Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: 2810713 on February 28, 2005, 12:11:31 AM
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Hi,
for a gas in a cylinder with a piston,
The P in PV=nRT is the P exerted by the gas on the
the piston.
But in the derivation of work done by the gas / on the gas , the external pressure is considered. I could not understand why this is so...
please help...
hrushikesh
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In the ideal case of reversible expansion and compression, the process takes place in a series of steps at equilibria where the external pressure is equal to internal pressure of the gas at any moment. Therefore, the gas law, which is the internal pressure, is applicable and the work is derived by integration.
In practical terms, work by compression or expansion is an irreversible (sudden) process where the gas pressure and external pressure don't equate instantaneously. Therefore, the gas law cannot be applied. Instead the difference between the initial and final ext pressure is used.
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Thank you very much , now the reversible case is clearer to me, thanks. :)
How can we use the difference between the initial and final pressure in the irreversible proocessBecause the final external P is equal to the internal P? .
hrushikesh
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oops, I had meant for the irreversible case a difference in volume is taken against a constant external final pressure, ie:
work = -Pext(deltaV) = -Pext x (Vfinal - Vinitial)
This is different than for a reversible case where the external and internal pressure varies gradually and continuously with the volume. In that case
work = -integral(PdV) where P = nRT/V can be substituted to get:
work = -integral(nRT/V)dV
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Yah, now both the cases are clear to me, thanks !!!
I enjoyed the conversations on this forum very much!
By the way , Demotivator don't feel U R low, U R helped people alot. Be positive :)
thanks again!
hrushikesh