How do you calculate isothermal expansion?
1 Answer
[I do it at the second half of this answer here.](https://socratic.org/questions/what-is-reversible-isothermal-expansion)
Copy-and-pasted, we get...
CALCULATION EXAMPLE
Calculate the work performed in a reversible isothermal expansion by
With the ideal gas law, we have that
color(green)(w_(rev)) = -int_(V_1)^(V_2) PdVwrev=−∫V2V1PdV
= -int_(V_1)^(V_2) (nRT)/VdV=−∫V2V1nRTVdV
= -nRTlnV_2 - (-nRTlnV_1)=−nRTlnV2−(−nRTlnV1)
= color(green)(-nRTln(V_2/V_1))=−nRTln(V2V1) ,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
color(blue)(w_(rev)) = -("1 mol")("8.314472 J/mol"cdot"K")("298.15 K")ln("45.4 L"/"22.7 L")wrev=−(1 mol)(8.314472 J/mol⋅K)(298.15 K)ln(45.4 L22.7 L)
== color(blue)(-"1718.3 J")−1718.3 J
So, the work involved the ideal gas exerting
cancel(DeltaU)^(0" for isothermal process") = q_(rev) + w_(rev)
=> color(blue)(q_(rev)) = -w_(rev) = color(blue)(+"1718.3 J")