# What does isothermal expansion mean?

Jun 28, 2016

Each word tells you exactly what it is.

Can you tell me what isothermal compression is? What about isobaric expansion (hint: pressure can be written in units of $\text{bar}$s)?

Isothermal - constant temperature

• "iso" = a prefix meaning "equal". In this case, it means the initial state of the variable is equal to the final state of the variable, i.e. $\Delta x = {x}_{f} - {x}_{i} = 0$.
• "thermal" = of or relating to heat, which relates very closely to temperature.

(No, it doesn't mean there is zero heat flow; that's an adiabatic process, where $q = 0$.)

So, in lieu of the definition being "no heat flow", isothermal implies $x = T$, and $\textcolor{b l u e}{\Delta T = 0}$.

$\therefore$ constant temperature.

Expansion - increase in volume

• "expand" is a word you should already know, meaning to get larger. So, expansion is its noun, meaning the enlargement of the substance (usually, the gas).

Of course, if something gets larger, its volume has increased.

$\therefore$ increase in volume

Combine those and you get an increase in volume at constant temperature.

For expansions, $w < 0$ if $\Delta U = q + w$.

Note: the original image could be confusing, so I revised it to the one you see above.

You would have learned this while talking about thermodynamics. You may have been taught that efficient, reversible work ${w}_{\text{rev}}$ is defined as

${w}_{\text{rev}} = - P {\int}_{{V}_{1}}^{{V}_{2}} \mathrm{dV}$

or

$\textcolor{b l u e}{{w}_{\text{rev}} = - P \left({V}_{2} - {V}_{1}\right)}$.

Well, in an isothermal expansion, that means...

• ${V}_{2} > {V}_{1}$
• The work is negative (we know that pressure can only be positive, hence $\left(- 1\right) \times \left(+\right) \times \left(+\right) = \left(-\right)$).

That makes sense because from the perspective of the gas, the gas does work, exerting energy.

From this, we can further note that for an ideal gas, the internal energy $\setminus m a t h b f \left(\Delta U = 0\right)$ at constant temperature.

Therefore, from the first law of thermodynamics, we have:

${\cancel{\Delta U}}^{0} = {q}_{\text{rev" + w_"rev}}$

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

So for ideal isothermal expansions, the expansion work done is due to the heat flow.