# What is reversible isothermal expansion?

##### 1 Answer

Well, take apart the terms:

**Reversible**means that in principle, the process is done*infinitely slowly*so that the microscopic reverse from the final state exactly regenerates the initial state. This requires an exact functional form of whatever term you are integrating.**Isothermal**just means constant temperature, i.e.#DeltaT = T_2 - T_1 = 0# .**Expansion**means an increase in volume...

*Hence, a reversible isothermal expansion is an infinitely-slow increase in volume at constant temperature.*

For an **ideal gas**, whose internal energy *only a function of temperature*, we thus have for the first law of thermodynamics:

#DeltaU = q_(rev) + w_(rev) = 0#

Thus,

This also means that...

*All the reversible isothermal PV work*

*done by an ideal gas to expand was possible by*

**reversibly absorbing heat***into the ideal gas.*

**CALCULATION EXAMPLE**

*Calculate the work performed in a reversible isothermal expansion by* *of an ideal gas from* *to* *at* *and a* * initial pressure.*

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 require the 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")# (however, one could use the ideal gas law to write

#ln(V_2/V_1) = ln(P_1/P_2)# in this constant-temperature situation.)

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")#