Question

Derive an expression for work done by the gas in an isothermal process.

Answer 1

Hint: Isothermal process is the process where the temperature of the system remains constant. Work done by the gas is actually the work done by that system against the external force or that done by the external agency. It is defined in terms of pressure and volume. Thus by using this definition and the ideal gas law, the expression for work done by the gas in an isothermal process can be derived.

Formula Used:
The work done by a gaseous system \[W\] against the external force is given by:\[dW = pdV\]
where, \[p\] is the pressure and \[V\]is the volume of the given gas.
The ideal Gas law is given as: \[pV = RT\]
where, \[p\] is the pressure, \[V\] is the volume, \[T\] is the temperature of the given gas and \[R\] is the gas constant.

Complete step by step answer:
There are three types of thermodynamic processes. They are Isothermal, Adiabatic and isobaric. Isothermal process is the process where the temperature of the system remains constant. The work done by a gaseous system \[W\] against the external force is given by
\[dW = pdV\]
where, \[p\] is the pressure and \[V\]is the volume of the given gas.
Let \[{V_1}\] be the initial volume and \[{V_2}\] be the final volume, then total work done will be
\[W = \int {dW = \int\limits_{{V_1}}^{{V_2}} {pdV} } \] \[ \to (1)\]
According to the ideal Gas law, \[pV = RT\]. Here, \[p\] is the pressure and \[V\]is the volume, \[T\] is the temperature of the given gas and \[R\] is the gas constant.

This equation can be rearranged and written in terms of pressure as follows
\[p = \dfrac{{RT}}{V}\]
Substituting this value of pressure in equation (1), we get
\[W = \int\limits_{{V_1}}^{{V_2}} {\left( {\dfrac{{RT}}{V}} \right)dV} \\
\Rightarrow W = RT\int\limits_{{V_1}}^{{V_2}} {\left( {\dfrac{{dV}}{V}} \right)} \\
\Rightarrow W = RT[\ln V]_{{V_1}}^{{V_2}} = RT[\ln {V_2} - \ln {V_1}] \\
\Rightarrow W = RT\ln \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)\]
This can also be written in terms of logarithm as
\[W = 2.303\,RT\,lo{g_{10}}\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)\]
This is the expression for work done by the gas in an isothermal process.
If there are \[n\] moles of gas, the expression for work done can also be written as
\[W = 2.303nRTlo{g_{10}}\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)\]
Also, \[{p_1}{V_1} = {p_2}{V_2}\] or \[\dfrac{{{V_2}}}{{{V_1}}} = \dfrac{{{p_1}}}{{{p_2}}}\]
Thus, the work done can also be written in terms of pressure as
\[\therefore W = 2.303\,nRT\,lo{g_{10}}\left( {\dfrac{{{p_1}}}{{{p_2}}}} \right)\]

Note:When an ideal gas expands freely, the temperature changes but the internal energy remains unchanged. That is, a part of the heat content of the system may also be used for doing work by the expanding gas. For isothermal processes, the gas should be contained in a good conductor such as a copper vessel. Also in isothermal processes, the pressure at the end of the process is double the pressure that was at the beginning.