Calculate the work in Joules by the Acetylene gas

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Homework Help Overview

The problem involves calculating the work done by Acetylene gas during a reaction with water at standard conditions. The reaction is represented by the equation CaC2 (s) + 2 H2O (l) --> Ca(OH)2 (aq) + C2H2 (g), and the context includes the use of thermodynamic principles related to gas expansion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate formula for calculating work in an isothermal gas expansion, with some suggesting the use of pressure ratios instead of volume ratios. There is uncertainty regarding the correct volumes or pressures to use in the calculations.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants have provided alternative formulas and questioned the original poster's understanding of the problem setup. There is no explicit consensus on the correct method yet.

Contextual Notes

Participants note that the reaction occurs at atmospheric pressure and that there may be confusion regarding the necessary parameters for the calculations, particularly the volumes or pressures involved.

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Homework Statement


A quantity of 4.50 g CaC2 is reacted with an excess of water at 298K and atmospheric pressure.
CaC2 (s) + 2 H2O (l) --> Ca(OH)2 (aq) + C2H2 (g)
Calculate the work done in Joules by the Acetylene gas against the atmospheric pressure.[/B]

Homework Equations


After asking my PChem professor, he said the I should use the work formula for an isothermic gas expansion, w= -nRT ln(V2/V1)
also for an isothermic expansion, q=-w[/B]

The Attempt at a Solution


P= 1 atm, n= 0.0702 mol R= 8.314 J/mol*K
I easily found the mole of the gas (0.0702), but my biggest problem is what are the volumes that are supposed to be used here? Using the moles of an ideal gas (22.4 L/mol), I found that the volume of the acetylene gas is 1.572 L, but I don't have a second volume to complete the equation. I feel like I'm missing something important, but I can't figure it out.[/B]
 
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ThymeSeer said:
at 298K and atmospheric pressure.
ThymeSeer said:
against the atmospheric pressure.
Can you recall another isothermal work process?
 
OH!
w= -nRT ln (P2/P1)

So the P1 is atmospheric pressure and for the P2 I use PV=nRT
Then I have both pressures and it's just plug and chug from there.
 
No, it's all at atmospheric.
 
Did your professor really say that you should use the work formula for an isothermic gas expansion, w= -nRT ln(V2/V1)
also for an isothermic expansion, q=-w?

Neither of these suggestions is correct.

Bystander is totally correct. Also, you weren't asked for the amount of heat transferred. But, if you did have to find it, you would be looking for the "heat of reaction," for the reaction under consideration.

Chet
 

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