Thermodynamics: Calculating work done in irreversible expansion

AI Thread Summary
The discussion focuses on calculating the heat, work, and entropy change for the isothermal expansion of 100g of ethane at 350°C, comparing reversible and irreversible processes. The user successfully solved the reversible case, determining work as -5738.79 J and entropy change as 9.209 J/K, with heat calculated as 5738.79 J. For the irreversible expansion, the user is uncertain about the approach, questioning whether to apply the same principles or if the work done is zero due to free expansion. It is noted that additional information, specifically the external pressure, is necessary to accurately solve the irreversible case.
Rocalvic
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Homework Statement



Suppose 100g of ethane (C2H6) expands isothermally at 350C from 50ml to 2L. Calculate the heat and work done by the system and the change in entropy if:

i)The process is via a reversible path.
ii)The process is non-reversible.

I don't know how to answer ii)

Homework Equations


w=-nRTln(v2/v1)

The Attempt at a Solution



I have managed to answer i)

Work = -NRTln (v2/v1) = (0.30026)(8.315)(623.15)ln(2/0.050)
Work = -5738.79 J.

Entropy change = nRln (v2/v1) = (0.30026)(8.3145)ln(2/0.050)
Entropy Change = 9.209 J.K

Heat: Q=-w, so Heat = 5738.79.

ii) I am unsure how to answer this question, do I apply the same principle? Or is this a result of free expansion and therefore the work done is 0?
Thanks
 
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Hi Rocalvic,

Welcome to Physics Forums!

You analyzed part i correctly, and you are also correct to note that there is not enough information provided to solve part ii (except for the change in entropy, of course). To do part ii, you need to know the pressure imposed by the surroundings during the irreversible change.
 

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