Isothermal reversible condensation

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SUMMARY

The discussion focuses on the thermodynamic principles involved in isothermal reversible condensation of 1.00 mol of H2O(g) at 100°C. The standard enthalpy of vaporization for water is 40.656 kJ mol-1, leading to a change in enthalpy (ΔH) equal to the negative of this value during condensation. The change in internal energy (ΔU) is zero due to the isothermal condition, but the work done (w) is calculated using the external pressure of 1 atm. The confusion arises from the application of ideal gas equations versus real liquid behavior.

PREREQUISITES
  • Understanding of thermodynamic concepts such as enthalpy, internal energy, and work.
  • Familiarity with the ideal gas law and its limitations.
  • Knowledge of isothermal and reversible processes in thermodynamics.
  • Ability to apply the first law of thermodynamics in practical scenarios.
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  • Learn about the differences between ideal gases and real liquids, particularly in thermodynamic contexts.
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  • Investigate the calculations of work done in different thermodynamic processes, including isothermal conditions.
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Students and professionals in chemistry and chemical engineering, particularly those studying thermodynamics and phase transitions. This discussion is beneficial for anyone looking to deepen their understanding of enthalpy changes during phase changes.

kido
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Homework Statement



A sample of 1.00 mol H20(g) is condensed isothermally and reversibly
to liquid water at 100°C. The standard enthalpy of vaporization of water at
100°C is 40.656 kJ mol-1. Find w, q, change in internal energy, and change in enthalpy for this process.

Homework Equations




The Attempt at a Solution



So since this is an isothermal reversible compression, change in temperature is equal to 0, so change in internal energy should also be equal to 0.
Because deltaH = deltaU + delta(pV), and for an ideal gas pV = nRT so deltaH = 0 + delta(nRT) = 0, change in enthalpy should also be equal to 0.

HOWEVER the solutions were quite different. Apparently deltaH = -standard enthalpy of vapourization, q = deltaH, w = -(external pressure) x deltaV, and deltaU = q + w.

What am I missing here? Why are the changes in internal energy and enthalpy not zero if the change in temperature is zero?

I know this is probably a really basic thermodynamics question but I just don't get it!
 
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Liquid water is not an ideal gas.
 
okay, so what are the expressions for deltaU and deltaH for a liquid?
 
kido said:
okay, so what are the expressions for deltaU and deltaH for a liquid?

Exactly what is given in the answer: the change in enthalpy is the enthalpy of vaporization. Condensation releases heat, so this value is negative. And by definition, the change in energy is the enthalpy without the work performed during the process.
 
Okay, i understand that now, thanks. There's one more thing that i am having trouble understanding, though. In the solutions, it says that because the condensation is done isothermally and reversibly, the external pressure is constant at 1atm.
I thought the whole point of a reversible process was that the pressure was not constant? Shouldn't the expression for work be
w = -nRTln(Vf/Vi),
not w = -(external pressure)*(Vf-Vi)?
 
Reversibility doesn't imply anything about the pressure being constant or not constant. And work is always \int P(V)\,dV. For ideal gases at constant temperature this expression can be reduced to the other expression you give, but again, we're not dealing with an isolated ideal gas here.
 
Ah, I see. I really must learn which expressions apply only to ideal gases!
So because this question mentions the standard enthalpy of vaporization, the external pressure is at standard conditions, which is 1atm?
Thank you very much for your help.
 
I am having the same problems in this question. Not sure if I did it correct.
I said "Delta H = -Enthalpy of Vaporization"
"W=0" because the final state (liquid) does not alter volume
"Delta H = q " this is true for isothermal processes
"Delta U= q + W, W=0 so Delta U=q"

is this thought correct?
 

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