Thermodynamics and Thermochemistry
    2.0 Thermodynamic Processes

2.0 Thermodynamic Processes

The process by which a system gets changed from one state to the other state, Macro properties get changed in this process, These are of the following types:-

(1) Isothermal Process - The process in which temperature of the system remains constant, In this process, the boundaries of the system are mobile, The energy released in the process is absorbed by the surroundings and the system takes the required energy back from the surroundings, while the temperature remains constant, Thus the change in temperature, $dT= 0$

Maximum work is done in the expansion of n moles of an ideal gas.

${W_{\max }} = - 2.303$ nR T log $\frac{{{V_2}}}{{{V_1}}}$

Where ${V_1}$= Initial volume of the gas

${V_2}$ = Final volume of the gas


(2) Adiabatic process: When there is no exchange of heat between the system and the surrounding then the process is called an adiabatic process i.e. $q = 0$

(3) Isobaric process: When the pressure on a system remains constant during an operation, then the process is termed isobaric i.e. $\Delta $P = 0

(4) Isochoric process: When there is no change in the volume of the system during an operation, the process is termed isochoric. i.e. $\Delta $V = 0

(5) Cyclic process: If a system after having undergone a change returns back to its initial state, the process is called a cyclic process. The path of such a process is called a cycle. Change in the value of any state function for a cyclic process is 0.

(6) Reversible Process: When a process is carried out so slowly such that the system and the surrounding are always in equilibrium, then the process is termed as a reversible process.

(7) Irreversible Process: A process which is carried out so rapidly that the system does not get a chance to attain equilibrium, then the process is termed as an irreversible process.



Graphical representation of various thermodynamic functions

Work done in isothermal reversible process

${\text{dq = dE }} - {\text{dW}}$

Consider, a reversible expansion of an ideal gas by volume dV then for isothermal process dE = 0

$\therefore $ ${\text{dq = }} - {\text{dW}}$

The total work done during expansion of gas from ${{\text{V}}_1}$ to ${{\text{V}}_2}$


$$\int {{\text{dW = }}\int\limits_{{{\text{v}}_2}}^{{{\text{v}}_2}} { - {\text{P}}{\text{.dV = - }}\int\limits_{{{\text{v}}_1}}^{{{\text{v}}_2}} {\frac{{{\text{nRT}}}}{{\text{V}}}} } } ,{\text{dV}}$$

$$ = - {\text{nRT ln }}\frac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}}$$

$${\text{W = - 2}}{\text{.303 nRT lo}}{{\text{g}}_{10}}\frac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}}$$

Since temperature is constant, therefore ${{\text{P}}_1}{{\text{V}}_1}{\text{ = }}{{\text{P}}_2}{{\text{V}}_2}$

$$\frac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}} = \frac{{{{\text{P}}_1}}}{{{{\text{P}}_2}}}$$

$${{\text{W}}_{{\text{rev}}}} = - 2.303{\text{ nRT lo}}{{\text{g}}_{10}}\left( {\frac{{{{\text{P}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}}} \right)$$

If the same change is carried out irreversibly work done by the system is given by the expression

$${{\text{W}}_{{\text{rev}}}} = {\text{ - }}{{\text{P}}_{{\text{ext}}}}\left( {{{\text{V}}_2} - {{\text{V}}_1}} \right)$$

Workdone in adiabatic reversible process

Consider a reversible expansion of an ideal gas by volume dV then from ${{\text{I}}^{{\text{st}}}}$ law of thermodynamics

$${\text{dq = 0}}$$

$${\text{du = dW}}$$

$${\text{du = n}}{{\text{C}}_{\text{V}}}{\text{ dT}}$$

$${C_P} - {C_V} = R(Mayer'srelation)$$


${W_{rev}} = \frac{{nR}}{{\gamma - 1}}\left[ {{T_2} - {T_1}} \right]$


$\frac{{{C_P}}}{{{C_V}}} - 1 = \frac{R}{{{C_V}}}$


$\gamma - 1 = \frac{R}{{{C_V}}}$ $\because $ ${\text{ }}{C_V} = \frac{R}{{\gamma - 1}}$


$dW = + \frac{R}{{\gamma - 1}}n.dT$


$\int {dW = \int\limits_{{T_1}}^{{T_2}} {\frac{{nR}}{{\gamma - 1}}dT} } $


${W_{rev}} = \frac{{nR}}{{\gamma - 1}}\left[ {{T_2} - {T_1}} \right]$


Where $\gamma = \frac{{{C_P}}}{{{C_V}}},\gamma $ is called Poisson's ratio.

Improve your JEE MAINS score
10 Mock Test
Increase JEE score
by 20 marks
Detailed Explanation results in better understanding
Exclusively for
JEE MAINS and ADVANCED
9 out of 10 got
selected in JEE MAINS
Lets start preparing
DIFFICULTY IN UNDERSTANDING CONCEPTS?
TAKE HELP FROM THINKMERIT DETAILED EXPLANATION..!!!
9 OUT OF 10 STUDENTS UNDERSTOOD