# Isothermal Process

Isothermal process is the thermodynamic process in which the pressure and volume of the system change but temperature remains constant. For an isothermal process to take place, the cylinder should have conducting walls and the gas should be allowed to expand or compress very slowly.

​Consider a cylinder having conducting walls and containing an ideal gas. If the gas is allowed to expand very slowly, cooling will occur very slowly. But the temperature of the gas does not fall, because before falling, heat is conducted into the cylinder through the conducting walls from the surroundings. If the gas is allowed to compress very slowly, heat will be produced very slowly. Before the temperature of the gas rises, the heat produced leaves to the surrounding though the conducting walls. In this way, isothermal expansion or contraction can take place.

## Equation

​For one mole of an ideal gas, $PV=RT$ If temperature remains constant, $PV=\text{constant}$

This equation is the equation of an isothermal process for an ideal gas. If $P_1$ and $V_1$ are the initial pressure and volume of the gas and $P_2$ and $V_2$ are the final pressure and volume of the gas, then, $P_1V_1=P_2V_2$

## Application of First Law of Thermodynamics in Isothermal Process

According to the first law of thermodynamics, $dQ=dU+PdV$ If the temperature does not change then there is no change in internal energy i.e. $dU=0$. $dQ=PdV$ Therefore,
1. When a gas expands isothermally, then, $dV$ is positive. So, $PdV$ and $dQ$ are also positive.
Thus, an amount of heat equivalent to the work done by the gas will have to be supplied in order to expand the gas isothermally.
2. When a gas compresses isothermally, then, $dV$ is negative. So, $PdV$ and $dQ$ are also negative.
Thus, an amount of heat equivalent to the work done on the gas will have to be removed in order to compress the gas isothermally.

## Work Done In Isothermal Process

​Consider one mole of an ideal gas contained in a cylinder having conducting walls. Thus, for one mole of the ideal gas, $PV=RT\text{ __(a)}$

When the system changes its state from initial pressure $P_1$ and initial volume $V_1$ to final pressure $P_2$ and final volume $V_2$, then, the amount of work done is, $W=\int_{V_1}^{V_2} P\;dV \text{ __(b)}$

From $(a)$ and $(b)$, $W=\int_{V_1}^{V_2} \frac{RT}{V} \; dV$ In isothermal process, temperature remains constant. $W=RT\int_{V_1}^{V_2} \frac{1}{V} \; dV$ $W=RT\left[\log_eV\right]_{V_1}^{V_2}$ $W=RT\left(\log_eV_2-\log_eV_1\right)$ $W=RT\log_e\left(\frac{V_2}{V_1}\right)$

Therefore, for $n$ mole of an ideal gas, $W=nRT\log_e\left(\frac{V_2}{V_1}\right)\text{ __(1)}$ also, $P_1V_1=P_2V_2$ $\frac{V_2}{V_1}=\frac{P_1}{P_2}$

So, equation $(1)$ becomes, $W=nRT\log_e\left(\frac{P_1}{P_2}\right)\text{ __(2)}$ The equations $(1)$ and $(2)$ give the work done during isothermal process.