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

Last updated date: 16th Apr 2024
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MVSAT 2024
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Hint: In an isothermal process, the temperature remains constant. When work is done, the volume expands thereby reducing the pressure. It is a thermodynamic process. In this process, the transfer of heat to the surroundings takes place to make the temperature constant.

Complete step by step solution:
Let us consider 1 mole of gas is enclosed in an isothermal container. Let ${P_1},$${V_1}$ and T be the initial pressure, initial volume, and temperature. As work is done, let the gas expand to ${P_2},$${V_2}$ where ${P_2}$ is the reduced pressure and ${V_2}$ is the expanded volume.
Since the process is an Isothermal Process, the temperature remains constant. We know that work done is given by,
$W = \int {dW} $
$ \Rightarrow W = \int_{{V_1}}^{{V_2}} {PdV} \_\_\_\_\_\_\_\_\left( 1 \right)$
We have the relation $PV = nRT$
$ \Rightarrow PV = RT$ $\left( {\because n = 1mole} \right)$ and R is the ideal gas constant.
$ \Rightarrow P = \dfrac{{RT}}{V}$
Substituting the value of P in equation 1, we get
$ \Rightarrow W = RT\int_{{V_1}}^{{V_2}} {\dfrac{{dV}}{V}} $
$ \Rightarrow W = RT\left[ {\ln V} \right]_{{V_1}}^{{V_2}}$
$ \Rightarrow W = RT\left[ {\ln {V_2} - \ln {V_1}} \right]$
\[ \Rightarrow W = RT\ln \dfrac{{{V_2}}}{{{V_1}}}\]
\[\therefore W = 2.303RT{\log _{10}}\dfrac{{{V_2}}}{{{V_1}}}\]
We know that for constant temperature,
\[\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{V_2}}}{{{V_1}}}\]
Thus, \[W = 2.303RT{\log _{10}}\dfrac{{{P_1}}}{{{P_2}}}\]
Thus, work done by the gas in an isothermal process is given by the expression, \[W = 2.303RT{\log _{10}}\dfrac{{{P_1}}}{{{P_2}}}\].

Note: 1. We can say that the work is positive when the force and the displacement are in the same direction and work is negative when the force and the displacement are opposite in direction. When the force and the displacement are perpendicular to each other, then the work done is zero or zero work.
2. During the isothermal process, both pressure and volume changes. Some heat engines like the Carnot Cycle are carried out by the isothermal process. Also, this process is of special interest to the ideal gases.
3. In the adiabatic process, the system does not exchange any heat with the surroundings, which is a contrast to the isothermal process.