Thermodynamics: Calculating work done in irreversible expansion

Click For Summary
SUMMARY

The discussion focuses on calculating work done during the irreversible expansion of 100g of ethane (C2H6) at 350°C from 50ml to 2L. For the reversible process, the work done is calculated as -5738.79 J, with an entropy change of 9.209 J/K. The heat exchanged is equal to the negative of the work done, resulting in 5738.79 J. For the irreversible process, additional information regarding the external pressure is required to determine the work done, as the principles of free expansion apply.

PREREQUISITES
  • Understanding of the ideal gas law and isothermal processes.
  • Familiarity with thermodynamic equations, specifically work and entropy calculations.
  • Knowledge of reversible and irreversible processes in thermodynamics.
  • Basic concepts of heat transfer and energy conservation.
NEXT STEPS
  • Study the concept of irreversible processes in thermodynamics.
  • Learn how to calculate work done in irreversible expansions using external pressure.
  • Explore the implications of free expansion on thermodynamic properties.
  • Review the derivation and application of the equation for entropy change in non-reversible processes.
USEFUL FOR

Students studying thermodynamics, particularly those tackling problems involving gas expansions, as well as educators and professionals seeking to deepen their understanding of reversible and irreversible processes.

Rocalvic
Messages
1
Reaction score
0

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
 
Physics news on Phys.org
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.
 

Similar threads

Chemistry No heat exchange with the surroundings in an irreversible expansion of an ideal gas?
  • · Replies 4 ·
Replies
4
Views
3K
Chemistry A few questions on an irreversible adiabatic expansion of an ideal gas
  • · Replies 1 ·
Replies
1
Views
3K
Chemistry How to explain the difference between recompression of ideal gas reversibly and irreversibly after initial irreversible expansion?
  • · Replies 5 ·
Replies
5
Views
3K
Chemistry Entropy in isolated composite system for irreversible process
  • · Replies 3 ·
Replies
3
Views
3K
Chemistry Calculation of thermodynamic variables for irreversible adiabatic process of ideal gas
  • · Replies 9 ·
Replies
9
Views
3K
Is the Gas Expansion Reversible or Irreversible?
  • · Replies 2 ·
Replies
2
Views
2K
Chemistry How does it work to mix two gases reversibly in this device?
  • · Replies 1 ·
Replies
1
Views
2K
Change in entropy of the surroundings in an irreversible process
  • · Replies 14 ·
Replies
14
Views
1K
Graduate Work done by irreversible and reversible processes
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Irreversible Isochoric Process in a Cycle
  • · Replies 8 ·
Replies
8
Views
4K