Understanding reversible, isothermic expansion

isochoric/isometric:
isothermal:
isobaric:

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Cameron Sasmor 1G
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Joined: Fri Sep 26, 2014 2:02 pm

Understanding reversible, isothermic expansion

Postby Cameron Sasmor 1G » Sun Jan 11, 2015 12:07 pm

I do not understand how the external pressure can always be equal to the pressure of the gas during a reversible, isothermic reaction and how the dw=-P(ext)(dV) equation is derived. Can someone explain this please?

Sanmeet Atwal 1D
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Joined: Fri Sep 26, 2014 2:02 pm

Re: Understanding reversible, isothermic expansion

Postby Sanmeet Atwal 1D » Sun Jan 11, 2015 9:52 pm

In the text they say:
"a reversible process is one that can be reversed by an infinitely small change in a variable (an “infinitesimal” change). For example, if the external pressure exactly matches the pressure of the gas in the system, then the piston moves in neither direction. If the external pressure is increased infinitesimally, the piston moves in. If, instead, the external pressure is reduced infinitesimally, the piston moves out."
They say that if initially the external and internal pressures are equivalent then changing either pressure by an extremely small amount will cause the piston to shift. If the external pressure increases by the smallest increment then the piston will move in and work will be done on the system, but if the external pressure is decreased by the smallest increment then the piston will move out and the system will do work. Furthermore, if the gas expands and does work then the internal pressure of the gas also falls because volume increases and in order to keep the internal and external pressures the same the external pressure must also be reduced. This is represented graphically in figure 7.6:
IMG_4239.jpg

As the volume increases then the internal pressure will decrease and since internal pressure=external pressure, the external pressure will also decrease and vice versa. The work is the area under the curve which can be derived using calculus.

We know that W= -P(Vf-Vi)
to find the area under the curve we have to integrate the work equation in respect to change in volume using derivatives:
dW= -PdV
we know from PV=nRt that P=nRt/v so plugging this formula in for the pressure we get:
dW= -(nRt/v)dV
In order to solve for V we have to use integrals:
∫dW= -∫nRt/v)dV
since -nRt is a constant we can pull it out in front of the integral to get:
W= -nRt ∫(1/v)dV
From integral rules we know that ∫(1/x)dx = ln(x) so
∫(1/v)dV = ln(v)
since the volume is changing the limits of the integration would be from Vfinal to Vinitial so
∫(1/v)dV from Vf to Vi = ln(Vf)-ln(Vi) which is also equal to ln(Vf/Vi) using log rules.
Therefore W= -nRt ln(Vf/Vi)
Hope this helps!!

Chem_Mod
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Re: Understanding reversible, isothermic expansion

Postby Chem_Mod » Mon Jan 12, 2015 1:24 am

The explanation of reversibility above is correct. Also, here is a more concrete example of a reversible vs nonreversible expansion:

Let P and Pext both be 2 atm, and initial volume Vi = 1 Liter. For the reversible expansion, the external pressure is slowly reduced from 2 atm to 1 atm. Along the entire path, the system is equilibrated with P = Pext, so PV=nRT is satisfied at every timepoint. The final volume Vf = 2 Liters. The work done is the area underneath the curve.

For the irreversible expansion, the external pressure is suddenly dropped from 2 atm to 1 atm. As the gas expands towards 2 Liters, P > Pext and PV=nRT is not true until the final state. The path is no longer that isothermal curve. It is a horizontal line at 1 atm, and the work done is the area of the rectangle underneath. This area is a lot smaller than the area underneath the entire isothermal curve, so the work done is smaller.


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