## Work of Expansion- reversible & isothermal

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Kaitlin Ross 3E
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### Work of Expansion- reversible & isothermal

How would you calculate the work of expansion for an ideal gas that expands reversibly and isothermally causing a change in volume and a change in pressure?

Chem_Mod
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### Re: Work of Expansion- reversible & isothermal

In an isothermal expansion, the temperature remains constant. Your question mentions that this is a reversible process, so very little to no heat is lost.

UInternal = Q + W
Reversible Process:
UInternal = W = -PdV
Since the pressure is dependent on the Volume, we use the ideal gas law to isolate volume change:
PV=nRT
P = nRT/V
There is no change in temperature (isothermal), or number of moles (free expansion of gas into a container). Therefore:
UInternal = W = -PdV = -nRT$nRT\int_{Vi}^{Vf} \frac{dV}{V}$

And we integrate and solve!

Wfinal-Winitial = -nRT*ln(Vfinal/Vinitial)