Calculating sublimation temperature

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Discussion Overview

The discussion revolves around calculating the normal sublimation temperature of carbon dioxide (CO2) at 1.00 atm pressure, using the standard enthalpy of sublimation and the Clausius-Clapeyron equation. The context includes theoretical and mathematical reasoning related to phase changes and thermodynamics.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Some participants note that the standard enthalpy of sublimation for dry ice is 6.03 kJ/mol and mention the triple point of CO2 at 5.1 atm and -59.7 °C.
  • One participant suggests that the problem resembles a Clausius-Clapeyron problem and expresses uncertainty about calculating delta entropy for sublimation at 1 atm.
  • Another participant states the relationship dP/dT = delta S / delta V and questions how to find delta V, proposing to use delta H = p times delta V at 5.1 atm.
  • One participant emphasizes the need to understand the derivation of the Clausius-Clapeyron equation to relate changes in entropy and volume between equilibrium states.
  • There is a discussion about using the ideal gas law to find the volume of the gas, with some participants agreeing that neglecting the specific volume of the liquid is a valid assumption.
  • One participant inquires about solving the Clausius relation to derive the equation P = Ce^-(L/RT), while another clarifies that it is essentially equivalent to the Clausius-Clapeyron equation.
  • Participants discuss using initial conditions at 5.1 atm to find the constant of integration C and subsequently solving for temperature at 1 atm pressure.
  • A later reply confirms that one participant arrived at the correct answer of 145 K.

Areas of Agreement / Disagreement

Participants generally agree on the approach involving the Clausius-Clapeyron equation and the use of initial conditions to solve for the sublimation temperature, though there are varying levels of confidence and clarity regarding specific steps in the calculations.

Contextual Notes

Some assumptions regarding the ideal gas behavior and neglecting specific volumes are mentioned, but the discussion does not resolve all uncertainties related to the derivation and application of the equations involved.

xdrgnh
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Homework Statement


The standard enthalpy of sublimation of dry ice (solid carbon dioxide) is 6.03 kJ/mol.
The triple point of CO2 is at 5.1 atm, -59.7 °C.

Calculate the normal sublimation temperature of CO2 at 1.00 atm pressure.



Homework Equations



Delta G= Delta H-T(delta S)

The Attempt at a Solution



I know delta G will be zero and and that delta H will be the same as the standard Enthalpy. I imagime I need to calculate delta entropy for sublimation at 1 atm but I have no idea how.
 
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xdrgnh said:

Homework Statement


The standard enthalpy of sublimation of dry ice (solid carbon dioxide) is 6.03 kJ/mol.
The triple point of CO2 is at 5.1 atm, -59.7 °C.

Calculate the normal sublimation temperature of CO2 at 1.00 atm pressure.



Homework Equations



Delta G= Delta H-T(delta S)

The Attempt at a Solution



I know delta G will be zero and and that delta H will be the same as the standard Enthalpy. I imagime I need to calculate delta entropy for sublimation at 1 atm but I have no idea how.
This looks like a Clausius-Clapeyron problem.

Chet
 
I know dP/dT= delta S/ delta V. Do I find delta V using delta H= p times delta v where p is pressure specifically 5.1 atm?
 
xdrgnh said:
I know dP/dT= delta S/ delta V. Do I find delta V using delta H= p times delta v where p is pressure specifically 5.1 atm?
No. You need to look up the derivation of the Clausius-Clapeyron equation. This will tell you how to get from your first equation to an equation you can actually use to solve this problem.

Think about this: The ΔS and the ΔV in your first equation refer to the changes in entropy and volume between two equilibrium states. What are those two equilibrium states? Is the volume in one of the equilibrium states much less than the volume in the other equilibrium state? How is the change in entropy between the two equilibrium states related to the change in enthalpy?

Chet
 
Last edited:
Enthalpy Over Temperature Gives Entropy. Can I Treat The Vapors As An Ideal Has And Use The Idea Gas Law To Find Volume Of The Gas?
 
xdrgnh said:
Enthalpy Over Temperature Gives Entropy. Can I Treat The Vapors As An Ideal Has And Use The Idea Gas Law To Find Volume Of The Gas?
If you mean enthalpy change over temperature gives entropy change, then yes. Regarding the ideal gas law, yes. You also neglect the specific volume of the liquid. These are the assumptions Clapeyron used.

Chet
 
Do I Need To Solve The ClaudiusRelation To Get P=Ce^-(L/RT)
 
xdrgnh said:
Do I Need To Solve The ClaudiusRelation To Get P=Ce^-(L/RT)

No. This is the essentially equivalent to the Clausius Clapeyron equation. All you need to do first is use your input data to get the value of the constant of integration C for your problem.

Chet
 
You Mean Yes For Solving The DE I Then Use The Initial Conditions At P= 5.1 Atm To Find C Once I Have C I Can Plug In 1 ATM For The Pressure And Solve For T Right?
 
  • #10
xdrgnh said:
You Mean Yes For Solving The DE I Then Use The Initial Conditions At P= 5.1 Atm To Find C Once I Have C I Can Plug In 1 ATM For The Pressure And Solve For T Right?
Excellent. Great job.

Chet
 
  • #11
I Got The Right Answer Of 145 K Thanks So Much.
 

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