Work done during expansion of a gas

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Homework Help Overview

The discussion revolves around the calculation of work done during the expansion of a gas, specifically in the context of the first law of thermodynamics, where internal energy, heat transfer, and work are related. Participants are examining a graph that represents pressure and volume changes during the process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to calculate the work done based on the area under the curve in a pressure-volume graph. There are discussions about the signs of work and heat, with some questioning the values obtained and the implications of negative work. Others express uncertainty about their calculations and seek confirmation of their reasoning.

Discussion Status

Some participants have provided calculations and reasoning, while others are seeking further assistance. There appears to be a shared understanding of the equations involved, but discrepancies in numerical results have led to requests for clarification. No consensus has been reached regarding the final answer.

Contextual Notes

Participants mention potential errors in their calculations and the need to verify results with a teacher. There is a reference to previous mistakes in similar problems, indicating a concern about the accuracy of the current approach.

tanzerino
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[PLAIN]http://img101.imageshack.us/img101/8176/24851738.png

it is definetaly negative as Eint=Q+W
and Q is smaller than the area under curve taking care of the scaling of pressure in atm.
but what i get is -1.5x10^6
 
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Einternal=Q+W
Q he said is given so +ve and work from the graph is -ve
calculated work was much greateer that's why i think the answer is -ve then wen i tried the numbers i get -1.5 which isn't an answer
 
any help?
 
You should show your work. But I get the same answer as you do.

\Delta Q = \Delta U + W

where W is the work done by the gas.

That work is the area under the graph. There appears to be an increase in volume of 5 m^3 and a drop in pressure from 7 to 3 atm. So the work done is (7+3)*5/2 = 25 atm-m^3 of work = 2.525e6 Joules.

So:

\Delta U = \Delta Q - W = 1.02e6 - 2.525e6 = -1.5e6 J

AM
 
thanks for your help.i will check for any mistakes with the teacher this is cengage and the second mistake already.shame!
 

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