Derive an expression for the work obtained in an isothermal reversible expansion of an ideal gas.

Hint Isothermal process is a thermodynamic process in which the temperature of the system will remain constant throughout the process. A reversible reaction is the kind of reaction that can return to its initial state by making some changes. Here we have an isothermal reversible expansion of an ideal gas.
Formula used
$PV = nRT$ Where, $P$ stands for the pressure of the system, $V$ stands for the volume of the system, $n$ stands for the number of moles of gas in the system, $R$ stands for the universal gas constant and $T$ stands for the temperature of the system.

Complete Step by step solution
Here we consider $n$ moles of ideal gas contained in a container fitted with a weightless and frictionless piston. Here the work done is in expanding the volume. Let $dV$ be the small change in volume against the external pressure.
Now the work done in increasing the volume by a small element $dV$ can be written as,
$dW = - PdV$
The total work done for the expansion of the gas from its initial volume ${V_1}$ to the final volume ${V_2}$ can be obtained by integrating the small work done
$W = - \int\limits_{{V_1}}^{{V_2}} {PdV} $
In the case of an ideal gas, we can write the ideal gas equation as,
$PV = nRT$
$ \Rightarrow P = \dfrac{{nRT}}{V}$
$W = - \int\limits_{{V_1}}^{{V_2}} {\dfrac{{nRT}}{V}} dV$
For isothermal expansion the temperature $T$is constant, hence we can write
$W = - nRT\int\limits_{{V_1}}^{{V_2}} {\dfrac{{dV}}{V}} $
$W = - nRT\left[ {\ln V} \right]_{{V_1}}^{{V_2}}$
Applying the limits we get,
$W = - nRT\left( {\ln {V_2} - \ln {V_1}} \right)$
$W = - nRT\ln \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)$
Converting into logarithm we get,
$W = - 2.303nRT\log \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)$………………………………………………………………………..equation
This is the total work done for the isothermal reversible expansion of an ideal gas.

Note
The work done can be expressed in terms of pressure also.
At constant temperature
${P_1}{V_1} = {P_2}{V_2}$
From this we get
$\dfrac{{{V_2}}}{{{V_1}}} = \dfrac{{{P_1}}}{{{P_2}}}$
Substituting in the above expression in equation
We get,
$W = - 2.303nRT\log \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right)$……………………………………………………………………equation
Equation and gives the expression for the work done by an isothermal reversible expansion of an ideal gas.