Work of Expansion- reversible & isothermal


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Kaitlin Ross 3E
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Work of Expansion- reversible & isothermal

Postby Kaitlin Ross 3E » Sat Jan 23, 2016 9:14 pm

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?

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Re: Work of Expansion- reversible & isothermal

Postby Chem_Mod » Sun Jan 24, 2016 9:41 am

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:
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

And we integrate and solve!

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

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