How do you calculate isothermal expansion?

1 Answer
Jul 4, 2017

[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 11 molmol of an ideal gas from 22.722.7 LL to 45.445.4 LL at 298.15298.15 KK and 11 ba rbar.

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

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 "1 bar"1 bar:

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/molK)(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 "1718.3 J"1718.3 J of energy to expand, expunging "1718.3 J"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")