How Does Reversible Expansion Calculate Work in Thermodynamics?

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Discussion Overview

The discussion revolves around a thermodynamics problem involving the calculation of work during a reversible expansion of a gas. Participants are attempting to understand the methodology used to arrive at specific numerical answers for work, heat transfer, and changes in internal and enthalpy energy.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents a problem involving a 10.0 mol sample of a perfect gas undergoing a reversible expansion from 5.00 L to 20.0 L at 5.00 atm, seeking clarification on the calculations provided by their professor.
  • Another participant calculates the work done using the formula w = -PdV, determining that the change in volume (dV) is 15.0 L and the pressure (P) is -5.00 atm, leading to a work value of -75.0 L atm, which they convert to Joules.
  • The same calculation is reiterated by another participant, confirming the work value of -75.0 L atm and its conversion to Joules, but they express uncertainty about the professor's answer.
  • A later reply expresses gratitude for the confirmation of their calculations, indicating agreement with the derived work value.

Areas of Agreement / Disagreement

Participants generally agree on the calculation of work as -75.0 L atm, but there is no consensus on how the professor arrived at a different answer, leaving the discussion unresolved regarding the professor's methodology.

Contextual Notes

There is uncertainty regarding the assumptions made in the problem, particularly in relation to the professor's calculations and any additional factors that may have influenced the results.

Puchinita5
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Homework Statement



Okay i feel like I'm going crazy. Here is a sample problem my professor gave me. But I cannot figure out for my life how he got his answer and it hsould be very simple.


A 10.0 mol sample of a perfect gas having Cv = 2.0R undergoes a reversible expanion form 5.00 L to 20.0 L at 5.00 atm of pressure. Calculate q, w, delta U, and delta H for this process..

Solving for work, he has...


w = -PdV = -5(20-5) = -22.5 L atm = -22.8 J

HOW ON EARTH does he get those numbers?







Homework Equations





The Attempt at a Solution

 
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dV is the change in volume, which is (20.0-5.00 L)=15.0 L
-P is the given pressure x(-1), -P = -5.00 atm
w = -PdV= -5.00*15.0 L atm = -75.0 L atm
Since L atm is a unit of work, it can be converted to J:
1 L atm = 101.325 J, so -75.0(101.325) J = -7.60x10^3 J = w

I'm not sure what else your professor did to get to his answer, but this is what I got based on what was given.
 
dV is the change in volume, which is (20.0-5.00 L)=15.0 L
-P is the given pressure x(-1), -P = -5.00 atm
w = -PdV= -5.00*15.0 L atm = -75.0 L atm
Since L atm is a unit of work, it can be converted to J:
1 L atm = 101.325 J, so -75.0(101.325) J = -7.60x10^3 J = w

I'm not sure what else your professor did to get to his answer, but this is what I got based on what was given.
 
thank you! I just needed confirmation that I wasn't missing something. That was exactly what I thought the answer should be
 

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