Inflating Balloon: Help Sergio Find the Correct Answer

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To calculate the work done by the balloon on the surrounding air, only the given excess pressure of 4.2e5 Pa should be used, without adding atmospheric pressure. The change in volume is 0.00082 m³. The correct formula for work is W = P x dV, leading to W = 4.2e5 Pa x 0.00082 m³. This results in a work value of 344.4 J, which is the correct answer for the problem.
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This is one of many postings from me this week unfortunately I'm just not getting the correct answers...anyway here I go :

A toy balloon is inflated with helium at a constant pressure that is 4.2e5 Pa in excess of atmospheric pressure. If the balloon inflates from a volume of 1.4e-4m3 to 9.6e-4m3, how much work is done by the balloon on the surrounding air? answer in J
this is my work

P= 4.2e5Pa + 101300Pa
= 521 300 Pa
dV = (Vf-Vi) = (9.6e-4 - 1.4e-4)
= 0.00082m3

W= 521 300Pa x (0.00082m3)
427.466 J (incorrect on homework server)

I'm guessing it has something to do with the work on the surrounding air...I have been looking in my books and course notes for this and I can't find anything...can anybody help me out??

Any help is appreciated,
Sergio
 
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Is this what the problem is supposed to be: A toy balloon is inflated with helium at a constant pressure that is 4.2e^5 Pa in excess of atmospheric pressure. If the balloon inflates from a volume of 1.4e^{-4}m^3 to 9.6e^{-4}m^3, how much work is done by the balloon on the surrounding air? answer in J

(?)
 
Yes that's exactly the question on my homework page :)
 
Hey we finally got the answer...for the pressure...do not add the atmospheric pressure...simply use the one in the question

THank you for your time

Sergio
 
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