# How do you calculate isothermal expansion?

Jul 4, 2017

I do it at the second half of this answer here.

Copy-and-pasted, we get...

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 $1$ $b a r$.

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 use the pressure of $\text{1 bar}$:

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

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

So, the work involved the ideal gas exerting $\text{1718.3 J}$ of energy to expand, expunging $\text{1718.3 J}$ of heat from itself:

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

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