Work done during expansion of a gas

AI Thread Summary
The discussion centers on calculating the work done during the expansion of a gas using the equation Eint = Q + W, where Q is positive and W is negative due to the graph's area under the curve. The calculated work, based on a volume increase of 5 m³ and a pressure drop from 7 to 3 atm, results in 2.525e6 Joules. Consequently, the change in internal energy is found to be ΔU = ΔQ - W, yielding a negative value of -1.5e6 J. Participants express frustration over potential errors in the textbook and the need for clarification with the teacher. The conversation highlights the complexities of thermodynamic calculations in gas expansion scenarios.
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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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