maximum work during reversible process??????

[adianadiadi]

I have seen maximum work is achievable when the process is carried out reversibly. Is it correct?

For example, if you look at the following problems,

1) Ten litres of an ideal gas at a pressure of 10 atm expands isothermally against a constant pressure of 1 atm until its total volume becomes 100 litres.The amount of heat absorbed in the expansion is

The answer is -90 L-atm
(I have used the formula w = -p(ext)*(v2-v1))

2) Ten litres of an ideal gas at a pressure of 10 atm expands isothermally under reversible conditions until the final volume becomes 100 litres.The amount of heat absorbed in the expansion is

The answer is -23.03 L-atm
(Here the formula used is w = -2.303*p(internal)*log v2/v1)

In the first problem, the process is irreversible and the process in second problem is reversible. Here the work done in case of irreversible process is greater than that in reversible process. (Just consider the absolute values and not the negative signs)

Could you please explain this? Am I wrong anywhere?

[Mapes]

Hi adianadiadi, welcome to PF. The second answer should be -230 L-atm:

\int -p\,dV=-nRT\int_{V_1}^{V_2} \frac{dV}{V}=-P_1V_1\ln 10=-230

Does this answer your question?

[adianadiadi]

ThanQ very much Mapes! Now I got the answer.

If you have time, could you please tell me whether the solution given to the problem 6.4 is correct or not? I think this is wrong. It should be then, 32.2 L-atm.

These problems are given in NCERT text books.
http://www.ncert.nic.in/book_publishing/CLASS%2011/Chemistry/Part%20I/6.pdf

Problem 6.2
Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres.How much heat is absorbed and how much work is done in the expansion ?
Solution
We have q = – w = pex (10 – 2) = 0(8) = 0
No work is done; no heat is absorbed.
Problem 6.3
Consider the same expansion, but this time against a constant external pressure of 1 atm.
Solution
We have q = – w = pex (8) = 8 litre-atm
Problem 6.4
Consider the same expansion, to a final volume of 10 litres conducted reversibly.
Solution
We have q = – w = 2.303 × 10 x log10/2 = 16.1 litre-atm


Thanking you once again.

[Mapes]

Looks good, but check 6.4 again.

[adianadiadi]

In 6.4, the value 10 is just the pressure. I think we should also take initial volume 2 lit.

I got another doubt. Whether the work done during compression is maximum under reversible conditions or in irreversible conditions?