Click to Chat
0120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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 = G_{2} – G_{1} = 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 kT (½ kT×2= kT) to its rotational kinetic energy. So the total rotational kinetic energy will be, E = kT But 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, substitute R/N_{A} for k in the equation E = kT, E = kT = (R/N_{A}) T For one mole of gas (N_{A} =1), the equation E =(R/N_{A})T will be, E = RT To find out the total rotational kinetic energy of all the molecules, substitute 25.0^{°} 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) (25.0^{°} 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} = 3/2 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, 250^{°} 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) (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 (ΔE_{int}) of the ideal gas is defined as the, ΔE_{int} = 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 ΔE_{int} = 3/2 nR ΔT, the increase in temperature ΔT will 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 mole n and 8.31 J/ mol. K for the gas constant R in the equation ΔT = 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} 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: 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 m_{w } and mass of final state m_{s}. So, m = m_{w} - m_{s} To obtain mass of water that changed into ice m, substitute 1.78 kg for mass of water m_{w} and 1.02 kg for mass of final state m_{s} in the equation m = 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 ΔS of the system during this process, substitute 0.76 kg for mass m, 333×10^{3} 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) (333×10^{3} 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. 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. 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 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 n is equal to the Avogadro’s number N_{A} divided by molecular weight M. So, n = N_{A}/M The average kinetic energy E_{av} of a diatomic molecule like H_{2}O is defined as, E_{av} = 3/2 kT Here k is the Boltzmann constant and T is the temperature. Solution: (a) As heat of vaporization L_{v} 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, L_{v} = εn Or, 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 for M and 6.03×10^{23}/mol for N_{A} in the equation ε = 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} J From the above observation we conclude that, the value of average energy ε would be 6.75×10^{-20} J. (b) As, E_{av} = 3/2 kT, therefore the ratio of ε to the average kinetic energy E_{av} of H_{2}O molecule will be, ε / E_{av} = ε / (3/2 kT) = 2 ε / 3 kT To obtain the ratio of ε to the average kinetic energy E_{av} of H_{2}O molecule, substitute 6.75×10^{-20} J for ε , 1.38×10^{-23} J/K for k and 32° C for T in the equation ε / E_{av}= 2 ε / 3 kT, we get, ε / E_{av}= 2 ε / 3 kT = 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 ε to the average kinetic energy E_{av} of H_{2}O molecule would be 10.7. To read more, Buy study materials of Thermodynamics comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.
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 = G_{2} – G_{1} = 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 kT (½ kT×2= kT) to its rotational kinetic energy.
So the total rotational kinetic energy will be,
E = kT
But 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, substitute R/N_{A} for k in the equation E = kT,
= (R/N_{A}) T
For one mole of gas (N_{A} =1), the equation E =(R/N_{A})T will be,
E = RT
To find out the total rotational kinetic energy of all the molecules, substitute 25.0^{°} C for temperature T and 8.31 J/ K for R for one mole of gas in the equation 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.
Internal energy (E_{int}) of the ideal gas is defined as the,
E_{int} = 3/2 nRT
Where n is the number of moles, R is the gas constant and T is the temperature.
To find out the internal energy of the ideal gas, substitute 1 mole for n, 250^{°} 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) (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?
We can assume neon is an ideal gas. The internal energy (ΔE_{int}) of the ideal gas is defined as the,
ΔE_{int} = 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 ΔE_{int} = 3/2 nR ΔT, the increase in temperature ΔT will be,
ΔT = 2/3 ΔE_{int}/nR
To obtain the increase in temperature ΔT, substitute 1.34 TeV for ΔE_{int}, 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 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} 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.
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.
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 m_{w } and mass of final state m_{s}.
So, m = m_{w} - m_{s}
To obtain mass of water that changed into ice m, substitute 1.78 kg for mass of water m_{w} and 1.02 kg for mass of final state m_{s} in the equation m = 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 ΔS of the system during this process, substitute 0.76 kg for mass m, 333×10^{3} 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) (333×10^{3} 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.
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.
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 H_{2}O molecules, assuming that the kinetic energy is related to temperature in the same way as it is for gases?
Number of molecules n is equal to the Avogadro’s number N_{A} divided by molecular weight M.
So, n = N_{A}/M
The average kinetic energy E_{av} of a diatomic molecule like H_{2}O is defined as,
E_{av} = 3/2 kT
Here k is the Boltzmann constant and T is the temperature.
(a)
As heat of vaporization L_{v} 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,
L_{v} = εn
Or,
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 for M and 6.03×10^{23}/mol for N_{A} in the equation ε = 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} J
From the above observation we conclude that, the value of average energy ε would be 6.75×10^{-20} J.
(b)
As, E_{av} = 3/2 kT, therefore the ratio of ε to the average kinetic energy E_{av} of H_{2}O molecule will be,
ε / E_{av} = ε / (3/2 kT)
= 2 ε / 3 kT
To obtain the ratio of ε to the average kinetic energy E_{av} of H_{2}O molecule, substitute 6.75×10^{-20} J for ε , 1.38×10^{-23} J/K for k and 32° C for T in the equation ε / E_{av}= 2 ε / 3 kT, we get,
ε / E_{av}= 2 ε / 3 kT
= 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 ε to the average kinetic energy E_{av} of H_{2}O molecule would be 10.7.
To read more, Buy study materials of Thermodynamics comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
THIRD LAW OF THERMODYNAMICS:- In all heat engines,...
Solved Examples on Thermodynamics:- Problem 1 :- A...
Enthalpy of Reaction It is the enthalpy change...
Enthalpy of a System The quantity U + PV is known...
Specific Heat Capacity or Specific Heat [c]:- It...
Thermodynamic change or Thermodynamic Process:-...
HESS’S LAW This law states that the amount...
Level 2 Objective Problems of Thermodynamics Level...
Work Done During Isothermal Expansion:- Consider...
Introduction to Thermodynamics:- Thermodynamics:-...
Application of Hess's Law 1. Calculation of...
Relationship between free Energy and Equilibrium...
The First Law of Thermodynamics:- The first law of...
Objective Questions of Thermodynamics and Answers...
Work Done During Adiabatic Expansion:- Consider...
Macroscopic Properties He properties associated...
Solved Problems on Specific Heat, Latent Heat and...
Solved Problems on Thermodynamics:- Problem 1:- A...
Thermodynamic State of a System and Macroscopic...
Second Law of Thermodynamics:- Entropy:- The...
Specific Heat Capacity and Its Relation with...
Application of bond energies (i) Determination of...
Miscellaneous Exercises Thermal Physics:- Problem:...
Reversible and Irreversible Process:- Reversible...
BOMB CALORIMETER The bomb calorimeter used for...