Isothermal reversible condensation

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Discussion Overview

The discussion revolves around the thermodynamic analysis of isothermal reversible condensation of water vapor to liquid water at 100°C. Participants explore the calculations of work (w), heat (q), change in internal energy (ΔU), and change in enthalpy (ΔH) for this process, addressing various thermodynamic principles and equations.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant asserts that since the process is isothermal, the change in internal energy should be zero, leading to the conclusion that ΔH is also zero.
  • Another participant points out that liquid water is not an ideal gas, suggesting that the assumptions made about ideal gas behavior may not apply.
  • A participant clarifies that the change in enthalpy during condensation is equal to the negative of the enthalpy of vaporization, indicating that condensation releases heat.
  • There is a question regarding the constancy of external pressure during a reversible process, with one participant suggesting that the work expression should involve a logarithmic term rather than a constant pressure term.
  • Another participant explains that reversibility does not imply constant pressure and that work is calculated using the integral of pressure over volume changes.
  • A participant expresses confusion about the application of expressions for internal energy and enthalpy for liquids, seeking clarification on their correct forms.
  • One participant summarizes their understanding of the problem, stating that ΔH equals the negative enthalpy of vaporization and that work is zero due to no change in volume in the final state.

Areas of Agreement / Disagreement

Participants exhibit a mix of agreement and disagreement. While some clarify and refine the understanding of thermodynamic principles, others express confusion about the application of these principles, particularly regarding the behavior of liquids versus gases and the implications of reversibility.

Contextual Notes

There are unresolved assumptions regarding the applicability of ideal gas laws to the condensation process and the definitions of work and pressure in reversible processes. The discussion also highlights the need for clarity on the differences between gases and liquids in thermodynamic calculations.

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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