4.3 Isothermal process

4.1 Isobaric Process:
4.2 Isochoric process
4.3 Isothermal process
4.5 Adiabatic process
4.6 Polytropic process

7.1 Second Law of Thermodynamics:
7.2 Carnot Engine

An isothermal process is a constant temperature process. In an isothermal process

$T$= constant or $\Delta T$=0

$P \propto \frac{1}{V}$ or $PV$= constant .

So $P-V$ graph is a rectangular hyperbola with $${P_i}{V_i} = {P_f}{V_f}$$

As $T$= constant, hence U=constant for an ideal gas, because $U$ is a function of $T$ only.$$\Delta U = 0$$$$\therefore Q = W$$

Work done in isothermal process

$$\begin{equation} \begin{aligned} W = \int\limits_{{V_i}}^{{V_f}} {PdV} \\ = \int\limits_{{V_i}}^{{V_f}} {\left( {\frac{{nRT}}{V}} \right)} dV\left ( {as\ P = \frac{{nRT}}{V}} \right) \\ = nRT\int\limits_{{V_i}}^{{V_f}} {\frac{{dV}}{V}} \left( {as\ T = constant} \right) \\ = nRT\ln \left( {\frac{{{V_f}}}{{{V_i}}}} \right) \\ = nRT\ln \left( {\frac{{{P_i}}}{{{P_f}}}} \right)\left( {as\ {P_i}{V_i} = {P_f}{V_f}} \right) \\\end{aligned} \end{equation} $$

For a process to be isothermal, any heat flow into or out of the system must occur slowly enough, so that thermal equilibrium is maintained.

Question 13. An ideal gas expands isothermally along $AB$ and does $700 J$ of work.

(a) How much heat does the gas exchange along $AB$?

(b) The then expands adiabatically along $BC$ and does $400 J$ of work. When the gas returns to $A$ along $CA$, it exhausts $100 J$ of heat to its surroundings. How much work is done on the gas along this path?

(a) $AB$ is an isothermal process. Hence,$$\begin{equation} \begin{aligned} \Delta {U_{AB}} = 0 \\ {Q_{AB}} = {W_{AB}} = 700\,J \\\end{aligned} \end{equation} $$

(b) BC is an adiabatic process. Hence$$\begin{equation} \begin{aligned} {Q_{BC}} = 0 \\ {W_{BC}} = 400\,J \\ \Delta {U_{BC}} = - {W_{BC}} = - 400\,J \\\end{aligned} \end{equation} $$ABC is a cyclic process and internal energy is a state function. Therefore,$${\left( {\Delta U} \right)_{whole\,\,cycle}} = 0 = \Delta {U_{AB}} + \Delta {U_{BC}} + \Delta {U_{CA}}$$and from first law of thermodynamics,$${Q_{AB}} + {Q_{BC}} + {Q_{CA}} = {W_{AB}} + {W_{BC}} + {W_{CA}}$$Substituting the values,$$\begin{equation} \begin{aligned} 700 + 0 - 100 = 700 + 400 + \Delta {W_{CA}} \\ \Delta {W_{CA}} = - 500\;J \\\end{aligned} \end{equation} $$Negative sign implies that work is done on the gas.

Table shows different values in different processes.