## Irreversible vs. Reversible Expansion (Question 8.11) [ENDORSED]

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

Timothy_Yu_Dis3A
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### Irreversible vs. Reversible Expansion (Question 8.11)

I understand how to solve question 11 for Chapter 8 by using the formulas for calculating the work for the irreversible and reversible expansion respectively. However, I recall Dr. Lavelle mentioning (though I'm not 100% sure) during lecture that reversible expansion always does more work. If this is true, could someone give me a general explanation as to why this is the case?

Atishay_Mathur_3L
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### Re: Irreversible vs. Reversible Expansion (Question 8.11)  [ENDORSED]

Hi!

An reversible and isothermal expansion does more work than an irreversible process, because in a reversible process the internal pressure and external pressure are equal at all points in time. However, infinitesimal pressure changes allow the gas to expand such that no energy is wasted in increasing entropy of the surroundings. In this way the process ideal, because it would have to take place over an infinitely long time scale, but in theory, could produce the maximum work.