# What is reversible isothermal expansion?

Jun 25, 2017

Well, take apart the terms:

• Reversible means that in principle, the process is done infinitely slowly so that the microscopic reverse from the final state exactly regenerates the initial state. This requires an exact functional form of whatever term you are integrating.
• Isothermal just means constant temperature, i.e. $\Delta T = {T}_{2} - {T}_{1} = 0$.
• Expansion means an increase in volume...

Hence, a reversible isothermal expansion is an infinitely-slow increase in volume at constant temperature.

For an ideal gas, whose internal energy $U$ is only a function of temperature, we thus have for the first law of thermodynamics:

$\Delta U = {q}_{r e v} + {w}_{r e v} = 0$

Thus, ${w}_{r e v} \equiv - \int P \mathrm{dV} = - {q}_{r e v}$, where work is done is from the perspective of the system and $q$ is heat flow.

This also means that...

All the reversible isothermal PV work ${w}_{r e v}$ done by an ideal gas to expand was possible by reversibly absorbing heat ${q}_{r e v}$ into the ideal gas.

CALCULATION EXAMPLE

Calculate the work performed in a reversible isothermal expansion by $1$ $m o l$ of an ideal gas from $22.7$ $L$ to $45.4$ $L$ at $298.15$ $K$ and a $1$ $b a r$ initial pressure.

With the ideal gas law, we have that $P V = n R T$, or $P = \frac{n R T}{V}$. So, the work is:

$\textcolor{g r e e n}{{w}_{r e v}} = - {\int}_{{V}_{1}}^{{V}_{2}} P \mathrm{dV}$

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

$= - n R T \ln {V}_{2} - \left(- n R T \ln {V}_{1}\right)$

$= \textcolor{g r e e n}{- n R T \ln \left({V}_{2} / {V}_{1}\right)}$,

negative with respect to the system.

We keep in mind that the pressure did change, but we don't have an idea of how, off-hand. The work thus does not require the use the pressure of $\text{1 bar}$:

$\textcolor{b l u e}{{w}_{r e v}} = - \left(\text{1 mol")("8.314472 J/mol"cdot"K")("298.15 K")ln("45.4 L"/"22.7 L}\right)$

$=$ $\textcolor{b l u e}{- \text{1718.3 J}}$

(however, one could use the ideal gas law to write $\ln \left({V}_{2} / {V}_{1}\right) = \ln \left({P}_{1} / {P}_{2}\right)$ in this constant-temperature situation.)

So, the work involved the ideal gas exerting $\text{1718.3 J}$ of energy to expand, due to the $\text{1718.3 J}$ of heat it absorbed into itself:

${\cancel{\Delta U}}^{0 \text{ for isothermal process}} = {q}_{r e v} + {w}_{r e v}$

$\implies \textcolor{b l u e}{{q}_{r e v}} = - {w}_{r e v} = \textcolor{b l u e}{+ \text{1718.3 J}}$