Question

How do you calculate isothermal expansion?

Answer 1

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` `mol` of an ideal gas from `22.7` `L` to `45.4` `L` at `298.15` `K` and `1` `ba r`.

With the ideal gas law, we have that `PV = nRT`, or `P = (nRT)/V`. So, the work is:

`color(green)(w_(rev)) = -int_(V_1)^(V_2) PdV`

`= -int_(V_1)^(V_2) (nRT)/VdV`

`= -nRTlnV_2 - (-nRTlnV_1)`

`= color(green)(-nRTln(V_2/V_1))`,

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 `"1 bar"`:

`color(blue)(w_(rev)) = -("1 mol")("8.314472 J/mol"cdot"K")("298.15 K")ln("45.4 L"/"22.7 L")`

`=` `color(blue)(-"1718.3 J")`

So, the work involved the ideal gas exerting `"1718.3 J"` of energy to expand, expunging `"1718.3 J"` of heat from itself:

`cancel(DeltaU)^(0" for isothermal process") = q_(rev) + w_(rev)`

`=> color(blue)(q_(rev)) = -w_(rev) = color(blue)(+"1718.3 J")`