JEE Main & Advanced Physics Thermodynamical Processes Isothermal Process

When a thermodynamic system undergoes a physical change in such a way that its temperature remains constant, then the change is known as isothermal changes.

(1) Essential condition for isothermal process

(i) The walls of the container must be perfectly conducting to allow free exchange of heat between the gas and its surrounding.

(ii) The process of compression or expansion should be so slow so as to provide time for the exchange of heat.

Since these two conditions are not fully realised in practice, therefore, no process is perfectly isothermal.

(2) Equation of state : In this process, P and V change but T = constant i.e. change in temperature \[\Delta T=0\].

Boyleís law is obeyed i.e. PV= constant \[\Rightarrow \,\,{{P}_{1}}{{V}_{1}}={{P}_{2}}{{P}_{2}}\]

(3) Example of isothermal process : Melting of ice (at \[{{0}^{o}}C\]) and boiling of water (at \[{{100}^{o}}C\]) are common example of this process.

(4) Indicator diagram : According to PV = constant, graph between P and V is a part of rectangular hyperbola. The graphs at different temperature are parallel to each other are called isotherms.

\[{{T}_{1}}<{{T}_{2}}<{{T}_{3}}\]

Two isotherms never intersect

(i) Slope of isothermal curve : By differentiating PV = constant. We get

\[P\,dV+V\,dP=0\]

\[\Rightarrow \] \[PdV=\,-\,VdP\]

\[\Rightarrow \] Slope =\[\tan \theta =\frac{dP}{dV}=-\frac{P}{V}\]

(ii) Area between the isotherm and volume axis represents the work done in isothermal process.

If volume increases \[\Delta W=+\] (Area under curve) and if volume decreases \[\Delta W=-\] (Area under curve)

(5) Specific heat : Specific heat of gas during isothermal change is infinite. As  \[C=\frac{Q}{m\Delta T}=\frac{Q}{m\times 0}=\infty \]  [As \[\Delta T=0\]]

(6) Isothermal elasticity \[({{E}_{\theta }})\]: For this process PV = constant.

\[\Rightarrow \] \[P\,dV=-V\,dP\]\[\Rightarrow \]\[P=\frac{dP}{-dV/V}=\frac{\text{Stress}}{\text{Strain}}={{E}_{\theta }}\]

\[\Rightarrow \] \[{{E}_{\theta }}=P\,\]i.e. isothermal elasticity is equal to pressure

At N.T.P., \[{{E}_{\theta }}=\] Atmospheric pressure \[=1.01\times {{10}^{5}}N/{{m}^{2}}\]

(7) Work done in isothermal process

\[W=\int_{\,{{V}_{i}}}^{\,{{V}_{f}}}{P\,dV}=\int_{\,{{V}_{i}}}^{\,{{V}_{f}}}{\frac{\mu RT}{V}dV}\]       [As \[PV=\mu RT\]]

\[W=\mu RT{{\log }_{e}}\left( \frac{{{V}_{f}}}{{{V}_{i}}} \right)=2.303\ \mu RT\ {{\log }_{10}}\left( \frac{{{V}_{f}}}{{{V}_{i}}} \right)\]

or \[W=\mu RT{{\log }_{e}}\left( \frac{{{P}_{i}}}{{{P}_{f}}} \right)=2.303\mu RT\ {{\log }_{10}}\left( \frac{{{P}_{i}}}{{{P}_{f}}} \right)\]

(8) FLOT in isothermal process : From \[\Delta Q=\Delta U+\Delta W\]

\[\because \] \[\Delta U=0\]  [As \[\Delta T=0\]]      \[\Rightarrow \]\[\Delta Q=\Delta W\]

i.e. heat supplied in an isothermal change is used to do work against external surrounding.

or if the work is done on the system than equal amount of heat energy will be liberated by the system.