31 J/ mol. K for the gas constantRin the equationΔT= 2/3 ΔE_{int}/nR,

ΔT= 2/3 ΔE_{int}/nR= 2/3 (1.34 TeV) (10

^{12}eV/1 TeV) (1.6×10^{-19}J/eV) / (0.120 mol) (8.31 J/ mol. K)= 1.43×10

^{-7}KFrom 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 Δ

Sfor a reversible isothermal process is defined as,Δ

S=Q/T= -

mL/THere

mis the mass,Lis the latent heat andTis the temperature.

Solution:

- 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

mwill be the difference of mass of waterm_{w }and mass of final statem_{s}.So,

m=m_{w}-m_{s}To obtain mass of water that changed into ice

m, substitute 1.78 kg for mass of waterm_{w}and 1.02 kg for mass of final statem_{s}in the equationm=m_{w}-m_{s},

m=m_{w}-m_{s}= 1.78 kg – 1.02 kg

= 0.76 kg

The change of water at 0

^{°}C to ice at 0^{°}C is isothermal.To obtain the change in entropy Δ

Sof the system during this process, substitute 0.76 kg for massm, 333×10^{3}J/kg for heat of fusion of waterLand 273 K forTin the equation ΔS= -mL/T,Δ

S= -mL/T= -(0.76 kg) (333×10

^{3}J/kg )/(273 K)= -927 J/K

From the above observation we conclude that, the change in entropy Δ

Sof the system during this process will be -927 J/K.

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

S= -(- 927 J/K)= 927 J/K

From the above observation we conclude that, the change in entropy Δ

Sof the system would be 927 J/K.

- In accordance to second law of thermodynamics, entropy change Δ
Sis always zero.The total change in entropy will be,

Δ

S= (-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 H_{2}O molecules, assuming that the kinetic energy is related to temperature in the same way as it is for gases?

Concept:Number of molecules

nis equal to the Avogadro’s numberN_{A}divided by molecular weightM.So,

n=N_{A}/MThe average kinetic energy

E_{av}of a diatomic molecule like H_{2}O is defined as,

E_{av}= 3/2kTHere

kis the Boltzmann constant andTis the temperature.

Solution:

(a)As heat of vaporization

L_{v}is approximately equal toεn, whereεis the average energy of the escaping molecules andnis the number of molecules per kilogram, so,

L_{v}=εnOr,

ε=L_{v}/n=

L_{v}/(N_{A}/M)=

L_{v}M/N_{A}To obtain the average energy

ε, substitute 2256×10^{3}J/kg for latent heat of vaporization of water molecule (H_{2}O)L_{v}, 0.018 kg/mol forMand 6.03×10^{23}/mol forN_{A}in the equationε=L_{v}M/N_{A}, we get,

ε=L_{v}M/N_{A}= (2256×10

^{3}J/kg)( 0.018 kg/mol)/( 6.03×10^{23})= 6.75×10

^{-20}JFrom the above observation we conclude that, the value of average energy

εwould be 6.75×10^{-20}J.

(b)As,

E_{av}= 3/2kT, therefore the ratio ofεto the average kinetic energyE_{av}of H_{2}O molecule will be,

ε/E_{av}=ε/ (3/2kT)= 2

ε/ 3kTTo obtain the ratio of

εto the average kinetic energyE_{av}of H_{2}O molecule, substitute 6.75×10^{-20}J forε, 1.38×10^{-23}J/K forkand 32° C forTin the equationε/E_{av}= 2ε/ 3kT, we get,

ε/E_{av}= 2ε/ 3kT= 2(6.75×10

^{-20}J)/3(1.38×10^{-23}J/K)( 32° 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

εto the average kinetic energyE_{av}of H_{2}O molecule would be 10.7.

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