# How do you calculate isothermal expansion?

##### 1 Answer

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* *of an ideal gas from* *to* *at* *and* *.*

With the ideal gas law, we have that

#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

#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

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

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