# What is isothermal expansion of an ideal gas?

May 26, 2016

By definition:

• Isothermal means the temperature does not change.
• Expansion means the volume has increased.

Therefore, isothermal expansion is the increase in volume under constant-temperature conditions.

In this situation, the gas does work, so the work is negatively-signed because the gas exerts energy to increase in volume.

During isothermal conditions, the change in internal energy $\Delta U$ is $0$ for only an ideal gas, so efficient work done is entirely transformed into efficient heat flow.

In other mathy words:

$\textcolor{b l u e}{{w}_{\text{rev}}} = - {\int}_{{V}_{1}}^{{V}_{2}} P \mathrm{dV}$

$= - {\int}_{{V}_{1}}^{{V}_{2}} \frac{n R T}{V} \mathrm{dV}$

$= - n R T {\int}_{{V}_{1}}^{{V}_{2}} \frac{1}{V} \mathrm{dV}$

$= \textcolor{b l u e}{- n R T \ln | \frac{{V}_{2}}{{V}_{1}} |}$

where:

• ${w}_{\text{rev}}$ is the most efficient work possible (reversible work) in $\text{J}$. It is as slow as possible to ensure that no energy is lost to the atmosphere.
• $P$ is the pressure in, say, $\text{bars}$, $\text{atm}$, etc.
• ${\int}_{{V}_{1}}^{{V}_{2}} \mathrm{dV}$ is the integral from the initial to the final volume, which basically gives you the result of performing infinitesimally slow work.
• $\mathrm{dV}$ is the differential volume; that is, it is an infinitesimally small change in the volume.
• $n R T$ is a constant for an isothermal situation, and each variable holds the same meaning as in the ideal gas law. This can be pulled out as a coefficient in the integral.

The integral of $\frac{1}{V}$ is $\ln | V |$ evaluated from $V = {V}_{1}$ to $V = {V}_{2}$, which turns out to be $\ln | {V}_{2} | - \ln | {V}_{1} | = \ln | \frac{{V}_{2}}{{V}_{1}} |$.

For expansion, ${V}_{2} > {V}_{1}$, so we know that $\ln | \frac{{V}_{2}}{{V}_{1}} | > 0$. Thus, for isothermal expansion, ${w}_{\text{rev}} < 0$.

During isothermal conditions, the internal energy from the first law of thermodynamics is

$\setminus m a t h b f \left(\Delta U = {q}_{\text{rev" + w_"rev}}\right) = 0 ,$

which means $\textcolor{b l u e}{{q}_{\text{rev" = -w_"rev}}}$.

As a brief comparison, isothermal contraction is when the volume decreases. It means work was done on the gas.

This makes the work positive because the gas absorbs the energy that was imparted into it to do work on it.

$\Delta U$ is still $0$, and ${q}_{\text{rev" = -w_"rev}}$ is still true, but here, ${V}_{2} < {V}_{1}$, thus $\ln | \frac{{V}_{2}}{{V}_{1}} | < 0$ and ${w}_{\text{rev}} > 0$.