We know that cyclic is given as:
(∂P / ∂V)S (∂V / ∂S)P (∂S / ∂P)V = -1
(∂P / ∂V)S = - [(∂S / ∂V) P]/[(∂S / ∂P)V] = [- (∂S / ∂T)P (∂T / ∂V)P]/[(∂S / ∂T) V (∂T / ∂P)V]
= -[ (CP / T) ...

ContinueComments Off on Using an appropriate cyclic rule and an appropriate Maxwell relation, show that (∂P / ∂V)_{S} = γ (∂P / ∂V)_{T } where γ = C_{P / CV}

ContinueComments Off on Three real heat engines have the same thermal efficiency and are connected in series. The first engine absorbs 2400 kJ of heat from a thermal reservoir at 1250 K and the third engine ejects its waste of 300 kJ to a sink at 150 K. determine the work output from each engine.

Lewis introduced a concept by making use of free energy function G to represent the actual behavior of real gases which is very much different from the concept of ideal gases. This concept is known as concept of Fugacity.
We know that the variation of free ...

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The variation of free energy change with change in temperature and pressure is discussed below:
Consider the following equation:
G = H – TS …………………………………….……. (1)
As H = U + PV
Substituting the value of H in equation (1) we will get:
G = U + PV – TS ...

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Physical Significance of Entropy:
1. Entropy as a measure of the Disorder of the system:
We know that all the spontaneous process in which heat is transferred through a finite temperature. Spontaneous process is also known as irreversible process because these processes take place at very fast ...

Given: temperature of the iron cube = 400o C = 400 + 273 = 673 K
Temperature of water = 10 kg
Temperature of water and cube after equilibrium = 50o C = 50 +273 = 323 K
Specific heat of water, cpw = 4186 J/kg K
Entropy changes ...

ContinueComments Off on An iron cube at a temperature of 400^{o} C is dropped into an insulated bath containing 10kg water at 25^{o} C. the water finally reaches a temperature of 50^{o} C at steady state. Given that the specific heat of water is equal to 4186 J / kg K. find the entropy changes for the iron cube and the water. Is the process reversible? If so why?

For one mole of an ideal gas:
dS = Cv dT / T + R dV / V
Integrating the above equation, assuming that Cv remains constant for an ideal gas, we have:
S = Cv ln T + R ln V + So
Where So represents the ...

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There are three types of processes in which entropy changes of an ideal gas. These three processes are:
1. Isothermal process
2. Isobaric process
3. Isochoric process
1. Isothermal process:
The process in which there is no change in temperature is known as Isothermal process. Entropy changes from S1 to ...

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