We know that if we have to take a thermodynamic system from initial to final state we have several paths that can be taken. In this article we will be discussing about those thermodynamic processes. Before that we will see what a quasi-static process is. It has been discussed that state variables are defined only when the thermodynamic system is in equilibrium with the surrounding. So a process in which at each moment the system is in thermodynamic equilibrium with the surrounding is known as a quasi-static process.

## The Thermodynamic Processes:

(a) Isothermal Process: It is a thermodynamic process in which temperature remains constant. We know,

\(~~~~~~~~~~~~~~W = ∫P dV\)

\(~~~~~~~~~~~~~~According~~ to ~~Gas ~~law,\)

\(~~~~~~~~~~~~~~PV = nRT\)

\(~~~~~~~~~~~~~~\)

Using this value of P in work done we get,

\(~~~~~~~~~~~~~~\)

\(~~~~~~~~~~~~~~\)

If \( V_B > V_A\)

Also we know internal energy only depends on temperature. As the temperature is constant hence ∆U = 0. So from first law of thermodynamics,

\(~~~~~~~~~~~~~~Q = W\)

(b)Adiabatic Process: It is a thermodynamic process in which no heat is exchanged between the system and the surrounding. So, Q = 0. Mathematically this process is represented as

\(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(~~~~~~~~~~~~~~W = ∫P dV\)

Substituting P we get,

\(~~~~~~\)

\(~~~~~~~~~~~~~~\)

\(~~~~~~~~~~~~~\)

For adiabatic process,

\(~~~~~~~~~~~~~~ ∆U = -W \)

So if work done is negative internal energy increases and vice versa.

(c)Isochoric Process: In isochoric process the change in volume of thermodynamic system is zero. As change in volume is zero so work done is zero. From First law,

\(~~~~~~~~~~~~~~ Q = ∆U \)

(d) Isobaric Process: The pressure remains constant during this process. So ,

\(~~~~~~~~~~~~~~\)

So if volume increases work done is positive else negative.

(e) Cyclic Process: It is a process in which the final state of the system is equal to initial state. As we know Change in internal energy is state function so in this case ∆U = 0.

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