Question

How to calculate work done by isothermal processes?

Answer 1

So first we must know what an isothermal process is. It is a thermodynamic process in which the temperature of the system remains constant. The transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained. The change of a substance, object or system at a particular constant temperature, is known as an Isothermal Process.

Complete answer:
So, as we know an isothermal process is the one in which the pressure and volume of the gas changes at constant temperature.
Hence, $\Delta T = 0$
The total work done by a gas in expanding from ${V_1}$ to ${V_2}$
$W = \int\limits_{{V_1}}^{{V_2}} {P(dv)}$ $equation1$
By ideal gas law
$PV = RT$
Therefore, $P = \dfrac{{RT}}{V}$
Hence by substituting the value of $P$ in $equation1$
$W = \int\limits_{{V_1}}^{{V_2}} {\dfrac{{RT}}{V}} dv$
Since ‘R’ is universal gas constant and ‘T’ is also constant because it is isothermal process, we can simplify the equation as,
$W = RT\int\limits_{{V_1}}^{{V_2}} {\dfrac{{dv}}{V}}$
Since, $\int {\dfrac{1}{x}} dx = {\log _e}x$ , the equation can be simplified further as,
$W = RT{[{\log _e}V]^{{V_2}}}{_{{V_1}}^{}}$
$W = RT[{\log _e}{V_2} - {\log _e}{V_1}]$
$W = RT{\log _e}\dfrac{{{V_2}}}{{{V_1}}}$
$W = 2.306RT{\log _{10}}\dfrac{{{V_2}}}{{{V_1}}}$
Since, ${P_1}{V_1} = {P_2}{V_2}$ from Boyle’s law
$W = 2.306RT{\log _{10}}\dfrac{{{V_2}}}{{{V_1}}}$

Note: The difference between isothermal and adiabatic processes is that an isothermal process occurs under constant temperature but other parameters like pressure and volume can change accordingly. While in adiabatic process there is heat transfer to keep the temperature constant and one of the main differences is that in adiabatic process the temperature keeps on varying while in isothermal process it's constant. The work done in an isothermal process is due to the change of heat in the system whereas in adiabatic process its due to the change in the internal energy.